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Introduction to Modern Algebra
May 7, 2018
Part IX. FactorizationVII.45. Unique Factorization Domains
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Table of contents
1 Lemma 45.9.
2 Lemma 45.10. The Ascending Chain Condition for a PID
3 Theorem 45.11.
4 Lemma 45.12.
5 Lemma 45.13.
6 Theorem 45.17.
7 Corollary 45.18 Fundamental Theorem of Arithmetic.
8 Lemma 45.23.
9 Lemma 45.25 Gauss’s Lemma.
10 Lemma 45.27.
11 Corollary 45.28.
12 Theorem 45.29.
13 Corollary 45.30.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj .
Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R. For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni . Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj . Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.
Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R. For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni . Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj . Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R.
For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni . Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj . Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R. For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni .
Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj . Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R. For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni . Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.9.
Lemma 45.9.
Lemma. 45.9. Let R be a commutative ring and let N1 ⊆ N2 ⊆ ... be anascending chain of ideals Ni in R. Then N = supi Ni is an ideal of R.
Proof Let a, b ∈ N. Then there are ideals Ni and Nj in the chain witha ∈ Ni and b ∈ Nj . WLOG, Ni ⊆ Nj and a, b ∈ Nj . Every ideal is anadditive subgroup, so a± b ∈ Nj . By the definition of ideal, ab ∈ Nj . Soa± b, ab ∈ N.Since 0 ∈ Ni for all i , it follows that for all b ∈ N, we have −b ∈ N and0 ∈ N. By Exercise 18.48, N is a subring of R. For a ∈ N and r ∈ R, wehave a ∈ Ni for some i and since Ni is an ideal, then da = ad ∈ Ni . Soad ∈ ∪iNi and da ∈ N. So N is an ideal of R.
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Lemma 45.10. The Ascending Chain Condition for a PID
Lemma 45.10. The Ascending Chain Condition for a PID
Lemma. 45.10. The Ascending Chain Condition for a PID. Let D bea PID. If N1 ⊆ N2 ⊆ ... is an ascending chain of ideals, then there exists apositive integer r such that Nr = Ns for all s ≥ r . Equivalently, everystrictly ascending chain of ideals in a PID is of finite length. Under suchconditions it is said that the ascending chain condition holds for ideals in aPID.
Proof. By Lemma 45.9, we have that N = ∪iNi is an ideal of D. Since Dis a PID then N is a principal ideal and so N = 〈c〉 for some c ∈ D.
SinceN = ∪iNi , then c ∈ Nr for some r ∈ N. For s ≥ r we have〈c〉 ⊆ Nr ⊆ Ns ⊆ N = 〈c〉. So Nr = Ns for all s ≥ r .
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Lemma 45.10. The Ascending Chain Condition for a PID
Lemma 45.10. The Ascending Chain Condition for a PID
Lemma. 45.10. The Ascending Chain Condition for a PID. Let D bea PID. If N1 ⊆ N2 ⊆ ... is an ascending chain of ideals, then there exists apositive integer r such that Nr = Ns for all s ≥ r . Equivalently, everystrictly ascending chain of ideals in a PID is of finite length. Under suchconditions it is said that the ascending chain condition holds for ideals in aPID.
Proof. By Lemma 45.9, we have that N = ∪iNi is an ideal of D. Since Dis a PID then N is a principal ideal and so N = 〈c〉 for some c ∈ D. SinceN = ∪iNi , then c ∈ Nr for some r ∈ N.
For s ≥ r we have〈c〉 ⊆ Nr ⊆ Ns ⊆ N = 〈c〉. So Nr = Ns for all s ≥ r .
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Lemma 45.10. The Ascending Chain Condition for a PID
Lemma 45.10. The Ascending Chain Condition for a PID
Lemma. 45.10. The Ascending Chain Condition for a PID. Let D bea PID. If N1 ⊆ N2 ⊆ ... is an ascending chain of ideals, then there exists apositive integer r such that Nr = Ns for all s ≥ r . Equivalently, everystrictly ascending chain of ideals in a PID is of finite length. Under suchconditions it is said that the ascending chain condition holds for ideals in aPID.
Proof. By Lemma 45.9, we have that N = ∪iNi is an ideal of D. Since Dis a PID then N is a principal ideal and so N = 〈c〉 for some c ∈ D. SinceN = ∪iNi , then c ∈ Nr for some r ∈ N. For s ≥ r we have〈c〉 ⊆ Nr ⊆ Ns ⊆ N = 〈c〉. So Nr = Ns for all s ≥ r .
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Lemma 45.10. The Ascending Chain Condition for a PID
Lemma 45.10. The Ascending Chain Condition for a PID
Lemma. 45.10. The Ascending Chain Condition for a PID. Let D bea PID. If N1 ⊆ N2 ⊆ ... is an ascending chain of ideals, then there exists apositive integer r such that Nr = Ns for all s ≥ r . Equivalently, everystrictly ascending chain of ideals in a PID is of finite length. Under suchconditions it is said that the ascending chain condition holds for ideals in aPID.
Proof. By Lemma 45.9, we have that N = ∪iNi is an ideal of D. Since Dis a PID then N is a principal ideal and so N = 〈c〉 for some c ∈ D. SinceN = ∪iNi , then c ∈ Nr for some r ∈ N. For s ≥ r we have〈c〉 ⊆ Nr ⊆ Ns ⊆ N = 〈c〉. So Nr = Ns for all s ≥ r .
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]
If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit.
Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit).
If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit.
As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....
By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉).
Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11.
