Let $F_1 \subseteq F_2$ be a field extension, with char($F_1$)$=0$, and let $f\in F_1[x]$. Let $K_i=$splitting field of $f$ over $F_i$.

We see that $K_1\subseteq K_2$ and that $K_1/F_1$ and $K_2/F_2$ are Galois extensions.

Question: What is the relationship between Gal($K_1/F_1$) and Gal$(K_2/F_2)$?

Are they isomorphic? Or is one a subgroup of the other? Or what?

  • $\begingroup$ If $K_1 \subseteq K_2$, and $f$ splits in $K_1$, then $K_1=K_2$, because splitting field is a smallest field in which $f$ splits. Are you sure it's what you are asking for? $\endgroup$ – mzg147 May 24 '18 at 17:06
  • $\begingroup$ How do you obtain $K_1=K_2$? I am a little skeptical. $\endgroup$ – Pascal's Wager May 24 '18 at 17:17
  • $\begingroup$ $K_1$ and $K_2$ are both splitting fields of the same polynomial. And splitting fields are unique. $\endgroup$ – mzg147 May 24 '18 at 17:19
  • $\begingroup$ But they are splitting fields of $f$ over different fields. I mean, $K_2$ must be an extension field of $F_2$, but it isn't clear to me that $K_1$ is also an extension field of $F_2$. $\endgroup$ – Pascal's Wager May 24 '18 at 17:20
  • $\begingroup$ Well, it's my first time seeing "splitting field over a field". It is always "splitting field of a polynomial", and that polynomial has *coefficients * in some field. So, the only field that really matters is $F_1$, because $f\in F_1[x]$. $\endgroup$ – mzg147 May 24 '18 at 17:27

First, $Gal(K_1/F_1)$ and $Gal(K_2/F_2)$ are, in general, not literally comparable as sets (or as subgroups) since $Gal(K_1/F_1)$ contains automorphisms $K_1 \to K_1$ and $Gal(K_2/F_2)$ contains automorphisms $K_2 \to K_2$.

Now, suppose that $r_1,\dots,r_n$ are the roots of $f$ in $K_2$. Then $K_2=F_2(r_1,\dots,r_n)$ and $K_1=F_1(r_1,\dots,r_n)$ ($K_1$ will be isomorphic to this, but we might as well work with this splitting field).

My answer is that there is an inclusion $Gal(K_2/F_2) \hookrightarrow Gal(K_1/F_1)$ (which sort of means that we can think of $Gal(K_2/F_2)$ as a subgroup of $Gal(K_1/F_1))$ but not necessarily a reverse inclusion.

Define $\phi:Gal(K_2/F_2) \to Gal(K_1/F_1)$ by $\phi(\alpha)=\alpha|_{K_1}$ for $\alpha \in \phi:Gal(K_2/F_2)$. Now, for $\alpha \in \phi:Gal(K_2/F_2)$, we know that $\alpha|_{K_1}:K_1 \to K_2$. But since for each $i$, $f(\alpha(r_i))=0$, $\alpha$ sends each $r_i$ to some $r_j$ and $\alpha|_{F_1}\equiv id|_{F_1}$, so $\alpha|_{F_1(r_1,\dots,r_n)}$ takes $F_1(r_1,\dots,r_n)$ into $F_1(r_1,\dots,r_n)$. Also, you can show that $\phi$ is injective by working with the $r_i$. Suppose that $\phi(\alpha_1)\equiv\phi(\alpha_2)$. Then for each $i$, $\alpha_1(r_i)=\alpha_2(r_i)$. Then $\alpha_1\equiv \alpha_2$. Thus $\phi:Gal(K_2/F_2) \hookrightarrow Gal(K_1/F_1)$.

As an example of when we might not have the reverse inclusion, consider $F_1=\Bbb{Q}$, $F_2 = \Bbb{Q}(\sqrt2)$ and $f = x^2 - 2$ which implies that $K_1=K_2=F_2$. Then $Gal(K_2/F_2)$ is trivial but $Gal(K_1/F_1)$ is not.

  • $\begingroup$ So $Gal(K_2/F_2)$ is isomorphic to $img( \phi)$, which is a subgroup of $Gal(K_1/F_1)$. By the way, thank you for emphasizing that $Gal(K_2/F_2)$ isn't really a subgroup of $Gal(K_2/F_2)$. +1 for clarity! $\endgroup$ – Pascal's Wager Jun 7 '18 at 4:43

This is the Theorem on Natural Irrationalities.

Let me change your notation slightly and call the small field $F$ and the big one $F'$. By adjoining all the roots $\{r_1,\cdots,r_n\}$ of a polynomial, you’re simply making the most general finite Galois extension of $F$ (since the characteristic is zero, the extension is automatically separable). So your two extensions are $K=F(\mathbf r)$, where I’m using $\mathbf r$ to stand for the $n$ roots of $f$, and $K'=F'(\mathbf r)$.

Changing the notation again slightly, I point out that $K'=F'K$, the compositum. So we have five fields, $F\subset K\cap F'\subset K\subset F'K$, fitting in with $F\subset K\cap F'\subset F'\subset F'K$. If you draw the standard Hasse diagram showing these relationships, you have a rhombus with a tail, kite-like. The Theorem says that $KF'$ is Galois over $F'$, and the associated Galois group is isomorphic to $\text{Gal}^K_{K\cap F'}$, via restriction to $K$. Therefore, to answer your question, the upper Galois group is a subgroup of the lower one.

I believe that you’ll find it easy enough to prove the Theorem for yourself, but nothing holds you back from looking a proof up. Notice that we make no hypotheses on the extension $F'\supset F$: it need not be finite or Galois or even algebraic. You can think of situations where $F'=\Bbb R$ and $KF'=\Bbb C$.

  • $\begingroup$ I think you are saying that $Gal(K_1/F_1)$ is isomorphic to some subgroup of $Gal(K_2/F_2).$ If so, what is the isomorphism? $\endgroup$ – Pascal's Wager May 31 '18 at 17:01
  • $\begingroup$ Not at all, @Pascal'sWager. The upper group is a subgroup of the lower group. That is Gal$(K_2/F_2)$ is a subgroup of Gal$(K_1/F_1)$, by restricting to the intersection $F_2\cap K_1$. For an example, take a cyclotomic polynomial for $f$, $F_1=\Bbb Q$, and $F_2=\Bbb R$. $\endgroup$ – Lubin Jun 1 '18 at 3:15

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