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Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?

($\cong$ denotes an isomorphism that is the identity on restriction to $K$).

Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!

maths
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  • Notice that in your example there is a (unique) automorphism $\varphi \colon \mathbb{Q}[X] \to \mathbb{Q}[X]$ with $\varphi(X) = X-1$ for which $\varphi(X^2-2) = x^2-2X-1$. If generally $\phi \colon K[X] \to K[X]$ is an automorphism with $\phi(f) = g$ then $\phi$ maps $(f)$ isomorphically into $(g)$, and thus induces an isomorphism $K[X]/(f) \to K[X]/(g)$, $P + (f) \mapsto \phi(P) + (g)$. But I don’t know if every isomorphism $K[X]/(f) \to K[X]/(g)$ lifts to an automorphism $K[X] \to K[X]$ which maps $f$ to $g$. I feel like this would be a rather strong statement. – Jendrik Stelzner Dec 31 '15 at 14:08
  • Related: https://math.stackexchange.com/questions/259406/if-ideal-quotients-of-a-ring-are-isomorphic-are-these-ideals-isomorphic – rogerl Dec 31 '15 at 14:24
  • @jendrik: hmm, I was wondering exactly the same thing...would be interested in an answer to that. – maths Jan 03 '16 at 17:30
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    @dcw: After i quick search I found this post, which shows that it is (sadly) not always possible to lift such an isomorphism. – Jendrik Stelzner Jan 03 '16 at 17:40
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    @jendrik Thanks. Interestingly, a map from $K(\alpha)$ to $K(\beta)$ does lift to a map from $\overline{K}$ to $\overline{K}$. – maths Jan 03 '16 at 17:58
  • Related: http://mathoverflow.net/questions/65109/checking-if-fx-i-is-isomorphic-to-fx-j – Watson Aug 17 '16 at 19:34
  • Related: https://math.stackexchange.com/questions/1831305/polynomial-ring-with-isomorphic-quotients – Watson Nov 22 '16 at 08:09

1 Answers1

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If $f,g$ are quadratic polynomials (and $K$ is any field of characteristic not 2), then by the quadratic formula the isomorphism classes are classified by the discriminant "$b^2 - 4ac$" (if $f = ax^2 + bx + c$), modulo squares. Ie, the isomorphism classes are in bijection with $K^\times/(K^\times)^2$.

In any other situation things become a lot more difficult.

If $K$ is $\mathbb{Q}$, then $\mathbb{Q}[X]/(f)\cong\mathbb{Q}[X]/(g)$ if and only if the action of $G_\mathbb{Q} := \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the roots of $f$ is isomorphic to the action of $G_\mathbb{Q}$ on the roots of $g$. Ie, if $X_f,X_g$ denote the sets of roots of $f,g$, then $\mathbb{Q}[X]/(f)\cong\mathbb{Q}[X]/(g)$ if and only if there is a bijection $\phi : X_f\stackrel{\sim}{\rightarrow}X_g$ such that $\phi(\sigma x) = \sigma\phi(x)$ for all $x\in X_f$, $\sigma\in G_\mathbb{Q}$.

This is a consequence of the Galois correspondence, which says that the association sending any finite extension $L := \mathbb{Q}[X]/(h)$ of $\mathbb{Q}$ (with $h$ irreducible) to $X_h$ (as a set with $G_\mathbb{Q}$-action) gives an equivalence of categories between the category of finite field extensions of $\mathbb{Q}$ and the category of finite sets with a transitive $G_\mathbb{Q}$-action.

The result follows from the fact that two finite extensions of $\mathbb{Q}$ are isomorphic as fields if and only if they are isomorphic as extensions of $\mathbb{Q}$ (any abstract isomorphism between them must fix their prime subfields).

The same result will also be true if $K$ is replaced by any $\mathbb{F}_p$ (though this situation is trivial since finite extensions of $\mathbb{F}_p$ are uniquely determined by degree). With suitable modifications the result is also true when $K$ is any finite extension of $\mathbb{Q}$, and with more care, even true when $K$ is an algebraic extension of $\mathbb{Q}$.

If $K$ is not an algebraic extension over its prime subfield, then things can get weird, as we move into the world of arithmetic geometry. For example, for any field $k$, you can set $K := k(t)$, then $K = K[X]/(X) = k(t)$ is isomorphic to $K[X]/(X^2-t)\cong k(\sqrt{t})$.

If $K$ is finite over $\mathbb{Q}$, then in certain cases you may also be able to use class field theory.

Though, I should mention that in practice it's rare that you would care about isomorphisms between fields (as abstract fields). You generally will want to restrict yourself to "nice" extensions of a fixed base field, in which case a number of the conditions above can be relaxed.

oxeimon
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