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. Let a ∈ D where ’a’ is neither 0 nor a unit. [We first show that ’a’has at least one irreducible factor.]If ’a’ itself is irreducible then we are done. If ’a’ is not irreducible, thena = a1b1 where neither a1 or b1 is a unit. Now 〈a〉 ⊂ 〈a1〉 by Note 1 Part(1) (if 〈a〉 = 〈a1〉 then by Note 1 Part (2) ’a’ and a1 would be associates,contradicting the fact that neither a1 nor b1 is a unit). If a1 is irreduciblethen a1 is an irreducible factor of ’a’. If not, write a1 = a2b2 where neithera2 nor b2 is a unit. As above, we have 〈a1〉 ⊂ 〈a2〉. Continue this processto form a strictly ascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ ....By Lemma 45.10, this chain terminates with some 〈ar 〉 and this ar mustbe irreducible (or else we would contruct 〈ar+1〉 with 〈ar 〉 ⊂ 〈ar+1〉). Wenow have a = b1b2...brar and so ar is an irreducible factor of ’a’.
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Theorem 45.11.
Theorem 45.11. (Continued)
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. (Continued) Now that we know ’a’ has an irreducible factor, weshow that it can be written as a product of irreducible factors.By above, we have that ’a’ (neither 0 nor a unit in D) is irreducible or ofthe form a = p1c1 for p1 an irreducible and c1 not a unit.
If c1 is not aunit (and of course it’s not 0) then by the argument of the first paragraphwe have 〈a〉 ⊂ 〈c1〉 and if c1 is not irreducible then c1 = p2c2 forirreducible p2 with c2 not a unit. Continuing we again get a strictlyascending chain of ideals 〈a〉 ⊂ 〈c1〉 ⊂ 〈c2〉 ⊂ ....By Lemma 45.10, this chain terminates with some cr = qr that isirreducible (as argued in the first paragraph). Then a = p − 1p2...prqr is aproduct of irreducibles.
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Theorem 45.11.
Theorem 45.11. (Continued)
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. (Continued) Now that we know ’a’ has an irreducible factor, weshow that it can be written as a product of irreducible factors.By above, we have that ’a’ (neither 0 nor a unit in D) is irreducible or ofthe form a = p1c1 for p1 an irreducible and c1 not a unit. If c1 is not aunit (and of course it’s not 0) then by the argument of the first paragraphwe have 〈a〉 ⊂ 〈c1〉 and if c1 is not irreducible then c1 = p2c2 forirreducible p2 with c2 not a unit.
Continuing we again get a strictlyascending chain of ideals 〈a〉 ⊂ 〈c1〉 ⊂ 〈c2〉 ⊂ ....By Lemma 45.10, this chain terminates with some cr = qr that isirreducible (as argued in the first paragraph). Then a = p − 1p2...prqr is aproduct of irreducibles.
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Theorem 45.11.
Theorem 45.11. (Continued)
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. (Continued) Now that we know ’a’ has an irreducible factor, weshow that it can be written as a product of irreducible factors.By above, we have that ’a’ (neither 0 nor a unit in D) is irreducible or ofthe form a = p1c1 for p1 an irreducible and c1 not a unit. If c1 is not aunit (and of course it’s not 0) then by the argument of the first paragraphwe have 〈a〉 ⊂ 〈c1〉 and if c1 is not irreducible then c1 = p2c2 forirreducible p2 with c2 not a unit. Continuing we again get a strictlyascending chain of ideals 〈a〉 ⊂ 〈c1〉 ⊂ 〈c2〉 ⊂ ....By Lemma 45.10, this chain terminates with some cr = qr that isirreducible (as argued in the first paragraph). Then a = p − 1p2...prqr is aproduct of irreducibles.
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Theorem 45.11.
Theorem 45.11. (Continued)
Lemma. 45.11. Let D be a PID. Every element that is neither 0 nor aunit of D is a product of irreducibles.
Proof. (Continued) Now that we know ’a’ has an irreducible factor, weshow that it can be written as a product of irreducible factors.By above, we have that ’a’ (neither 0 nor a unit in D) is irreducible or ofthe form a = p1c1 for p1 an irreducible and c1 not a unit. If c1 is not aunit (and of course it’s not 0) then by the argument of the first paragraphwe have 〈a〉 ⊂ 〈c1〉 and if c1 is not irreducible then c1 = p2c2 forirreducible p2 with c2 not a unit. Continuing we again get a strictlyascending chain of ideals 〈a〉 ⊂ 〈c1〉 ⊂ 〈c2〉 ⊂ ....By Lemma 45.10, this chain terminates with some cr = qr that isirreducible (as argued in the first paragraph). Then a = p − 1p2...prqr is aproduct of irreducibles.
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Lemma 45.12.
Lemma 45.12.
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. Let 〈p〉 be a maximal ideal of D, a PID. Suppose p = ab in D.Then by Note 1 Part(1) 〈p〉 ⊆ 〈a〉.
If 〈p〉 = 〈a〉 then by Note 1 Part(2) ’a’and p are associates and b is a unit. If 〈p〉 6= 〈a〉 then since 〈p〉 is maximalit must be that 〈a〉 = D. From the definition of ”ideal in D” we haveD = 〈1〉, so in this case 〈a〉 = 〈1〉 and by Note 1 Part(2), ’a’ and 1 areassociates and hence ’a’ is a unit. Thus, if p = ab then either ’a’ is a unitor b is a unit; that is, p is irreducible.
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Lemma 45.12.
Lemma 45.12.
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. Let 〈p〉 be a maximal ideal of D, a PID. Suppose p = ab in D.Then by Note 1 Part(1) 〈p〉 ⊆ 〈a〉. If 〈p〉 = 〈a〉 then by Note 1 Part(2) ’a’and p are associates and b is a unit. If 〈p〉 6= 〈a〉 then since 〈p〉 is maximalit must be that 〈a〉 = D.
From the definition of ”ideal in D” we haveD = 〈1〉, so in this case 〈a〉 = 〈1〉 and by Note 1 Part(2), ’a’ and 1 areassociates and hence ’a’ is a unit. Thus, if p = ab then either ’a’ is a unitor b is a unit; that is, p is irreducible.
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Lemma 45.12.
Lemma 45.12.
