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Though seemed to be a common question, I couldn’t find any information on that.

I can’t imagine the Euclidean function would be like, if there exists one, so my guess is not. But how to prove it?

fantasie
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    If it were, it would be a PID. But this is a famous example.... – Randall Oct 29 '20 at 14:00
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    No, $\Bbb Z[X]$ is not a Euclidean domain as it is not PID, to prove the last assertion consider the ideal $\langle 2, X\rangle$ in $\Bbb Z[X]$. – Sumanta Oct 29 '20 at 14:01
  • By the dupes, Euclidean domains are PIDs, but $\Bbb Z[x],$ is not a PID (we have many posts on such topics that can be located by search). – Bill Dubuque Oct 30 '20 at 01:02

3 Answers3

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No. Since $\langle 2, x \rangle$ is not an ideal generated by one element, $\mathbb Z[x]$ is not even a PID, let alone an Euclidean ring.

fantasie
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This ring has Krull dimension $2$ but Euclidean domains necessarily have Krull dimension no greater than $1$.

rschwieb
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  • How do you show $\mathbb Z[x]$ has Krull dimension $1$? – JonathanZ Oct 29 '20 at 14:51
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    @JonathanZsupportsMonicaC, the claim is that it doesn't. – LSpice Oct 29 '20 at 14:52
  • Ooops, my bad. I wanted to ask how one shows it has Krull dimension 2. – JonathanZ Oct 29 '20 at 14:54
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    One can for example show that for noetherian rings $R$, one has $\text{dim}(R[x]) = \text{dim}(R) + 1$. By induction one also gets the analogous result for more variables then. This reduces the problem to computing the Krull dimension of $\mathbb{Z}$, which is $1$, since it is a pid (and not a field). – Con Oct 29 '20 at 14:57
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    But also it is entirely obvious that ${0}\subseteq (x)\subseteq (2)$ is a proper chain, making the dimension at least $2$. – rschwieb Oct 29 '20 at 15:03
  • Indeed. I just prefer to see the general statement behind it. – Con Oct 29 '20 at 15:05
  • @TMO Wasn't directed at you. I was about to say what you said plus what I said, anyhow. – rschwieb Oct 29 '20 at 15:07
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The other answers are already great. Let me nevertheless state the following fact, which is a nice exercise:

$R[x]$ is euclidean if and only if $R$ is a field.

One direction is usually proven in standard algebra courses/books. Namely, if $k$ is a field, then $k[x]$ is euclidean with the degree as euclidean function.

On the other hand (to give you a hint): We can show that if $R[x]$ is a pid (euclidean domains are pid's), then $R$ has to be a field. For this consider the ideal $(a,x)$ for some non-zero $a \in R$. Now we can find some $f \in R[x]$ with $(a,x) = (f)$. Using the notion of degree and evaluation morphisms, try to show that $a$ has an inverse in $R$.

Con
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