# Minimal polynomials and Galois extensions

I'm back again with a question, but this time I am only curious about one thing. Here's how it goes:

Let $$K$$ be a Galois extension of a field $$F$$. By the theorem of the primitive element, we know that $$K = F(\alpha_1)$$ for some $$\alpha_1 \in K.$$ Suppose that $$f(X)$$ is the minimal polynomial of $$\alpha_1$$ over $$F$$. Now $$K$$ is the splitting field for $$f(X)$$ as $$K$$ is separable and normal. We also know that $$F(\alpha_1, \alpha_2, \ldots, \alpha_n)$$ for the distinct roots $$\alpha_i$$ of $$f(X)$$ is a splitting field for $$f(X)$$. This means that $$F(\alpha_1, \alpha_2, \ldots, \alpha_n) = F(\alpha_1)$$. Since $$f(X)$$ has $$n = deg f(X)$$ distinct roots it must be the case that all roots of $$f(X)$$ are linear combinations of $$\alpha_1$$. This also means that $$F(\alpha_k) = F(\alpha_d)$$ for some $$k, d \leq n$$.

However, I am not sure if my arguments are correct. I am also quite afraid of drawing my own conclusions as I am currently self-studying Galois theory with basically no prior knowledge of algebra. I hope that someone can correct me if I am wrong!

• All roots are polynomials in $\alpha_1$ and $\alpha_1$ is a polynomial in every other root. The interesting thing is that there is a matrix giving $\alpha_j^0,\ldots,\alpha_j^{n-1}$ in term of $\alpha_1^0,\ldots,\alpha_1^{n-1}$, this is the canonical representation of the Galois group, and the normal basis theorem is that it is also the regular representation. Dec 30, 2020 at 1:38
• You should specify your extension is finite; there are infinite Galois extensions, and of course those cannot be generated by a single element. Dec 30, 2020 at 2:46

This is all good, except that $$F(\alpha_1)$$ will be the set of all linear combinations of powers $$\{\alpha_1^j\}_{j=0}^{n-1}$$ where $$n = \deg(f)$$, not just linear combinations of $$\alpha_1$$ itself.

You can see that this set of powers is linearly independent over $$F$$ because if there did exist a nontrivial linear combination of these equaling zero$$^\dagger$$—i.e. if we had $$\displaystyle \sum_{k=0}^{n-1} c_k\alpha_1^k = 0$$ for some collection of not-all-zero coefficients $$c_k$$ from $$F$$—then the polynomial $$g(x) = \displaystyle \sum_{k=0}^{n-1} c_kx^k$$ would be of smaller degree than $$f$$ despite having $$\alpha_1$$ as a root. But this can't be since $$f$$ was the minimal polynomial of $$\alpha_1$$.

Moreover, this set of powers must be an exhaustive list of the generating elements for $$F(\alpha_1)$$ because the degree of the extension $$[F(\alpha_1):F]$$ is equal to the degree of the minimal polynomial of $$\alpha_1$$, and the degree of the extension is literally how many "basis vectors" $$F(\alpha_1)$$ has when viewed as a vector space over $$F$$.

Finally, we'd actually have $$F(\alpha_1) \cong F(\alpha_j)$$ for any $$1 \leq j \leq n$$. To see this, first recall that $$F(\alpha_1) \cong F[\alpha_1]$$ because $$\alpha_1$$ is algebraic over $$F$$ (see my post here). Thinking of $$F(\alpha_1)$$ as $$F[\alpha_1]$$, we can show that $$F[\alpha_1] \cong F[\alpha_j]$$ via different homomorphisms out of $$F[x]$$:

For each $$j$$, define a homomorphism $$\phi_j:F[x] \to F[\alpha_j]$$ to be the map $$g(x) \mapsto g(\alpha_j)$$; that is, we evaluate each polynomial at $$\alpha_j$$. You can prove for yourself (for instance, using the fact that univariate polynomial rings over fields are principal ideal domains) that the kernel of each $$\phi_j$$ must be the principal ideal generated by the minimal polynomial of $$\alpha_j$$, namely $$f$$. Applying the first isomorphism theorem for rings to each homomorphism, we see that $$F[\alpha_j] \cong F[x] / \langle f \rangle$$ for every $$j$$ between $$1$$ and $$n$$. By transitivity of congruence, then, we must have $$F[\alpha_1] \cong F[\alpha_j]$$ for every $$j$$.

$$^\dagger$$ Just applying the fact that a set of vectors $$\{ \mathbf{v}_k \}_{k=1}^n$$ in a vector space $$V$$ over a field $$F$$ is linearly dependent $$\iff$$ there exists elements $$c_k \in F$$, not all zero, such that $$\displaystyle \sum_{k=1}^n c_k \mathbf{v}_k = \mathbf{0}$$.