Why is the polynomial ring of more than one variable not a PID?

Question

Let $R$ be a field. Show that the polynomial ring $R[x_{1},...,x_{n}]$ is not a PID if $n>1$.

How do I show this?

So far, I've shown that $(2)+(x)$ wouldn't be principal in $\mathbb{Z}[x]$ so I was thinking of showing that $(x_1)+(x_2)$ would not be principal.

Answer 1

Let $R$ be a domain which is not a field. Then $R[x]$ is not a PID.

Let $r\in R$, $r\ne0$ and $r$ not invertible. We want to prove that $(r,x)$ is not principal.

Suppose $r=f(x)g(x)$, for some $f$ and $g$. Then both $f$ and $g$ are constant. Thus a generator for $(r,x)$ must be a constant, $a$. Then $x=ah(x)$, so $h$ must have degree $1$, but this implies $a$ is invertible, by comparing coefficients.

(This is the same proof as for $\mathbb{Z}[x]$.)

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Since the ring of polynomials in one indeterminate is not a field, an easy induction proves the claim.