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. Let 〈p〉 be a maximal ideal of D, a PID. Suppose p = ab in D.Then by Note 1 Part(1) 〈p〉 ⊆ 〈a〉. If 〈p〉 = 〈a〉 then by Note 1 Part(2) ’a’and p are associates and b is a unit. If 〈p〉 6= 〈a〉 then since 〈p〉 is maximalit must be that 〈a〉 = D. From the definition of ”ideal in D” we haveD = 〈1〉, so in this case 〈a〉 = 〈1〉 and by Note 1 Part(2), ’a’ and 1 areassociates and hence ’a’ is a unit. Thus, if p = ab then either ’a’ is a unitor b is a unit; that is, p is irreducible.
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Lemma 45.12.
Lemma 45.12.
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. Let 〈p〉 be a maximal ideal of D, a PID. Suppose p = ab in D.Then by Note 1 Part(1) 〈p〉 ⊆ 〈a〉. If 〈p〉 = 〈a〉 then by Note 1 Part(2) ’a’and p are associates and b is a unit. If 〈p〉 6= 〈a〉 then since 〈p〉 is maximalit must be that 〈a〉 = D. From the definition of ”ideal in D” we haveD = 〈1〉, so in this case 〈a〉 = 〈1〉 and by Note 1 Part(2), ’a’ and 1 areassociates and hence ’a’ is a unit. Thus, if p = ab then either ’a’ is a unitor b is a unit; that is, p is irreducible.
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Lemma 45.12.
Lemma 45.12. (Continued)
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. (Continued) Conversely, suppose that p is an irreducible in D. If〈p〉 ⊆ 〈a〉 then by Note 1 Part(1) we must have p = ab for some b in D.If ’a’ is a uit, then ’a’ and 1 are associates and by Note 1 Part(2), we have〈a〉 = 〈1〉 = D and 〈a〉 is a maximal ideal. If ’a’ is not a unit, then b mustbe a unit (since p is irreducible) so there exists u ∈ D such that bu = 1.
Then pu = abu = a and by Note 1 Part(1) 〈a〉 ⊆ 〈p〉 and since p and ’a’are associates, by Note 1 Part (2) we have 〈a〉 = 〈p〉. We have now shownthat if 〈p〉 ⊆ 〈a〉 then either 〈a〉 = D (if ’a’ is a unit) or 〈a〉 = 〈p〉 (if ’a’ isnot a unit).
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Lemma 45.12.
Lemma 45.12. (Continued)
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. (Continued) Conversely, suppose that p is an irreducible in D. If〈p〉 ⊆ 〈a〉 then by Note 1 Part(1) we must have p = ab for some b in D.If ’a’ is a uit, then ’a’ and 1 are associates and by Note 1 Part(2), we have〈a〉 = 〈1〉 = D and 〈a〉 is a maximal ideal. If ’a’ is not a unit, then b mustbe a unit (since p is irreducible) so there exists u ∈ D such that bu = 1.Then pu = abu = a and by Note 1 Part(1) 〈a〉 ⊆ 〈p〉 and since p and ’a’are associates, by Note 1 Part (2) we have 〈a〉 = 〈p〉.
We have now shownthat if 〈p〉 ⊆ 〈a〉 then either 〈a〉 = D (if ’a’ is a unit) or 〈a〉 = 〈p〉 (if ’a’ isnot a unit).
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Lemma 45.12.
Lemma 45.12. (Continued)
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. (Continued) Conversely, suppose that p is an irreducible in D. If〈p〉 ⊆ 〈a〉 then by Note 1 Part(1) we must have p = ab for some b in D.If ’a’ is a uit, then ’a’ and 1 are associates and by Note 1 Part(2), we have〈a〉 = 〈1〉 = D and 〈a〉 is a maximal ideal. If ’a’ is not a unit, then b mustbe a unit (since p is irreducible) so there exists u ∈ D such that bu = 1.Then pu = abu = a and by Note 1 Part(1) 〈a〉 ⊆ 〈p〉 and since p and ’a’are associates, by Note 1 Part (2) we have 〈a〉 = 〈p〉. We have now shownthat if 〈p〉 ⊆ 〈a〉 then either 〈a〉 = D (if ’a’ is a unit) or 〈a〉 = 〈p〉 (if ’a’ isnot a unit).
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Lemma 45.12.
Lemma 45.12. (Continued)
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. (Continued) Conversely, suppose that p is an irreducible in D. If〈p〉 ⊆ 〈a〉 then by Note 1 Part(1) we must have p = ab for some b in D.If ’a’ is a uit, then ’a’ and 1 are associates and by Note 1 Part(2), we have〈a〉 = 〈1〉 = D and 〈a〉 is a maximal ideal. If ’a’ is not a unit, then b mustbe a unit (since p is irreducible) so there exists u ∈ D such that bu = 1.Then pu = abu = a and by Note 1 Part(1) 〈a〉 ⊆ 〈p〉 and since p and ’a’are associates, by Note 1 Part (2) we have 〈a〉 = 〈p〉. We have now shownthat if 〈p〉 ⊆ 〈a〉 then either 〈a〉 = D (if ’a’ is a unit) or 〈a〉 = 〈p〉 (if ’a’ isnot a unit).
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Lemma 45.12.
Lemma 45.12. (Continued)
Lemma. 45.12. An ideal 〈p〉 in a PID is maximal if and only if p isirreducible.
Proof. (Continued) So there is no proper ideal of D which properlycontains 〈p〉 (of course all ideals of D are principal). That is, 〈p〉 is amaximal ideal.
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Lemma 45.13.
Lemma 45.13.
Lemma. 45.13. In a PID, if an irreducible p divies ab then either p | a orp | b.
Proof. Let D be a PID and suppose that for an irreducible p ∈ D we havep | ab. Then ab ∈ 〈p〉 (since 〈p〉 consists of all multiples of p).
Since p isirreducible, by Lemma 45.12 〈p〉 is a maximal ideal in D. By Corollary27.16, every maximal ideal is a prime ideal, so 〈p〉 is a prime ideal. Thenab ∈ 〈p〉 implies that either a ∈ 〈p〉 or b ∈ 〈p〉. That is, by Note 1 Part(1), either p | a or p | b.
