# Counterexample PID [duplicate]

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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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• Did you already try something? We can help you more if you show what you've tried – Math Girl Dec 18 '18 at 11:12
• Did you try any example at all? Because you can really just pick any example. – Tobias Kildetoft Dec 18 '18 at 11:17
• I tried with the polynomial ring over Z but without success. – MathIsLove Dec 18 '18 at 11:18
• 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? – Tobias Kildetoft Dec 18 '18 at 11:20
• Yes but I don’t remember the proof. – MathIsLove Dec 18 '18 at 11:22

## 2 Answers

$$\mathbf Z[X]$$ 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.

• Thank you very much! – MathIsLove Dec 18 '18 at 11:26

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 $$A$$.

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