# Proof of Order of Galois Group equals Degree of Extension

I am struggling to understand this proof that for a Galois extension, the order of the Galois group equals the degree of extension. Here's a summary:

1. Let $f(x)$ be a polynomial in $F[x]$ with no repeated roots and suppose that $E$ is the splitting field for $f(x)$ over $F$.
2. This result is trivial for $n=1$, so do induction on $n$. Assume the result holds true for all $f(x)$ such that $0 \leq k < n$.
3. Let $p(x)$ be an irreducible factor of $f(x)$ with $r$ and one of the roots being $\alpha$ so that $F \subset F(\alpha) \subset E$. Thus, $[E : F(\alpha)]=n/r$ and $[F(\alpha) : F]=r$.
4. From the induction hypothesis, $[E : F(\alpha)]=\lvert G(E/F(\alpha)) \rvert$.
5. There are exactly $r$ isomorphisms from that maps $F(\alpha)$ to a $F(\beta)$ for some $\beta$ is some root of $p(x)$.
6. Consequently, there are $[E : F]=[E : F(\alpha)][F(\alpha) : F]=n$ possible automorphisms of $E$ that fix $F$, or $\lvert G(E/F) \rvert=[E : F]$.

I'm having trouble understanding how 4 and 5 come together to imply 6. How are we supposed to combine an isomorphism from $F(\alpha)$ to $F(\beta)$ with an automorphism of $E$ that fixes $F(\alpha)$? If someone could please explain to me how these two automorphisms can be combined to get a new automorphism of $E$ that fixes $F$, that would be great. Thanks!

• The main step is considering a tower $E/F(\alpha)/F$ and saying that $\sigma \in Gal(E/F) \implies \sigma = \rho \circ \phi$ where $\phi$ is an isomorphism $F(\alpha) \to \sigma(F(\alpha)) = F(\sigma(\alpha))$ fixing $F$ and $\rho \in Gal(E/F(\sigma(\alpha)))$ Commented Oct 22, 2016 at 23:36

First just to point out that this is when $E$ is a finite separable normal extension of $F$. At the steps you ask about $\alpha$ and $\beta$ have the same min poly over $F$ and so an earlier result shows that $F(\alpha)$ is isomorphic to $F(\beta)$ and there is an isomorphism $\psi_{j}$ say that acts as the identity on $F$ and maps $\alpha$ to $\beta$. But now the splitting field $E$ sits on top of both these fields and (essentially) using the result that proves splitting fields are isomorphic you can construct an isomorphism $\phi_{j}$ say from $E$ to $E$ such that its restriction to $F(\alpha )$ is $\psi_{j}$. Now there is a $\phi_{j}$ for each root and there are $r$ of these and $\phi_{j}$ is in Gal($E$/$F$). Let $\theta_{s}$ be the elements of Gal($E$/$F(\alpha))$ and so there are $n/r$ such $\theta_{s}$. Now look at the maps $\phi_{j}\theta_{s}$. They are distinct, there are $n$ of them and you show they exhaust Gal($E$/$F$).

Hope this helps

• Thanks! In order to understand the part where you said "essentially," I had to look at a different theorem, but after re-reading that proof and your answer, this all makes much more sense. Commented Oct 23, 2016 at 1:09
• Glad you have understood. Just to let you know that Galois theory is a great bit of maths but does contain some complex results that most people take a bit of time to get on top of. There are two major results to get over before you can do "the fundamental theorem of Galois theory". One is the result you were asking about and the other is " If $G$ is a finite subgroup of Aut $F$ then $\mid G \mid = [ F: \mbox{ Fix}G]$," sometimes called "Artin's Theorem". So you are almost there. Commented Oct 23, 2016 at 9:37
• Thanks for the tip! I will keep that in mind when I read the next section. Commented Oct 23, 2016 at 13:35

This is a rather old post now, but I think it's worth it for me to fill in the gaps in this book's proof that the order of Galois group equals the degree of extension now that I have a better understanding of Galois theory.

Let $$f(x)$$ be a polynomial in $$F[x]$$ with no repeated roots and suppose that $$E$$ is the splitting field for $$f(x)$$ over $$F$$. Let $$n=[E : F]$$.

