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

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.

• Show that $\;(x_1,\,x_2)\;$ is not principal, for example. – DonAntonio Mar 12 '16 at 16:53
• I've shown that (2)+(x) wouldn't be principal in Z[x] so I was thinking of showing that ($x_{1}) + (x_{2}$) would not be principal – quinan2000 Mar 12 '16 at 16:55
• @quinan2000 Yes, that's the way. It's actually better proving the more general result that if $R$ is a domain and not a field, then $R[x]$ is not a PID, so an easy induction proves the claim for $n>2$ variables. – egreg Mar 12 '16 at 16:55

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 $$\mathbb{Z}[x]$$.)

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

• There is some more work to be done. Once you show that $a$ is invertible (i.e $a$ is a unit in $R[x]$) then $R[x] = (a) = (r, x)$. Hence there are polynomials $p, q\in R[x]$ such that $pr + qx = 1$ and putting $x = 0$ (or comparing constant terms on both sides) we see that $r$ is invertible. – Paramanand Singh Apr 10 '17 at 6:08
• this actually proves that if $R[x]$ is PID then $R$ is a field. – Paramanand Singh Apr 10 '17 at 6:09