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Lemma 45.13.
Lemma 45.13.
Lemma. 45.13. In a PID, if an irreducible p divies ab then either p | a orp | b.
Proof. Let D be a PID and suppose that for an irreducible p ∈ D we havep | ab. Then ab ∈ 〈p〉 (since 〈p〉 consists of all multiples of p). Since p isirreducible, by Lemma 45.12 〈p〉 is a maximal ideal in D. By Corollary27.16, every maximal ideal is a prime ideal, so 〈p〉 is a prime ideal.
Thenab ∈ 〈p〉 implies that either a ∈ 〈p〉 or b ∈ 〈p〉. That is, by Note 1 Part(1), either p | a or p | b.
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Lemma 45.13.
Lemma 45.13.
Lemma. 45.13. In a PID, if an irreducible p divies ab then either p | a orp | b.
Proof. Let D be a PID and suppose that for an irreducible p ∈ D we havep | ab. Then ab ∈ 〈p〉 (since 〈p〉 consists of all multiples of p). Since p isirreducible, by Lemma 45.12 〈p〉 is a maximal ideal in D. By Corollary27.16, every maximal ideal is a prime ideal, so 〈p〉 is a prime ideal. Thenab ∈ 〈p〉 implies that either a ∈ 〈p〉 or b ∈ 〈p〉. That is, by Note 1 Part(1), either p | a or p | b.
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Lemma 45.13.
Lemma 45.13.
Lemma. 45.13. In a PID, if an irreducible p divies ab then either p | a orp | b.
Proof. Let D be a PID and suppose that for an irreducible p ∈ D we havep | ab. Then ab ∈ 〈p〉 (since 〈p〉 consists of all multiples of p). Since p isirreducible, by Lemma 45.12 〈p〉 is a maximal ideal in D. By Corollary27.16, every maximal ideal is a prime ideal, so 〈p〉 is a prime ideal. Thenab ∈ 〈p〉 implies that either a ∈ 〈p〉 or b ∈ 〈p〉. That is, by Note 1 Part(1), either p | a or p | b.
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Theorem 45.17.
Theorem 45.17.
Theorem. 45.17. Every PID is a UFD
Proof. Theorem 45.11 shows that every PID satisfies the first property ofa UFD and gives for a in a PID D where ’a’ is neither 0 nor a unit, afactorization a = p1p2 · · · pr into irreducibles. property 2 of a UFD saysthat such a factorization is unique (in terms of associates).
Leta = q1q2 · · · qs be another factorization of ’a’ into irreducibles. Then wehave p1 | (q1q2 · · · qs). By Corollary 45.14, p1 | qj for some j . Reorder theq’s such that qj becomes q1. Then q1 = p1u1 where u1 is a unit. Then p1
and q1 are associates. Then p1p2 · · · pr = (p1u1)q1q2 · · · qs . Bycancellation in integral domain D (Theorem 19.5)p2p3 · · · pr = u1q1q2 · · · qs .
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Theorem 45.17.
Theorem 45.17.
Theorem. 45.17. Every PID is a UFD
Proof. Theorem 45.11 shows that every PID satisfies the first property ofa UFD and gives for a in a PID D where ’a’ is neither 0 nor a unit, afactorization a = p1p2 · · · pr into irreducibles. property 2 of a UFD saysthat such a factorization is unique (in terms of associates). Leta = q1q2 · · · qs be another factorization of ’a’ into irreducibles. Then wehave p1 | (q1q2 · · · qs). By Corollary 45.14, p1 | qj for some j . Reorder theq’s such that qj becomes q1.
Then q1 = p1u1 where u1 is a unit. Then p1
and q1 are associates. Then p1p2 · · · pr = (p1u1)q1q2 · · · qs . Bycancellation in integral domain D (Theorem 19.5)p2p3 · · · pr = u1q1q2 · · · qs .
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Theorem 45.17.
Theorem 45.17.
Theorem. 45.17. Every PID is a UFD
Proof. Theorem 45.11 shows that every PID satisfies the first property ofa UFD and gives for a in a PID D where ’a’ is neither 0 nor a unit, afactorization a = p1p2 · · · pr into irreducibles. property 2 of a UFD saysthat such a factorization is unique (in terms of associates). Leta = q1q2 · · · qs be another factorization of ’a’ into irreducibles. Then wehave p1 | (q1q2 · · · qs). By Corollary 45.14, p1 | qj for some j . Reorder theq’s such that qj becomes q1. Then q1 = p1u1 where u1 is a unit. Then p1
and q1 are associates. Then p1p2 · · · pr = (p1u1)q1q2 · · · qs . Bycancellation in integral domain D (Theorem 19.5)p2p3 · · · pr = u1q1q2 · · · qs .
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Theorem 45.17.
Theorem 45.17.
Theorem. 45.17. Every PID is a UFD
Proof. Theorem 45.11 shows that every PID satisfies the first property ofa UFD and gives for a in a PID D where ’a’ is neither 0 nor a unit, afactorization a = p1p2 · · · pr into irreducibles. property 2 of a UFD saysthat such a factorization is unique (in terms of associates). Leta = q1q2 · · · qs be another factorization of ’a’ into irreducibles. Then wehave p1 | (q1q2 · · · qs). By Corollary 45.14, p1 | qj for some j . Reorder theq’s such that qj becomes q1. Then q1 = p1u1 where u1 is a unit. Then p1
and q1 are associates. Then p1p2 · · · pr = (p1u1)q1q2 · · · qs . Bycancellation in integral domain D (Theorem 19.5)p2p3 · · · pr = u1q1q2 · · · qs .
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Theorem 45.17.
