I'm currently studying for my abstract algebra exam and came along this question in one of the reviews:

If F[$x_1, x_2, \ldots, x_n$] is PID, then F must be a field

I know that if F[x] is a PID, then F is a field. So if it's true that $\frac{F[x_1, x_2, \ldots, x_n]}{<x_n>} \cong F[x_1, x_2, \ldots, x_{n-1}]$, I could do an induction argument to do the proof. I just don't know if this statement is true (although it does seem to true). I've tried to come up with a ring homomorphism:

$\phi: {F[x_1, x_2, \ldots, x_n]} \rightarrow F[x_1, x_2, \ldots, x_{n-1}]$

where ker $\phi = <x_n>$ so I could apply the first isomorphism theorem, but haven't had any luck.

Thank you!

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    $\begingroup$ The property can't be true: otherwise apply it to $F[x]$: $F[x,y]$ is a PID implies that $F[x]$ is a field, which is absurd. As for the isomorphism, it's pretty clear : $x_i \mapsto x_i$ for $i<n$ and $x_n \mapsto 0$ $\endgroup$ – Max Dec 11 '17 at 22:09
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    $\begingroup$ @Max: It can be true if $n=0$ or $n=1$. Fortunately, contradictions prove anything, so the claimed statement is true without putting a restriction on $n$. (if that hurts your head, think of it in terms of "If $F[x_1, \ldots, x_n]$ is a PID, then $n \leq 1$") $\endgroup$ – Hurkyl Dec 11 '17 at 23:07
  • $\begingroup$ @Max But F[x,y] isn't a PID, so that counter example doesn't work/isn't relevant? Or am I confusing what you're saying? $\endgroup$ – klamont15 Dec 12 '17 at 0:27
  • $\begingroup$ @Hurkyl : yeah ok, it doesn't hurt my head it just feels like a really weird way of putting it; I would have said "$F[x_1,...,x_n]$ is never a PID, except if $n\leq 1$ and $F$ is a field" (which I know means the same in the end, but feels better). $\endgroup$ – Max Dec 12 '17 at 6:52

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