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Sorry if this question is too basic. It is from Fröhlich and Taylor's "Algebraic Number Theory".

Let $E/F$ be a finite Galois extension of fields, with $G=Gal(E/F)$, and let $K$ and $L$ be two subfields of $E$, containing $F$, such that $K/F$ and $L/F$ are both Galois. Let $M=Gal(E/K)$ and $N=Gal(E/L)$ be normal subgroups of $G$. Suppose ${\gamma_1,\ldots,\gamma_n}$ is a transversal for $MN$ in $G$, with $n=[G:MN]$. If $C$ is the compositum $KL$ in $E$, how can I show the map $$ k\otimes l\mapsto (k^{\gamma_1}l,\ldots,k^{\gamma_n}l) $$

induces an isomorphism between $K\otimes_F L$ and $\prod_{i=1}^n C$?

It is clear to me that this map is an $F$-algebra homomorphism, and that they both have the same dimension over $F$. Thus surjectivity, or injectivity, would be enough. I have not been able to figure out what the idempotents of $\prod_{i=1}^n C$ should look like in $K\otimes_F L$, so I have not been able to show surjectivity. Meanwhile, I think injectivity should be easier to show, because if we have $$ k_1\otimes l_1 + \cdots + k_m\otimes l_m\mapsto 0,$$ then we get a system of equations $$ k_1l_1 + \cdots + k_ml_m=0$$ $$ \cdots$$ $$ k_1^{\gamma_n}l_1 + \cdots k_m^{\gamma_n}l_m=0.$$ Summing up the columns, I get $$ \sum_{i=1}^m(\sum_{j=1}^n k_i^{\gamma_j})l_i=0 $$ and all this is happening in $L$. But I can't seem to finish this argument. Any help would be greatly appreciated.

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    To identify the idenpotents of $K\otimes_F L$, choose a primitive element for $L/F$ (which is separable, of course) and write $L=F[x]/(f)$ where $f$ is the minimal poly of that element. Then $K\otimes_F L = K[x]/(f)$, so you can see the idempotents correspond to the irreducible factors of $f$ in $K[x]$.2010-08-28

1 Answers 1

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To expand on Sam's awesome comment:

We can write $K$ as $F(\alpha)$, where the minimal polynomial of $\alpha$ is $f(x)$. Then we have a map from $L[x]$ to $K\otimes_F L$ given by $$ a_0+a_1x+\cdots+a_nx^n\mapsto 1\otimes a_0 + \alpha\otimes a_1+\cdots + \alpha^n\otimes a_n.$$ This induces an isomorphism between $L[x]/fL[x]$ and $K\otimes_F L$. Composing with the map I've given above, we then want to show the map $$ a(x)\mapsto (a(\alpha^{\gamma_1}),\ldots,a(\alpha^{\gamma_n}))$$ from $L[x]$ to $\prod_{i=1}^n C$, is surjective.

Set $$ f_i=\prod_{n\in N} (x-\alpha^{\gamma_in});$$ this is a polynomial in $L[x]$; now let $$ g_i=\prod_{j\neq i} f_j; $$ then $g_i$ maps to $(0,\ldots,g_i(\alpha^{\gamma_i}),0,\ldots)$. Since $g_i(\alpha^{\gamma_i})$ is a non-zero element of $K$, we can write $g_i(\alpha^{\gamma_i})^{-1}=h_i(\alpha^{\gamma_i})$, and so $g_ih_i$ maps to $(0,\ldots,1,\ldots)$, where the $1$ is in the $i$th place. Since $C=L(\alpha)$, this map is surjective.