A polynomial ring in more than one variable is not a PID

Let R be a commutative ring with 1. Prove that a polynomial ring in more than one variable over R is not a principal ideal domain.

[Short, Nice Proof]

If R[x_1,\ldots,x_n,x_{n+1}] = R[x_1,\ldots,x_n][x_{n+1}] is a PID, then by Corollary 8 on page 281 of D&F, R[x_1,\ldots,x_n] is a field- a contradiction.

[Long, Complicated Proof]

We begin with a lemma.

Lemma: Let R be a commutative ring with 1. If R[x] is a principal ideal domain, then every element of R is either 0, a unit, or a zero divisor. Proof: Let r \in R be nonzero. The ideal (r,x) = (d) is principal. Since r = dp for some p, the constant coefficient of d is nonzero. Now x = dq for some q. Let q_0, d_0, and p_0 be the constant terms of q, d, and p, respectively. If q_0 \neq 0, then d_0 is a zero divisor since d_0q_0 = 0. Thus r = d_0p_0 is a zero divisor. If q_0 = 0, then letting d_1 and q_1 be the linear coefficients of d and q, respectively, we have d_0q_1 + d_1q_0 = 1, so that d_0q_1 = 1. In particular, d_0 is a unit. Thus we in fact have (r,x) = (1 + xe(x)) for some e \in R[x]. Now (1 + xe(x))t(x) = x for some t, so that t(x) + xt(x)e(x) = x. Note that xe(x)t(x) has no constant term, so that comparing coefficients, the constant term t_0 of t is zero. Write t(x) = xt^\prime(x). Then xt^\prime(x) + x^2e(x)t^\prime(x) = x, so that x(t^\prime(x) + xe(x)t^\prime(x)) = x. Note that x is not a zero divisor, so that t^\prime(x)(1 + xe(x)) = 1. That is, we have (r,x) = R[x]. Using a lemma proved in this previous exercise, we have 0 \cong R[x]/((r)+(x)) \cong R/(r), so that (r) = R, and thus r is a unit. \square

Now let R be a commutative ring with 1 and consider R[x,y]. If R[x,y] = (R[x])[y] is a principal ideal domain, then every element of R[x] is either zero, a unit, or a zero divisor. However, x \in R[x] is not zero and not a zero divisor. Moreover, x is not a unit since xp(x) has constant term 0 for all p. Every polynomial ring in two or more variables is isomorphic to R[x,y] for some R, and thus the result is proved.

This is a handy lemma that we ended up not using here, but don’t want to throw out.

Lemma: Let R be a commutative ring with 1, and suppose a \in R is irreducible and not a unit. Then (a,x) \subseteq R[x] is not a principal ideal. Proof: Suppose to the contrary that (a,x) = (d). Since a = du for some u and a is irreducible, there are two cases. If u is a unit, then in fact x = ap for some p \in R[x]. Comparing degrees, we see that p(x) = c_0 + c_1x, so that ac_1 = 1. This is a contradiction since a is not a unit. If d is a unit, then (a,x) = R[x]. In particular, ap + xq = 1 for some p,q. Comparing constant coefficients, we have that a is a unit, a contradiction. Thus (a,x) is not principal. \square.

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  • mike  On February 7, 2011 at 12:26 am

    I understand why the constant coefficient d_0 is a unit. But I dont understand why that means that (r,x)=(1+x*e(x))for some e(x)in R[x]

    any help would be greatly appreciated
    Thank you for your time

    • nbloomf  On February 7, 2011 at 12:48 am

      We have (r,x) = (d). Since the constant term of d is a unit, we can multiply d by d_0^{-1} without changing the generated ideal. Say d(x) = \sum_{i=0}^m a_ix^i; then d_0^{-1}d(x) = 1 + x\left( \sum_{i=1}^m d_0^{-1}a_ix^{i-1} \right).

  • Julie  On February 13, 2011 at 12:29 am

    So, since we know that R[x] is a PID all we need to show is that the only elements in R are 0, zero divisors, or units. I know from our text that whatever is a unit in R is definitely a unit in R[x]. And, obviously 0 in R is the same as 0 in R[x]. However, I don’t understand the connection for zero divisors. If c is a zero divisor in R does that automatically imply that c is a zero divisor in R[x]? Or, maybe, is it vice versa?

    • nbloomf  On February 13, 2011 at 8:44 am

      The zero divisors in R are zero divisors in R[x] for the same reason that units in R are units in R[x]– If ab = 0 for some nonzero a,b \in R, then the equation still holds in R[x].

  • Peter  On December 3, 2011 at 9:28 pm

    Couldn’t you just say: Suppose R[x1,…xn] is a principal domain. Then by corollary 8 (chapter 8.2), R[x1,…,xn-1] is a field. Contradiction.

    • nbloomf  On December 5, 2011 at 10:49 pm

      Oh my… that’s much better than my solution. 🙂


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