## 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. $\square$.

• 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

• 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. 🙂

Thanks!