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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 $f = g$?

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  • $\begingroup$ No. If $k$ is a finite field, then if $f,g$ are irreducible polynomials over $k$ then $k[X]/f\cong k[X]/g$ if and only if $g$ and $f$ are the same degree. $\endgroup$ – Thomas Andrews Oct 12 '14 at 4:23
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    $\begingroup$ Not even true for characteristic zero, I guess - $k=\mathbb R$ and $f(x)=x^2+1$ and $g(x)=x^2+2$, for example. $\endgroup$ – Thomas Andrews Oct 12 '14 at 4:25
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    $\begingroup$ Sure, they are both $\mathbb C$. The only non-trivial algebraic extension of $\mathbb R$ is $\mathbb C$. $\endgroup$ – Thomas Andrews Oct 12 '14 at 4:26
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    $\begingroup$ More philosophically: isomorphism classes don't differentiate between choices of coordinates. $\endgroup$ – RghtHndSd Oct 12 '14 at 4:35
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    $\begingroup$ Yes, $g(x)=f(x+a)$ is certainly a very simple case, I suppose. $\endgroup$ – Thomas Andrews Oct 12 '14 at 4:59
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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 $f(X^k - X^l)$.

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  • $\begingroup$ By an isomorphism over $k$, you mean that that fixes $k$ pointwise? $\endgroup$ – Pteromys Oct 12 '14 at 20:21
  • $\begingroup$ @Pteromys That's right. $\endgroup$ – Slade Oct 12 '14 at 22:52

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