Clearly, if $$[E : F]=1$$, then $$E=F$$, so $$G(E/F)=\{id\}$$. Thus, $$|G(E/F)|=[E : F]=1$$.

Now, do induction on $$n$$: Assume that for all fields $$F$$, for any splitting field $$E$$ of some polynomial $$f(x)\in F[x]$$ such that $$[E : F] < n$$, $$|G(E/F)|=[E : F]$$.

At this point, I'm going to restate two theorems from the book below:

Theorem 21.32: Let $$\phi: E \rightarrow F$$ be an isomorphism of fields. Let $$K$$ be an extension field of $$E$$ and $$\alpha \in K$$ be algebraic over $$E$$ with minimal polynomial $$p(x)$$. Suppose that $$L$$ is an extension field of $$F$$ such that $$\beta \in L$$ is the root of the polynomial in $$F[x]$$ obtained from $$p(x)$$ under the image of $$\phi$$. Then, $$\phi$$ extends to a unique isomorphism $$\bar \phi : E(\alpha) \rightarrow F(\beta)$$ such that $$\bar \phi(\alpha)=\beta$$ and $$\bar \phi$$ agrees with $$\phi$$ on $$E$$.

Theorem 21.33: Let $$\phi: E \rightarrow F$$ be an isomorphism of fields and let $$p(x)$$ be a nonconstant polynomial in $$E[x]$$ and $$q(x)$$ be the corresponding polynomial under the isomorphism $$\phi$$. If $$K$$ is a splitting field of $$p(x)$$ and $$L$$ is a splitting field of $$q(x)$$, then $$\phi$$ extends to an isomorphism $$\psi: K \rightarrow L$$.

Let $$p(x)$$ be an irreducible factor of $$f(x)$$ and let $$r$$ be the degree of $$p(x)$$. Since $$p(x)$$ splits in $$E$$, there are $$r$$ roots of $$p(x)$$ in $$E$$. Label these roots $$\alpha_1, \alpha_2, ..., \alpha_r$$. By Theorem 21.32, for any $$1 \leq i \leq r$$, there is a unique isomorphism $$\phi_i : F(\alpha_1) \rightarrow F(\alpha_i)$$ such that $$\phi_i$$ fixes $$F$$ and $$\phi_i(\alpha_1)=\alpha_i$$. Moreover, by Theorem 21.33, there exists an isomorphism $$\psi_i : E \rightarrow E$$ such that $$\psi_i$$ agrees with $$\phi_i$$ on $$F(\alpha_1)$$.

Now, take some arbitrary automorphism $$\sigma \in Gal(E/F)$$. I will now restate another theorem from the book:

Proposition 23.5: Let $$E$$ be a field extension of $$F$$ and $$f(x)$$ be a polynomial in $$F[x]$$. Then, any automorphism in $$G(E/F)$$ defines a permutation of the roots of $$f(x)$$ which lie in $$E$$.

By Proposition 23.5, $$\sigma(\alpha_1)=\alpha_i$$ for some $$1\leq i \leq r$$. Moreover, $$\sigma(F)=F$$ since $$\sigma$$ must fix $$F$$ by the definition of $$Gal(E/F)$$. Therefore, $$\sigma(F(\alpha_1))$$ contains both $$F$$ and $$\alpha_i$$, so $$\sigma(F(\alpha_1)) \subset F(\alpha_i)$$.

Now, consider $$\sigma^{-1}(F(\alpha_i))$$. Clearly, $$\sigma^{-1}(F)=F$$, since $$\sigma$$ fixes $$F$$. Also, because $$\sigma(\alpha_1)=\alpha_i$$, $$\sigma^{-1}(\alpha_i)=\alpha_1$$. Thus, $$\sigma^{-1}(F(\alpha_i))$$ contains both $$F$$ and $$\alpha_1$$, so $$\sigma^{-1}(F(\alpha_i)) \subset F(\alpha_1)$$. In other words, $$F(\alpha_i)\subset \sigma(F(\alpha_1))$$. This suffices to show that $$\sigma(F(\alpha_1))=F(\alpha_i)$$.

