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This question already has an answer here:

We know that if F is a field, then the polynomial ring over F is a PID. Do you have a counterexample that shows that if F isn’t a field than the polynomial ring over F isn’t a PID?

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marked as duplicate by Dietrich Burde abstract-algebra Dec 18 '18 at 12:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Did you already try something? We can help you more if you show what you've tried $\endgroup$ – Math Girl Dec 18 '18 at 11:12
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    $\begingroup$ Did you try any example at all? Because you can really just pick any example. $\endgroup$ – Tobias Kildetoft Dec 18 '18 at 11:17
  • $\begingroup$ I tried with the polynomial ring over Z but without success. $\endgroup$ – MathIsLove Dec 18 '18 at 11:18
  • $\begingroup$ Great, that is a good place to start. Are you familiar with the fact that the ideal generated by an irreducible element is maximal in a PID? $\endgroup$ – Tobias Kildetoft Dec 18 '18 at 11:20
  • $\begingroup$ Yes but I don’t remember the proof. $\endgroup$ – MathIsLove Dec 18 '18 at 11:22
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$\mathbf Z[X]$ is such a counter-example: for any prime number $p$, the ideal $(p, X)$ is not principal for degree reasons in polynomials over an integral domain: $p$ should be a multiple of a generator, which therefore should be a constant, and this constant can be only $1$, $p$ or $-p$. You can easily check it can't be any of them.

Note that $(p,X)$ is a maximal ideal in $\mathbf Z[X]$ since $$ \mathbf Z[X]/(p,X)\simeq (\mathbf Z/p\mathbf Z)[X]/(X) \simeq \mathbf Z/p\mathbf Z$$ is a field.

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  • $\begingroup$ Thank you very much! $\endgroup$ – MathIsLove Dec 18 '18 at 11:26
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You can even show, extending slightly the argument in the other answer, the following.

Suppose $A$ is a domain. Then $A[x]$ is a PID iff $A$ is a field.

If $A$ is a field, then $A[x]$ is Euclidean, and thus a PID.

Suppose now $A[x]$ is a PID. Let $0 \ne a \in A$. We want to show that $a$ is a unit in $A$.

Consider the ideal $(a, x)$ of $A[x]$. Note that $(a, x)$ is the ideal of $A[x]$ of polynomials whose costant term is a multiple of $a$.

By assumption, $(a, x) = (c)$ for some $c \in A[x]$.

$c \mid a$, thus $c$ has to be a constant. But the only constants that divide $x$ are the units of $A$. (If $c (b_{0} + b_{1} x + \dots) = x$, then $c b_{1} = 1$.)

Therefore $c$ is a unit, and then $(a, x) = (c) = A[x]$, so that $1 \in (a, x)$, that is, $1$ is a multiple of $a$, that is, $a$ is a unit in $A$.

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  • $\begingroup$ Thank you. It is very clear! $\endgroup$ – MathIsLove Dec 18 '18 at 12:08

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