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?
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.