Theorem 45.17. (Continued)
Theorem. 45.17. Every PID is a UFD
Proof. (Continued) Repeating the process we have1 = u1u2 · · · urqr+1 · · · qs (WLOG s ≥ r). But if s > R and we have someq still present on the right-hand side, say qr+1, then the other elements ofthe right-hand side are an inverse of the q, for example(qr+1)
−1 = u1u2 · · · urqr+2qr+3 · · · qs . But this contradicts the fact thatthe q’s are irreducible and so (by definition) not units. So there are no q’sremaining on the right-hand side and r = s. So pi = uiqi for i = 1, 2, ..., rand such pi is an associate of qi . This is Property 2 in the definition of aUFD and so D is a UFD.
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Theorem 45.17.
Theorem 45.17. (Continued)
Theorem. 45.17. Every PID is a UFD
Proof. (Continued) Repeating the process we have1 = u1u2 · · · urqr+1 · · · qs (WLOG s ≥ r). But if s > R and we have someq still present on the right-hand side, say qr+1, then the other elements ofthe right-hand side are an inverse of the q, for example(qr+1)
−1 = u1u2 · · · urqr+2qr+3 · · · qs . But this contradicts the fact thatthe q’s are irreducible and so (by definition) not units. So there are no q’sremaining on the right-hand side and r = s. So pi = uiqi for i = 1, 2, ..., rand such pi is an associate of qi . This is Property 2 in the definition of aUFD and so D is a UFD.
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Corollary 45.18 Fundamental Theorem of Arithmetic.
Corollary 45.18 Fundamental Theorem of Arithmetic.
Corollary. 45.18 Fundamental Theorem of Arithmetic. The integraldomain Z is a UFD.
Proof. We know that Z is a PID (see the note after Definition 45.7). Soby Theorem 45.17, Z is a UFD.
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Lemma 45.23.
Lemma 45.23.
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. Let f (x) ∈ D[x ] be given where f (x) is a nonconstant polynomialwith coefficients a0, a1, ..., an. Let c be a gcd of the ai . Then for each i ,we have ai = cgi for some gi ∈ D.
We have f (x) = cg(x). Now there isno irreducible dividing all of the gi (if so, say the irreducible in b, then cbdivides all ai , but cb - c so in this case c is not a gcd of the ai ). So a gcdof the gi must be a unit and have an associate of 1. So 1 is a gcd of thegi and g(x) is a primitive polynomial.
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Lemma 45.23.
Lemma 45.23.
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. Let f (x) ∈ D[x ] be given where f (x) is a nonconstant polynomialwith coefficients a0, a1, ..., an. Let c be a gcd of the ai . Then for each i ,we have ai = cgi for some gi ∈ D. We have f (x) = cg(x). Now there isno irreducible dividing all of the gi (if so, say the irreducible in b, then cbdivides all ai , but cb - c so in this case c is not a gcd of the ai ). So a gcdof the gi must be a unit and have an associate of 1.
So 1 is a gcd of thegi and g(x) is a primitive polynomial.
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Lemma 45.23.
Lemma 45.23.
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. Let f (x) ∈ D[x ] be given where f (x) is a nonconstant polynomialwith coefficients a0, a1, ..., an. Let c be a gcd of the ai . Then for each i ,we have ai = cgi for some gi ∈ D. We have f (x) = cg(x). Now there isno irreducible dividing all of the gi (if so, say the irreducible in b, then cbdivides all ai , but cb - c so in this case c is not a gcd of the ai ). So a gcdof the gi must be a unit and have an associate of 1. So 1 is a gcd of thegi and g(x) is a primitive polynomial.
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Lemma 45.23.
Lemma 45.23.
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. Let f (x) ∈ D[x ] be given where f (x) is a nonconstant polynomialwith coefficients a0, a1, ..., an. Let c be a gcd of the ai . Then for each i ,we have ai = cgi for some gi ∈ D. We have f (x) = cg(x). Now there isno irreducible dividing all of the gi (if so, say the irreducible in b, then cbdivides all ai , but cb - c so in this case c is not a gcd of the ai ). So a gcdof the gi must be a unit and have an associate of 1. So 1 is a gcd of thegi and g(x) is a primitive polynomial.
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Lemma 45.23.
Lemma 45.23. (Continued)
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. (Continued) For uniqueness, if f (x) = dh(x) also for some h ∈ Dand h(x) ∈ D[x ] with h(x) primitive, then each irreducible factor of cmust divide d and each irreducible factor of d must divide c (or else, as inthe first paragraph, 1 is not a gcd of the respective coefficients of g or hand hence g or h is not primitive).By setting cg(x) = dh(x) (since both equal f (x)) and cancellingirreducible factors of c into d (Theorem 19.5), we arrive at ug(x) = vh(x)for a unit u ∈ D. But then v must be a unit of D or we would be able tocancel irreducible factors of v into u.
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Lemma 45.23.
Lemma 45.23. (Continued)
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. (Continued) For uniqueness, if f (x) = dh(x) also for some h ∈ Dand h(x) ∈ D[x ] with h(x) primitive, then each irreducible factor of cmust divide d and each irreducible factor of d must divide c (or else, as inthe first paragraph, 1 is not a gcd of the respective coefficients of g or hand hence g or h is not primitive).By setting cg(x) = dh(x) (since both equal f (x)) and cancellingirreducible factors of c into d (Theorem 19.5), we arrive at ug(x) = vh(x)for a unit u ∈ D. But then v must be a unit of D or we would be able tocancel irreducible factors of v into u.
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Lemma 45.23.
Lemma 45.23. (Continued)
Lemma. 45.23. If D is a UFD then for every nonconstant f (x) ∈ D[x ]we have f (x) = cg(x) where c ∈ D, g(x) ∈ D[x ] and g(x) is a primitive.The element c is unique up to a unit factor in D and is the content off (x). Also g(x) is unique up to a unit factor in D.
Proof. (Continued) So u and v are both units and c is unique up to aunit factor (here, d = v−1uc). Since f (x) = cg(x), then the primitivepolynomial g(x) is also unique up to a unit factor.
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma.