Thus, $$\sigma$$ restricted to $$F(\alpha_1)$$ is an isomorphism from $$F(\alpha_1)$$ to $$F(\alpha_i)$$ such that $$\sigma$$ fixes $$F$$ and $$\sigma(\alpha_1)=\alpha_i$$. However, by Theorem 21.32, there is only one isomorphism which satisfies these conditions, and $$\phi_i$$, as defined above, is also an isomorphism such that $$\phi_i$$ fixes $$F$$ and $$\phi_i(\alpha_1)=\alpha_i$$. This means that $$\sigma$$ must agree with $$\phi_i$$ on $$F(\alpha_1)$$.

Now, consider $$\psi_i^{-1}\sigma$$. As shown above, both $$\sigma$$ and $$\psi_i$$ agree with $$\phi_i$$ on $$F(\alpha_1)$$. Thus, for any $$x\in F(\alpha_1)$$:

$$\psi_i^{-1}\sigma(x)=\psi_i^{-1}(\phi_i(x))=\phi_i^{-1}(\phi_i(x))=x$$

Thus, $$\psi_i^{-1}\sigma$$ is an automorphism of $$E$$ which fixes $$F(\alpha_1)$$. In other words, $$\psi_i^{-1}\sigma$$ is a member of $$G(E/F(\alpha_1))$$.

We now restate another theorem:

Theorem 21.17: If $$E$$ is a finite extension of $$F$$ and $$K$$ is a finite extension of $$E$$, then $$K$$ is a finite extension of $$F$$ and $$[K : F]=[K : E][E : F]$$.

Thus:

$$[E : F]=[E : F(\alpha_1)][F(\alpha_1) : F]\implies [E : F(\alpha_1)]=\frac{[E : F]}{[F(\alpha_1) : F]}=\frac n r$$

Thus, $$[E : F(\alpha_1)] < n$$, so the induction hypothesis applies: $$|G(E/F(\alpha_1))|=[E : F(\alpha_1)]=n/r$$. We can thus label the elements of $$|G(E/F(\alpha_1))|$$ as $$\theta_1, \theta_2, ..., \theta_{n/r}$$.

As discussed earlier, $$\psi_i^{-1}\sigma$$ is a member of $$G(E/F(\alpha_1))$$. Thus, $$\psi_i^{-1}\sigma=\theta_j$$ for some $$1 \leq j \leq n/r$$. This means that, for any $$\sigma \in G(E/F)$$, $$\sigma=\psi_i\theta_j$$ for some $$1\leq i\leq r$$ and some $$1 \leq j \leq n/r$$. Since there are $$r$$ possibilities for $$\psi_i$$ and $$n/r$$ possibilities for $$\theta_j$$, there are $$r\cdot(n/r)=n$$ possibilities for $$\sigma$$. Therefore, $$|G(E/F)|=n=[E : F]$$. Q.E.D.

Now, here's a fun exercise which is relevant to the end of the proof:

Let $$1 \leq i,i_2 \leq r$$ and $$1 \leq j,j_2 \leq n/r$$. Prove that if $$\psi_i\theta_j=\psi_{i_2}\theta_{j_2}$$, then $$i=i_2$$ and $$j=j_2$$.

• In the proof there is stated that $\psi_i$ agrees with $\phi_i$ on $F(\alpha_1)$. However, in the proof you have used equality $\psi^{-1}_i(\phi_i(x))=\phi^{-1}_i(\phi_i(x))$. However, $\phi_i(x)$ is not a member of $F(\alpha_1)$, but $F(\alpha_i)$ instead. How does one show they agree on $F(\alpha_i)$ as well? Commented Jan 21, 2020 at 14:13
• @Machinato Hopefully, you understand why $\psi_i$ and $\phi_i$ agree on $F(\alpha_1)$. Then, by definition of $\phi_i$, we have $\phi_i(F(\alpha_1))=F(\alpha_i)$. Therefore, $\psi_i^{-1}$ and $\phi_i^{-1}$ agree on $F(\alpha_i)$, which is why I am able to say $\psi_i^{-1}(\phi_i(x))=\phi_i^{-1}(\phi_i(x))$. Commented Mar 24, 2020 at 19:55