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Question: Let $F$ be a commutative ring with identity. Is it true that $F$ is a field if and only if $F[X]$ is an Euclidean domain?

If $F$ is a field, clearly one can do division algorithm to prove that $F[X]$ is an Euclidean domain. What puzzles me is the converse, that is, if $F[X]$ is an Euclidean domain, can we conclude that $F$ is a field?

My attempt: Let $u\in F$ be a nonzero element. Consider $1,u\in F[X].$ Since $F[X]$ is an ED, there exist $f(X),r(X)\in F[X]$ such that $$1 = uf(X) + r(X)$$ where $\deg(r(X))<\deg(f(X)).$ Since the constant polynomial $1$ has degree $0,$ it follows that $\deg(f(X)) = 0.$ Denote $f(X) = v\in F.$ Since $\deg(r(X))<\deg(f(X)) = 0,$ it implies that $r(X) = 0.$ Therefore, we have $$1 = uv.$$ Hence, $u$ is a unit and thus $F$ is a field.

Is my attempt above correct?

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    $\begingroup$ You are assuming a lot more than "$F[x]$ is a Euclidean domain". You are assuming that its Euclidean function is the degree. Instead, consider the ideal $(u,X)$ in $F[X]$. $\endgroup$ – Angina Seng Nov 24 '18 at 4:11
  • $\begingroup$ @LordSharktheUnknown You are right. The existence of Euclidean function cannot be chosen explicitly. Since $F[X]$ is PID, the ideal $(u,X)$ is principal, say $(u,X) = (f)$ for some $f\in F[X].$ Then we can show that $f$ is a constant function which has multiplicative inverse, thus concluding that $1 \in (u,X).$ By comparing constant terms, we can conclude that $u$ is a unit. $\endgroup$ – Idonknow Nov 24 '18 at 5:01
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    $\begingroup$ There is an old article in the American Math. Monthly showing that a Euclidean domain with a unique quotient and remainder has to be F[x] for some field F. (The integers don't fit, since remainders could be negative.) The article title and year escape me at the moment... $\endgroup$ – KCd Nov 24 '18 at 6:37
  • $\begingroup$ Well, @KCd, someone gives two references for this fact in math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf (proof of Theorem 2.2) :) $\endgroup$ – darij grinberg Nov 26 '18 at 0:53
  • $\begingroup$ @darij grinberg спасибо за напоминание. I have been traveling over the last couple of days and was going to look up the reference when the trip ends. You've saved me the need to do that. $\endgroup$ – KCd Nov 26 '18 at 0:58
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No, that's not right. You're using Euclidean Division wrong. For one thing, you would have to conclude that $\deg(r(x)) < \deg(u)$, not $\deg(f(x))$. This would actually imply that $r(x) = 0$, and thus that $f(x)$ was an inverse for $u$, and thus would have to be in $F$, implying that $F$ is a field. But, as Lord Shark the Unknown points out, this depends on assuming that the degree is the Euclidean function. Since we can't do this, we have to find some other approach.

It's actually easier to prove a stronger version of the statement, namely that if $F[x]$ is a PID then $F$ is a field. If $F[x]$ is a PID, $F$ must be an integral domain as well, and so $F[x]/\langle x \rangle \cong F$ implies that $\langle x \rangle$ is prime. Since nonzero prime ideals are maximal in a PID, it follows that the quotient $F[x]/\langle x \rangle \cong F$ is actually a field, and so we are done.

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    $\begingroup$ So for example $x+1$ is constant? You need to rephrase this - "has non-zero constant term" springs to mind... $\endgroup$ – David C. Ullrich Nov 25 '18 at 14:17
  • $\begingroup$ @David C. Ullrich You are right - even as I was typing that something seemed off about it, but for some reason the more I keep thinking about it the more I became convinced that it had to be true. I have fixed the argument accordingly. $\endgroup$ – silvascientist Nov 25 '18 at 22:56
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    $\begingroup$ $F[x]$ is a domain, hence $F$ is a domain; since $F[x]/\langle x\rangle\cong F$, it follows that $\langle x\rangle$ is a prime ideal. $\endgroup$ – egreg Nov 25 '18 at 23:13
  • $\begingroup$ @egreg You’re right, that is a bit slicker. The answer has been updated. $\endgroup$ – silvascientist Nov 25 '18 at 23:59

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