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. Let f (x) = a0 + a1x + a2x2 + ... + anx
n andg(x) = b0 + b1x + b2x
2 + ... + bmxm be primitives in D[x ] and leth(x) = f (x)g(x). Let p be an irreducible in D. Then p does not divide allai and p does not divide bj∗ (or else a multiple of p is a gcd of the ai andof the bj and 1 is not a gcd since all gcd’s are associates). [since f (x) andg(x) are primitive.]
Let ar be the first coefficient (i.e., r is the smallest value) of f (x) notdivisible by p; that is, p | ai for 0 ≤ i < r but p - ar . Similarly, let p | bj for0 ≤ i < s but p - bs .
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma.
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. Let f (x) = a0 + a1x + a2x2 + ... + anx
n andg(x) = b0 + b1x + b2x
2 + ... + bmxm be primitives in D[x ] and leth(x) = f (x)g(x). Let p be an irreducible in D. Then p does not divide allai and p does not divide bj∗ (or else a multiple of p is a gcd of the ai andof the bj and 1 is not a gcd since all gcd’s are associates). [since f (x) andg(x) are primitive.]Let ar be the first coefficient (i.e., r is the smallest value) of f (x) notdivisible by p; that is, p | ai for 0 ≤ i < r but p - ar . Similarly, let p | bj for0 ≤ i < s but p - bs .
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma.
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. Let f (x) = a0 + a1x + a2x2 + ... + anx
n andg(x) = b0 + b1x + b2x
2 + ... + bmxm be primitives in D[x ] and leth(x) = f (x)g(x). Let p be an irreducible in D. Then p does not divide allai and p does not divide bj∗ (or else a multiple of p is a gcd of the ai andof the bj and 1 is not a gcd since all gcd’s are associates). [since f (x) andg(x) are primitive.]Let ar be the first coefficient (i.e., r is the smallest value) of f (x) notdivisible by p; that is, p | ai for 0 ≤ i < r but p - ar . Similarly, let p | bj for0 ≤ i < s but p - bs .
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma. (Continued)
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. (Continued) The cofficient of x r+s in h(x) = f (x)g(x) is (we arein a commutative ring):
cr+s = (a0br+s + a1br+s−1 + ... + ar−1bs+1) + arbs
+ (ar+1bs−1 + ar+2bs−2 + ... + ar+sb0)(1)
Now p | ai for 0 ≤ i < r implies thatp | (a0br+s + a1br+s−1 + ... + ar−1bs+1) and p | bj for 0 ≤ j < s impliesthat p | (ar+1bs−1 + ar+2bs−2 + ... + ar+sb0).
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma. (Continued)
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. (Continued) The cofficient of x r+s in h(x) = f (x)g(x) is (we arein a commutative ring):
cr+s = (a0br+s + a1br+s−1 + ... + ar−1bs+1) + arbs
+ (ar+1bs−1 + ar+2bs−2 + ... + ar+sb0)(1)
Now p | ai for 0 ≤ i < r implies thatp | (a0br+s + a1br+s−1 + ... + ar−1bs+1) and p | bj for 0 ≤ j < s impliesthat p | (ar+1bs−1 + ar+2bs−2 + ... + ar+sb0).
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Lemma 45.25 Gauss’s Lemma.
Lemma 45.25 Gauss’s Lemma. (Continued)
Lemma. 45.25 Gauss’s Lemma. If D is a UFD, then a product of twoprimitive polynomials in D[x ] is again primitive.
Proof. (Continued) But p does not divide ar or bs , so p does not dividearbs and consequently p does not divide cr+s . So we have that anyirreducible p ∈ D does not divide some coefficient of f (x)g(x). So the gcdof the coefficients of f (x)g(x) is 1 and f (x)g(x) is primitive.
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Lemma 45.27.
Lemma 45.27.
Lemma. 45.27. Let D be a UFD and let F be a field of quotients of D.Let f (x) ∈ D[x ] where (degree f (x)) > 0. If f (x) is an irreducible in D[x ],then f (x) is also an irreducible in F [x ]. Also, if f (x) is primitive in D[x ]and irreducible in F [x ], then f (x) is irreducible in D[x ].
Proof. We prove the contrapositive of the first claim. Suppose that anonconstant f (x) ∈ D[x ] factors into polynomials of lower degree in F [x ];that is f (x) = r(x)s(x) for r(x), s(x) ∈ F [x ]. Then since F is a field ofquotients of D, each coefficient in r(x) and s(x) is of the form a/b forsome a, b ∈ D, b 6= 0.
By ”clearing the denominators” (i.e. multiplyingthrough by a common multiple of the denominator) we can getdf (x) = r1(x)s1(x) for d ∈ D and r1(x), s1(x) ∈ D[x ] where the degrees ofr1(x) and s1(x) equal the degrees of r(x) and s(x), respectively.
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Lemma 45.27.
Lemma 45.27.
Lemma. 45.27. Let D be a UFD and let F be a field of quotients of D.Let f (x) ∈ D[x ] where (degree f (x)) > 0. If f (x) is an irreducible in D[x ],then f (x) is also an irreducible in F [x ]. Also, if f (x) is primitive in D[x ]and irreducible in F [x ], then f (x) is irreducible in D[x ].
Proof. We prove the contrapositive of the first claim. Suppose that anonconstant f (x) ∈ D[x ] factors into polynomials of lower degree in F [x ];that is f (x) = r(x)s(x) for r(x), s(x) ∈ F [x ]. Then since F is a field ofquotients of D, each coefficient in r(x) and s(x) is of the form a/b forsome a, b ∈ D, b 6= 0. By ”clearing the denominators” (i.e. multiplyingthrough by a common multiple of the denominator) we can getdf (x) = r1(x)s1(x) for d ∈ D and r1(x), s1(x) ∈ D[x ] where the degrees ofr1(x) and s1(x) equal the degrees of r(x) and s(x), respectively.
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Lemma 45.27.
Lemma 45.27.
