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I am trying to understand this problem. Part of it says

Let $K/F$ be any finite extension and let $\alpha \in K$. Let $L$ be a Galois extension of $F$ containing $K$ and let $H \leq Gal(L/F)$ be the subgroup corresponding to $K$. Define the norm of $\alpha$ from $K$ to $F$ to be $N_{K/F}(\alpha) = \prod_\sigma \sigma(\alpha)$, where the product is taken over all the embeddings of $K$ into an algebraic closure of $F$ (so over a set of coset representatives for $H$ in $Gal(L/F)$ by the Fundamental Theorem of Galois Theory). This is a product of Galois conjugates of $\alpha$. In particular, if $K/F$ is Galois this is $\prod_{\sigma \in Gal(K/F)} \sigma(\alpha)$.

(1) In the Fundamental Theorem of Galois Theory they say "the isomorphisms of $K$ (where $F \leq K \leq L$ and $L$ is Galois over $F$) into a fixed algebraic closure of $F$ containing $L$) which fix $F$ are in one-to-one correspondence with the cosets $\{\sigma H\}$ of $H$ in $G$." Are these isomorphisms part of the automorphism group of $L$?

(2) I am confused about the last sentence. They say you are taking the product over $\sigma \in Gal(K/F)$. However, aren't those all part of the same coset, namely $H$?


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up vote 2 down vote accepted

Let me address (1):

No - the automorphism group of $K$ consists of the field isomorphisms from $K$ to itself (hence the "auto" prefix). The fact that automorphisms are maps that both begin and end at $K$ is what allow them to be composed, which is the operation of the automorphism group - if I have two automorphisms of $K$, $\sigma:K\rightarrow K$ and $\tau:K\rightarrow K$, then $(\sigma\tau):K\rightarrow K$ is given by $\sigma\circ\tau$, i.e. the composition $K\xrightarrow{\tau}K\xrightarrow{\sigma}K$.

However, when $K/F$ is Galois, then every field embedding $f:K\rightarrow \bar{F}$, where $\bar{F}$ is an algebraic closure of $F$, actually has image exactly equal to $K$. Thus, a field embedding $f:K\rightarrow\bar{F}$ can actually be turned into a field isomorphism from $K$ to itself by forgetting about the rest of $\bar{F}$, i.e. restricting the codomain of $f$ to be its image. Thus, when (and only when) $K/F$ is Galois, the set of field embeddings $\{f:K\rightarrow\bar{F}\}$ can be identified with the group of field automorphisms $\{\sigma:K\rightarrow K\}$.

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Thank you! That helps. – badatmath May 12 '11 at 17:12
  1. Because $L$ is algebraic, any embedding of $K$ into an algebraic closure of $F$ that contains $L$ will extend to an embedding of $L$; and because $L$ is normal, this embedding will be an automorphism of $L$. So the isomorphisms of $K$ into the algebraic closure extend to automorphisms of $L$.

  2. If $G$ is Galois over $F$, then $\mathrm{Gal}(K/F)$ is isomorphic to $\mathrm{Gal}(L/F)/\mathrm{Gal}(L/K)$. The elements of $\mathrm{Gal}(K/F)$ are in one-to-one correspondence with the cosets of $\mathrm{Gal}(L/K)$, as per the usual isomorphism theorem. Note that $\mathrm{Gal}(L/K)$ is precisely the subgroup $H$ in the problem.

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Fair enough, thanks! – badatmath May 12 '11 at 17:13

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