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Let $k$ be a field and $f, g \in k[x]$ irreducible. How can I tell whether the field extensions $k[x] / (f)$ and $k[x] / (g)$ are isomorphic as extensions of $k$? Is there any necessary and sufficient condition?

Of course, a necessary condition is $\deg(f) = \deg(g)$.

A sufficient condition is that $f(x) = g(ax + b)$ for some $a \in k^\times$ and $b \in k$, or in other words that $f$ and $g$ are related by an automorphism of $k[x]$. However, this condition is not sufficient: for example, we have $\mathbb{Q}(\sqrt{2} + \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(2 \sqrt{2} + \sqrt{3})$, where the minimal polynomials of the generators are $$ f = x^4 - 10 x + 1, \qquad g = x^4 - 22x + 25, $$ respectively. These polynomials are clearly not related by an automorphism of $\mathbb{Q}[x]$, yet they both define extensions of $\mathbb{Q}$ isomorphic to $\mathbb{Q}(\sqrt{2}, \sqrt{3})$.

Tobias Fritz
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    If $k$ happens to be finite, then the degree is all you need. – Arthur May 22 '24 at 12:24
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    But, for example, $\Bbb Q[X]/(X^2-2)$ is not isomorphic to $\Bbb Q[X]/(X^2-3)$, see here. – Dietrich Burde May 22 '24 at 12:24
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    Maybe not totally satisfying but $k[X]/f$ and $k[X]/g$ for irreducible $f,g$ of the same degree are isomorphic iff there exists $h \in k[X]$ (can be chosen of degree strictly less than $f$) such that $g(h(x)) = 0~\mathrm{mod}~ f$. So similar to but somewhat weaker than the sufficient condition with the automorphism of $\mathbb{Q}[x]$ you mention. – Sebastian Schoennenbeck May 22 '24 at 12:32
  • That sounds like a start @SebastianSchoennenbeck, thanks! – Tobias Fritz May 22 '24 at 12:59
  • Show that a generator of the first field, $x+(f)$, is an element of the second field, and a generator of the second field is an element of the first. – Gerry Myerson May 22 '24 at 13:11
  • What are you commenting on @GerryMyerson? If you're referring to @SebastianSchoennenbeck's statement, then that's obvious enough. But I'm still hoping for a better answer to the question. – Tobias Fritz May 22 '24 at 14:01
  • I expect this problem is already complicated for cubic extensions. I haven't found a reference giving a complete solution. – lhf May 22 '24 at 15:26
  • I'm trying to give a pathway to an answer to your question, Tobias. – Gerry Myerson May 22 '24 at 22:58

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