Lemma. 45.27. Let D be a UFD and let F be a field of quotients of D.Let f (x) ∈ D[x ] where (degree f (x)) > 0. If f (x) is an irreducible in D[x ],then f (x) is also an irreducible in F [x ]. Also, if f (x) is primitive in D[x ]and irreducible in F [x ], then f (x) is irreducible in D[x ].
Proof. We prove the contrapositive of the first claim. Suppose that anonconstant f (x) ∈ D[x ] factors into polynomials of lower degree in F [x ];that is f (x) = r(x)s(x) for r(x), s(x) ∈ F [x ]. Then since F is a field ofquotients of D, each coefficient in r(x) and s(x) is of the form a/b forsome a, b ∈ D, b 6= 0. By ”clearing the denominators” (i.e. multiplyingthrough by a common multiple of the denominator) we can getdf (x) = r1(x)s1(x) for d ∈ D and r1(x), s1(x) ∈ D[x ] where the degrees ofr1(x) and s1(x) equal the degrees of r(x) and s(x), respectively.
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Lemma 45.27.
Lemma 45.27. (Continued)
Lemma. 45.27. Let D be a UFD and let F be a field of quotients of D.Let f (x) ∈ D[x ] where (degree f (x)) > 0. If f (x) is an irreducible in D[x ],then f (x) is also an irreducible in F [x ]. Also, if f (x) is primitive in D[x ]and irreducible in F [x ], then f (x) is irreducible in D[x ].
Proof. (Continued) By Lemma 45.23 f (x) = cg(x) , r1(x) = c1r2(x),and s1(x) = c2s2(x) for primitive polynomials g(x), r2(x) and s2(x) inD[x ] and c , c1, c2 ∈ D. Then dcg(x) = c1r2(x)c2s2(x) = c1c2r2(x)s2(x)and by Lemma 45.25 the product r2(x)s2(x) is primitive. By theuniqueness part of Lemma 45.23, c1c2 = dcu for some unit u in D. Butthen dcg(x) = dcur2(x)s2(x) and so f (x) = cg(x) = cur2(x)s2(x) wherecu ∈ D and r2(x), s2(x) ∈ D[x ].So f (x) factors nontrivially into polynomials of the same degree in D[x ] asthe degree of the polynomial factors of f (x) in F [x ].A nonconstant f (x) ∈ D[x ] that is primitive in D[x ] and irreducible in F [x ]is also irreducible in D[x ] since D[x ] ⊆ F [x ].
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Lemma 45.27.
Lemma 45.27. (Continued)
Lemma. 45.27. Let D be a UFD and let F be a field of quotients of D.Let f (x) ∈ D[x ] where (degree f (x)) > 0. If f (x) is an irreducible in D[x ],then f (x) is also an irreducible in F [x ]. Also, if f (x) is primitive in D[x ]and irreducible in F [x ], then f (x) is irreducible in D[x ].
Proof. (Continued) By Lemma 45.23 f (x) = cg(x) , r1(x) = c1r2(x),and s1(x) = c2s2(x) for primitive polynomials g(x), r2(x) and s2(x) inD[x ] and c , c1, c2 ∈ D. Then dcg(x) = c1r2(x)c2s2(x) = c1c2r2(x)s2(x)and by Lemma 45.25 the product r2(x)s2(x) is primitive. By theuniqueness part of Lemma 45.23, c1c2 = dcu for some unit u in D. Butthen dcg(x) = dcur2(x)s2(x) and so f (x) = cg(x) = cur2(x)s2(x) wherecu ∈ D and r2(x), s2(x) ∈ D[x ].So f (x) factors nontrivially into polynomials of the same degree in D[x ] asthe degree of the polynomial factors of f (x) in F [x ].A nonconstant f (x) ∈ D[x ] that is primitive in D[x ] and irreducible in F [x ]is also irreducible in D[x ] since D[x ] ⊆ F [x ].
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Corollary 45.28.
Corollary 45.28.
Corollary. 45.28. If D is a UFD and F is a field of quotients of D, then anonconstant f (x) ∈ D[x ] factors into a product of two polynomials oflower degrees r and s in F [x ] if and only if it has a factorization intopolynomials of the same degrees r and s in D[x ].
Proof. In the proof of Lemma 45.27, if f (x) factors in F [x ] intof (x) = r(x)s(x) where r(x) and s(x) are of degrees smaller than thedegree of f (x), then f (x) = cur2(x)s2(x) in D[x ] where the degrees ofr(x) and r2(x) are the same and the degrees of s(x) and s2(x) are thesame. The converse holds since D[x ] ⊆ F [x ].
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Corollary 45.28.
Corollary 45.28.
Corollary. 45.28. If D is a UFD and F is a field of quotients of D, then anonconstant f (x) ∈ D[x ] factors into a product of two polynomials oflower degrees r and s in F [x ] if and only if it has a factorization intopolynomials of the same degrees r and s in D[x ].
Proof. In the proof of Lemma 45.27, if f (x) factors in F [x ] intof (x) = r(x)s(x) where r(x) and s(x) are of degrees smaller than thedegree of f (x), then f (x) = cur2(x)s2(x) in D[x ] where the degrees ofr(x) and r2(x) are the same and the degrees of s(x) and s2(x) are thesame. The converse holds since D[x ] ⊆ F [x ].
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Theorem 45.29.
Theorem 45.29.
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. Let f (x) ∈ D[x ] where f (x) is neither 0 nor a unit. If f (x) is ofdegree 0, we are done since D is a UFD.
Suppose (degree f (x)) > 0. Letf (x) = g1(x)g2(x) · · · gr (x) be a factorization of f (x) in D[x ] having thegreatest number r of factors of positive degree (so no gi (x) is a constantpolynomial). There is such a greatest number of such factors since rcannot exceed the degree of f (x).
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Theorem 45.29.
Theorem 45.29.
