If $k[X]/f = k[X]/g$, does $f = g$? Let $k$ be a field and $f, g$ be irreducible monic polynomials in $k[X]$.  Suppose $k[X]/f \stackrel{\sim}{=} k[X]/g$.  Then does $f = g$?
If so, how can this be generalized?  Otherwise, how should I modify the  condition so that $f = g$?
 A: First, we should be explicit that we are talking about an isomorphism of fields over $k$, rather than just an isomorphism of fields.  Otherwise, there are strange examples like $\mathbb{C}(t)[X]/(X^2-t) \cong \mathbb{C}(\sqrt{t}) \cong \mathbb{C}(t) \cong \mathbb{C}(t)[X]/(X)$.
With this assumption, $k[X]/(f(X)) \cong k[X]/(g(X))$ implies that $f$ and $g$ are the same degree.
There exists a map of $k$-algebras $k[X]/(f(X)) \to k[X]/(g(X))$ if and only if there is some $h\in k[X]$ such that $g(X) \mid f(h(X))$.  Since any morphism of fields is an injection, when these two fields have the same degree over $k$, such a map must be an isomorphism.
The conclusion is that $k[X]/(f(X)) \cong k[X]/(g(X))$ as fields over $k$, exactly when $\deg{f}=\deg{g}$ and $g(X) \mid f(h(X))$ for some $h\in k[X]$.
This condition is maybe not so useful*, but perhaps it provides some concreteness to RghtHndSd's comment that the issue here involves "choices of coordinates".

* But here's a use: we can verify that, when $k$ is a finite field, $k[X]/(f(X)) \cong k[X]/(g(X))$ for all $f,g$ of the same degree: There exist some $k,l\geq 0$ with $X^k \equiv X^l\pmod{g}$, by the pigeonhole principle.  Then $g$ divides $f(X^k - X^l)$.
