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?