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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.

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    $\begingroup$ Show that $\;(x_1,\,x_2)\;$ is not principal, for example. $\endgroup$ – DonAntonio Mar 12 '16 at 16:53
  • $\begingroup$ 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 $\endgroup$ – quinan2000 Mar 12 '16 at 16:55
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    $\begingroup$ @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. $\endgroup$ – egreg Mar 12 '16 at 16:55
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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.

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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Apr 10 '17 at 6:08
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    $\begingroup$ this actually proves that if $R[x]$ is PID then $R$ is a field. $\endgroup$ – Paramanand Singh Apr 10 '17 at 6:09

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