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Let $R$ be a commutative ring with unity such that $R[X]$ is a UFD. Denote the ideal $\langle X\rangle $ by $I$.

Prove that

  • If $I$ is maximal, then $R[X]$ is a PID.
  • If $R[X]$ is a Euclidean Domain then $I$ is maximal.
  • If $R[X]$ is a PID then it is a ED.

We know that $R[X]/\langle X\rangle \cong R$. Since $I$ is maximal then $R$ is a field. Also every field is a Euclidean Domain and hence a PID and also a UFD.

By Gauss Lemma $R[X]$ is a UFD.

But these facts are not helping anything in these statements to prove.

Please give some hints.

user26857
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  • Have you tried using the main property of an Euclidean domain, i.e. the Euclidean algorithm? It'll be slightly harder since you don't know a priori that the norm is the degree (and it might not be, norms are quite "flexible" in some well-defined sense) but it still allows you to come to the same conclusion. – Patrick Da Silva Jul 22 '16 at 06:25
  • A hint for $(3)$: try proving $R$ is a field, whence the result follows. To do so, consider the ideal $I_{r} := \langle r, X \rangle$ for any nonzero element $r \in R$; try to use the fact that $I_{r}$ is principal to prove that $r$ is invertible. – Alex Wertheim Jul 22 '16 at 06:47

2 Answers2

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The first part is easy. We have $R[X]/I=R[X]/\langle X\rangle\cong R$. We are given that $I$ is maximal. So $R[X]/I$ is a field. Hence $R$ is a field. So $R[X]$ is an ED, and hence a PID.

learning_math
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These results are somewhat straightforward with this result:

Let $R$ be a commutative ring with $1 \neq 0$. The following are equivalent.

  1. $R$ is a field.
  2. $R[X]$ is a Euclidean domain.

  3. $R[X]$ is a P.I.D.

    1. $\Rightarrow$ 2. is a standard result.

    2. $\Rightarrow$ 3. follows from the result that says "Euclidean domains are P.I.D.'s"

    3. $\Rightarrow$ 1.: Suppose $R[X]$ is a P.I.D; the ideal $(X)$ is a prime ideal in $R[X]$ because $R[X]/(X) \cong R$ is an integral domain ($R$ is subring of the integral domain $R[X]$). Moreover, nonzero prime ideals in a P.I.D. are maximal, so $(X)$ is maximal in $R[X]$. Thus $R[X]/(X) \cong R$ is a field.

Now your statements should follow immediately.

[Note 3. $\Rightarrow$ 1. can be proved using the definition of a field, as hinted by @Alex Wertheim in the comments, but it is a little longer...]

cat
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