WABASH COLLEGE, DEPARTMENT OF MATHEMATICS

MAT332 Ring Theory Assignment

Mason Keller

April 18, 2016

1 A CONDITION ON MULTIPLYING IDEALS IN P.I.D.S

First, define the product and of two ideals I , J as

I J :={

n∑k=1

ik jk : n ∈N, ik ∈ I , jk ∈ J

},

I + J := {i + j : i ∈ I , j ∈ J

}

We wish to show that if I , J are ideals in a P.I.D. R, then I J = I⋂

J if and only if I + J = R.Suppose I = (a) and J = (b). Our definition of I J is easily manipulated to show that I J = (ab).Now, I

⋂J is the set of all multiples of both a and b, so each element is a multiple of lcm(a,b).

I + J = (a,b) = R = (1), which is to say that "a and b are coprime" is equivalent to I + J = R.l cm(a,b) = ab if and only if g cd(a,b) = 1, and the desired result follows.

2 EXAMPLE OF RING ISOMORPHISM THEOREMS

We wish to discover some properties of Z[x]/(5, x2 −3) = R. First we note that the ideals (5)and (x2 −3) are both subsets of (5, x2 −3), and conclude that R ∼= (Z[x]/(5))/((5, x2 −3)/(5)).Further, Z[x]/(5) ∼=Z/(5)[x]. We have now moved from Z[x]/(5, x2−3) to Z5[x]/(5, x2−3)/(5).The ’denominator’ quotient is the set of remainders from the operation a(x)5+b(x)(x2−3)/5.These remainders are simply polynomials with a factor of x2 − 3 with coefficients from Z5.But this is just the ideal (x2 −3) in Z5. So we have R = Z5[x]/(x2 −3). We can now move toR = Z5[x̄] where x̄2 ≡ 3. x̄ has precisely the same numerical properties as 2

p3, and moving

1

from Z5 back to Z/(5) gives us the desired isomorphism to Z[ 2p

3]/(5), which is a field of 25elements.

3 A SUFFICIENT CONDITION FOR A FACTORIAL DOMAIN TO BE A P.I.D.

Suppose R is a factorial domain. We wish to show that it is sufficient for every prime ideal tobe maximal to guarantee that R is a P.I.D.. Consider the set of non-principal ideals. This is aposet, ordered by set containment. Now, each totally ordered subset has a maximal elementbecause the union of the subset elements (which is really a union of a chain of ideals) is alsoan ideal. Said maximal element surely cannot be principal, since no element in the chain isprincipal. By Zorn’s lemma we can arrive at a maximal element for our whole poset, say x. Aprime ideal will be principal (by virtue of R being factorial), so x must not be a prime ideal.Since x is not prime, there are a1 and a2 so that neither ai ∈ x but a1a2 ∈ x. Also, x + (ai ) willbe a principal ideal, since it is bigger than x. But

[x + (a1)]∗ [x + (a2)] = x + (a1 ∗a2)+ [x ∗ (a1)+x ∗ (a2)] = x + (a1 ∗a2) = x,

the leftmost side of which is principal. So x must be principal, which is a contradiction. Ourposet must have been empty – all ideals were principal to begin with.

4 FACTORING IN F[X,Y ]

Let F be an infinite field, and let f ∈ F [x, y]. Suppose f (a, a) = 0 for all a ∈ F . We wish to showthat x − y is a factor of f (x, y). Because x − y has a leading coefficient in y which is a unit, wecan do polynomial long division without passing to F (x)[y]. The remainder additionally, hasto be constant in y , and so is also an element of F [x]. It has roots everywhere though, so itmust be 0, otherwise it could only have a finite number of roots. This means that f (x, y) =q(x, y)(x − y), which is what we desired.

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