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I have a question about the notation in an assertion in Lang's Algebra, chapter 6 §1, corollary 1.16:

Let $K/k$ be finite Galois with group $G$, and assume that $G$ can be written as a direct product $$G=G_1 \times \cdots \times G_n.$$ Let $K_i$ be the fixed field of $G_1 \times \cdots \times \{1 \} \times \cdots \times G_n$ where the group with $1$ element occurs in the $i$th place. Then $K_i/k$ is Galois, and $K_{i+1} \cap (K_1 \cdots K_i) = k$...

Very well then; suppose $n=3$ in the above situation. Then $K_1/k$ is Galois and $K_2 \cap K_1 = k$. Also, $K_3/k$ is Galois and... Well, what would $K_{3+1}$ be? In the case $K_1$ I suppose is meant that $K_2 \cap K_1 K_3 = k$, but in the case $K_3$ I have no idea. What does he mean? Thank you.

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The index $i$ should be restricted to $1 \le i < n$ (so there is no $i=3$ in the case $n=3$).

In the case $n=3$, the two cases are $i=1$ and $i=2$. If $i=1$, the theorem is saying that $K_2 \cap K_1 = k$. If $i=2$, the theorem is saying that $K_3 \cap K_1 K_2 = k$.

The result is a consequence of the correspondence theorem between subgroups of the Galois group of $K/k$, and subfields of $K$ containing $k$. If subgroups $H, H'$ have fixed fields $K, K'$ then the intersection $H \cap H'$ has fixed field $KK'$ and the subgroup $\langle H, H' \rangle$ has fixed field $K \cap K'$.

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  • $\begingroup$ Thank you, I didn't notice your answer till now. $\endgroup$ – Erik Vesterlund Jun 18 '13 at 18:17

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