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. Let f (x) ∈ D[x ] where f (x) is neither 0 nor a unit. If f (x) is ofdegree 0, we are done since D is a UFD. Suppose (degree f (x)) > 0. Letf (x) = g1(x)g2(x) · · · gr (x) be a factorization of f (x) in D[x ] having thegreatest number r of factors of positive degree (so no gi (x) is a constantpolynomial). There is such a greatest number of such factors since rcannot exceed the degree of f (x).
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Theorem 45.29.
Theorem 45.29.
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. Let f (x) ∈ D[x ] where f (x) is neither 0 nor a unit. If f (x) is ofdegree 0, we are done since D is a UFD. Suppose (degree f (x)) > 0. Letf (x) = g1(x)g2(x) · · · gr (x) be a factorization of f (x) in D[x ] having thegreatest number r of factors of positive degree (so no gi (x) is a constantpolynomial). There is such a greatest number of such factors since rcannot exceed the degree of f (x).
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) Now factor each gi (x) in the form gi (x) = cih(x)where ci is the content of gi (x) (by Lemma 45.23, c is a gcd of thecoefficients of gi (x)) and hi (x) is a primitive polynomial. Also, each hi (x)must be irreducible; if an hi (x) could be factored then the correspondingfactorization of gi (x) (described in the proof of Lemma 45.27) would givea factorization of f (x) with more than r factors, contradicting the choiceof r .
Thus we now have f (x) = c1h1(x)c2h2(x) · · · crhr (x) where thehi (x) are irreducible in D[x ]. If we now factor the ci into irreducibles in D(since D is a UFD), we obtain a factorization of f (x) into a product ofirreducibles in D[x ].
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) Now factor each gi (x) in the form gi (x) = cih(x)where ci is the content of gi (x) (by Lemma 45.23, c is a gcd of thecoefficients of gi (x)) and hi (x) is a primitive polynomial. Also, each hi (x)must be irreducible; if an hi (x) could be factored then the correspondingfactorization of gi (x) (described in the proof of Lemma 45.27) would givea factorization of f (x) with more than r factors, contradicting the choiceof r . Thus we now have f (x) = c1h1(x)c2h2(x) · · · crhr (x) where thehi (x) are irreducible in D[x ]. If we now factor the ci into irreducibles in D(since D is a UFD), we obtain a factorization of f (x) into a product ofirreducibles in D[x ].
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) Now factor each gi (x) in the form gi (x) = cih(x)where ci is the content of gi (x) (by Lemma 45.23, c is a gcd of thecoefficients of gi (x)) and hi (x) is a primitive polynomial. Also, each hi (x)must be irreducible; if an hi (x) could be factored then the correspondingfactorization of gi (x) (described in the proof of Lemma 45.27) would givea factorization of f (x) with more than r factors, contradicting the choiceof r . Thus we now have f (x) = c1h1(x)c2h2(x) · · · crhr (x) where thehi (x) are irreducible in D[x ]. If we now factor the ci into irreducibles in D(since D is a UFD), we obtain a factorization of f (x) into a product ofirreducibles in D[x ].
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) The factorization of f (x) ∈ D[x ] where f (x) hasdegree 0 is unique since D is a UFD. If f (x) has degree greater than 0,then any factorization of f (x) into irreducibles in D[x ] corresponds to afactorization in F [x ] into units (the factors in D; the constant factors)and, by Lemma 45.27, irreducible polynomials in F [x ]. By Theorem 23.20,these irreducible polynomials are unique, except for possible constantfactors in F . But as an irreducible in D[x ], each polynomial of degree > 0appearing in the factorization of f (x) in D[x ] is primitive (or else theconstant gcd of the coefficients could be factored out).
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) The factorization of f (x) ∈ D[x ] where f (x) hasdegree 0 is unique since D is a UFD. If f (x) has degree greater than 0,then any factorization of f (x) into irreducibles in D[x ] corresponds to afactorization in F [x ] into units (the factors in D; the constant factors)and, by Lemma 45.27, irreducible polynomials in F [x ]. By Theorem 23.20,these irreducible polynomials are unique, except for possible constantfactors in F . But as an irreducible in D[x ], each polynomial of degree > 0appearing in the factorization of f (x) in D[x ] is primitive (or else theconstant gcd of the coefficients could be factored out).
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) By the uniqueness part of Lemma 45.23, theseirreducible polynomial factors are unique in D[x ] up to unit factors (thatis, unique up to being associates). The product of the irreducibles in D inthe factorization of f (x) (that is, the constant factors) is the content off (x), which is unique up to a unit facotr by Lemma 45.23. Thus allirreducibles in D[x ] appearing in the factorization are unique up to orderand associates.
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Theorem 45.29.
Theorem 45.29. (Continued)
Theorem. 45.29. If D is a UFD, then D[x ] is a UFD.
Proof. (Continued) By the uniqueness part of Lemma 45.23, theseirreducible polynomial factors are unique in D[x ] up to unit factors (thatis, unique up to being associates). The product of the irreducibles in D inthe factorization of f (x) (that is, the constant factors) is the content off (x), which is unique up to a unit facotr by Lemma 45.23. Thus allirreducibles in D[x ] appearing in the factorization are unique up to orderand associates.
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Corollary 45.30.
Corollary 45.30.
Corollary. 45.30. If F is a field and x1, x2, ..., xn are indeterminates, thenF [x1, x2, ..., xn] is a UFD.
Proof. By Theorem 23.20, F [x ] is a UFD. By Corollary 45.30 andinduction, F [x1, x2], F [x1, x2, x3],...,F [x1, x2, ..., xn] are UFDs.
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Corollary 45.30.
Corollary 45.30.
Corollary. 45.30. If F is a field and x1, x2, ..., xn are indeterminates, thenF [x1, x2, ..., xn] is a UFD.
Proof. By Theorem 23.20, F [x ] is a UFD. By Corollary 45.30 andinduction, F [x1, x2], F [x1, x2, x3],...,F [x1, x2, ..., xn] are UFDs.
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