# Same minimal polynomial gives isomorphism

Let L and P be two field extensions of K. For $$a \in L$$ and $$b\in P$$ algebraic over K with the same minimal polynomial $$f_a=f_b$$ then there exists an isomorphism $$w: K(a) \rightarrow K(b)$$ with $$w(a)=b$$.

Proof attempt: Since $$f$$ is irreducible and has $$a$$ as root then $$K(a)$$ is isomorphic to $$K[x]/(f)$$ with the isomorphism $$X+f\rightarrow a$$. And we can do the same for $$b$$ and we get $$K(b)$$ is isomorphic to $$K[x]/(f)$$. Therefore $$K(a)$$ is isomorphic with $$K(b)$$.

My Questions:

1.What do I do now?

2."$$f$$ is irreducible and has $$a$$ as root then $$K(a)$$ is isomorphic to $$K[x]/(f)$$" we took it for granted in the lecture but why does this hold?

They might be trivial questions but I would really like to get an understanding of it.

Thanks in advance for the help.

• $K(a) \cong K[x]/(f) \cong K(b)$
– lhf
Nov 23, 2021 at 21:24
• but why w(a)=b? why is this isomorphism unique? Nov 23, 2021 at 21:27
• We have $a \mapsto x \bmod f \mapsto b$.
– lhf
Nov 23, 2021 at 21:28
• Trace the morphisms explicitly and verify that the composition fixes $K$ and sends $a$ to $b$. Nov 23, 2021 at 21:30
• the composition of only the two morphisms? Nov 23, 2021 at 21:36

For the first question, we just need to check the last condition $$w(a) = b$$. But just looking at the definition of the two isomorphisms in the composite $$w : K(a) \to K[x]/(f) \to K(b),$$ the first maps $$a$$ to $$x + (f)$$, and the second maps $$x + (f)$$ to $$b$$. So $$w(a) = b$$, as we desire.

For the second question, we start by considering the evaluation $$K$$-algebra homomorphism $$\phi : K[x] \to K(a)$$ which maps $$x$$ to $$a$$. Then since $$f(a) = 0$$, every $$g \in (f)$$ is mapped to zero by $$\phi$$. (Just note that $$\phi(g) = g(a)$$.) Thus $$\phi$$ descends to a $$K$$-algebra homomorphism $$\widetilde{\phi} : K[x] / (f) \to K(a)$$. To show that $$K(a) \cong K[x]/(f)$$, it just remains to show that $$\widetilde{\phi} : K[x] / (f) \to K(a)$$ is an isomorphism. Indeed, being a map from a field, $$\widetilde{\phi}$$ is automatically injective.

Finally, to check that $$\widetilde{\phi}$$ is surjective, we just need a characterization of an arbitrary element of $$K(a)$$. We'll use that $$K(a) = K[a]$$, which is proved in many places (including I'm sure on this site). Given this there is essentially nothing to do: any element $$v$$ of $$K[a]$$ is just a polynomial in $$a$$ with cofficients in $$K$$, say $$v = 3 a^2 + 42 a^3 + 6$$, and replacing each occurrence of $$a$$ with $$x$$ gives a polynomial (e.g. $$3 x^2 + 42 x^3 + 6$$) which is mapped to $$v$$ by $$\phi$$. This completes the proof.

Remark. You might ask: "Hang on, this seems to make sense but how did we use anywhere that $$f$$ was a minimal polynomial?". Indeed, so long as $$f$$ was any polynomial over $$K$$ with the root $$a$$ we could still construct the map $$\widetilde{\phi} : K[x]/(f) \to K(a)$$ and see that this map is surjective. The problem is that we lose injectivity: $$f$$ will in general of course not be irreducible (equivalently, up to a scalar the minimal polynomial of $$a$$), so $$K[x]/(f)$$ won't be a field.

To expand on lhf's comments: the polynomial ring $$K[X]$$ has the property that for any commutative ring extension of the form $$K[\alpha]$$, there is a unique homomorphism $$K[X]\to K[\alpha]$$ fixing $$K$$ and sending $$X\mapsto\alpha$$. This is just a general property of polynomial rings. Now, if $$\alpha$$ is algebraic over $$K$$, then the kernel of this homomorphism is the ideal $$(f_\alpha)$$ generated by its minimal polynomial, and by the fundamental theorem on homomorphisms we then have $$K[\alpha]\cong K[X]/(f_\alpha)$$, and the isomorphism fixes $$K$$ and sends $$\alpha\mapsto \overline X$$.

Now if $$a,b$$ are algebraic with the same minimal polynomial $$f$$, the corresponding ring extensions $$K[a]$$ and $$K[b]$$ are isomorphic to $$K[X]/(f)$$, both via an isomorphism sending $$a\mapsto\overline X$$ or $$b\mapsto\overline X$$ and fixing $$K$$. Now just compose one of the isomorphisms with the inverse of the other one to get an isomorphism between $$K[a]$$ and $$K[b]$$ which fixes $$K$$ and maps $$a\mapsto b$$.

Now finally note that $$K[\alpha]=K(\alpha)$$ for algebraic $$\alpha$$, so the isomorphism also applies to the field extensions in question.

Consider the homomorphism \begin{align} K[X]&\longrightarrow L,\\ P(X)&\longmapsto P(a). \end{align} Its image is $$K(a)$$ and its kernel is the ideal $$(f)$$, whence, by the First Isomorphism theorem, $$K[X]/(f)\simeq K(a),$$ and similarly for $$K(b)$$.