Question is to prove that every Galois group $\text{Gal}(F)$ is a profinite group.

Galois group of a field $F$ is Galois group of extension $F^s$ (maximal seperable extension) over $F$.

Profinite group is a closed subgroup of product of finite groups (discrete topology) with product topology.

We want to see $\text{Gal}(F)$ as subgroup of product of finite groups.

Consider finite galois extensions $K/F$, we have $F\subseteq K\subseteq F^s$. As $K/F$ is finite, Galois group is finite. Consider product of (topological) groups $P_F=\prod_{F\subset K} \rm{Gal}(K/F)$ and the map $$\rm{Gal}(F)\rightarrow\prod_{F\subseteq K} \rm{Gal}(K/F)$$ with $\sigma\mapsto (\sigma|_K)$. Idea is to prove that $\text{Gal}(F)$ is intersection of closed subsets and those closed subsets comes from inverse image of graph of some continuous functions.

Fix a galois extension $F\subseteq K\subseteq L$ consider $\pi_{L,K}:P_F\rightarrow \text{Gal}(L/F)\times \text{Gal}(K/F)$ with projection on first coordinate and restriction on another coordinate i.e., $(\sigma)\mapsto (\sigma,\sigma|_K)$ where $(\sigma)$ is tuple in $P_F$ and $\sigma$ is the component corresponding to extension $L$ and $\sigma|_K$ is the restriction of this map to $K$.

This map is continuous and let $\Gamma_{L/K}$ be the graph of this map. Then the instructor writes

$$\bigcap_{F\subseteq K\subseteq L}\pi_{L,K}^{-1}(\Gamma_{L/K})=\text{Gal}(F)$$

I am not able to fill this gap. Is it atleast true that what i understood is correct? Intersection is over all extensions where $K,L$ both varies?

  • $\begingroup$ I thought $Gal(F)$ could be larger than $P_F$. How do you know it is $\subset$ ? $\endgroup$
    – reuns
    Dec 13, 2016 at 19:46
  • $\begingroup$ Given an element $\sigma\in \text{Gal}(F)$ i.e., $\sigma: F^s\rightarrow F$ consider $\sigma|_K: K\rightarrow K $. In this way we have map $\sigma\mapsto (\sigma|_K)$. Now, suppose $\sigma|_K$ is identity for all $K$ then we need to have $\sigma=Id$. So, we have an inclusion from absolute galois group to $P_F$ and so $P_F$ is bigger.. $\endgroup$
    – user87543
    Dec 13, 2016 at 19:55
  • 1
    $\begingroup$ I'm a bit confused by what $\Gamma_{L/K}$ is supposed to be. The graph of $\pi_{L/K} : P_F\to \operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)$ is a subset of $P_F\times\operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)$ by definition, so it doesn't make sense to take the preimage of an element of this under $\pi_{L/K}$. If you mean $\Gamma_{L/F}$ to be the image of $P_F$ inside of $\operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)$, then the intersection in question is all of $P_F$. Either $\Gamma_{L/F}$ is supposed to be something else or there's something missing. $\endgroup$
    – Stahl
    Dec 13, 2016 at 20:04
  • $\begingroup$ @Stahl yes.. you are absolutely correct.. graph has to be what you have said and not what I have written... in any case, can you help me from that part where I misunderstood.. $\endgroup$
    – user87543
    Dec 13, 2016 at 20:10

1 Answer 1


Let me reformulate what's going on, since as stated, I believe there's a mistake. Let $\Gamma_{L/K} := \{(\rho,\rho')\in\operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)\mid \rho' = \left.\rho\right|_K\}$ for $K\subseteq L$ Galois extensions of $F$ contained in $F^s$, (so $\Gamma_{L/K}$ is the same set as it is intended in the statement (I think?)), but let $\pi_{L/K} : P_F\to\operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)$ be the map $(\sigma_E)_{F\subseteq E\subseteq F^s}\mapsto(\sigma_L,\sigma_K)$. Now, I believe the correct statement is that $$ \operatorname{Gal}(F^s/F) = \bigcap_{F^s/L/K/F}\pi_{L/K}^{-1}(\Gamma_{L/K}) $$ (with the maps and sets as I just defined them). Note that again $\pi_{L/K}$ is continuous (it is projection onto two coordinates), and that $\Gamma_{L/K}$ is closed for all suitable $L/K$ (because we give $\operatorname{Gal}(L/F)\times\operatorname{Gal}(K/F)$ the discrete topology).

Let $\sigma\in G_F := \operatorname{Gal}(F^s/F)$. We want to show that $\sigma\in I := \bigcap_{F\subseteq K\subseteq L}\pi^{-1}_{L/K}(\Gamma_{L/K})$. That is, we need to show that $\sigma\in \pi^{-1}_{L/K}(\Gamma_{L/K})$ for all $K\subseteq L$ Galois extensions of $F$. This means that $\pi_{L/K}(\sigma)\in\Gamma_{L/K}$ (again for all Galois extensions $K\subseteq L$ of $F$). To that end, fix $F\subseteq K\subseteq L\subseteq F^s$ Galois extensions. $\sigma$ embeds into $P_F$ as $(\left.\sigma\right|_E)_{F\subseteq E\subseteq F^s}$. Then $\pi_{L/K}(\sigma) = (\left.\sigma\right|_L,\left.\sigma\right|_K)$, but $\left.\sigma\right|_K = \left.\left(\left.\sigma\right|_L\right)\right|_K$ (restricting multiple times is the same as restricting to the smallest field in question). Thus, we have $\pi_{L/K}(\sigma) = (\left.\sigma\right|_L,\left.\left(\left.\sigma\right|_L\right)\right|_K)$, which is visibly in $\Gamma_{L/K}$. This shows that $G_F\subseteq I$.

Conversely, let $(\sigma_E)_{F\subseteq E\subseteq F^s}\in I$. Then we need to construct an automorphism $\sigma$ of $F^s$ that fixes $F$ such that $\left.\sigma\right|_E = \sigma_E$ for all finite Galois extensions $E/F$. To that end, let $\alpha\in F^s$. Then $\alpha$ is separable over $F$, and as such lives in some Galois extension $E_\alpha/F$. Define $\sigma(\alpha) := \sigma_{E_\alpha}(\alpha)$. The field which $\alpha$ lives in is not necessarily unique; we need to confirm that $\sigma $ is independent of choice of field containing $\alpha $. Say $K $ and $K'$ are any finite Galois extensions of $F $ containing $\alpha $. Then as $(\sigma_E)\in I$, we have $$ \sigma_K (\alpha) = \sigma_{K \cap K'}(\alpha) =\sigma_{K'} (\alpha), $$ so $\sigma$ is well-defined (it did not depend on the choice of finite Galois extension containing $\alpha$). Thus, $(\sigma_E)\subseteq P_F$ is the image of $\sigma\in G_F$, so that $I\subseteq G_F$, and we are finished (I leave it to you to check any lingering details about $\sigma$ actually being an element of $G_F$.)

  • $\begingroup$ I did not said much because I have nothing to say.. this is what I was looking for.. thanks $\endgroup$
    – user87543
    Dec 14, 2016 at 12:53

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