Decomposing an abelian extension into cyclic extensions

Question

I have a question regarding abelian and cyclic extensions on number fields.

Let $L/K$ be an extension of number fields which is abelian (i.e., its Galois group is commutative). By the fundamental theorem of abelian groups, we can decompose Gal$(L/K)$ into a product of cyclic groups, namely

$$\mbox{Gal}(L/K) \cong C_1 \times C_2 \times ... \times C_r,$$

for some $C_i$ cyclic. (Note that all of these groups are finite of course).

How can i decompose $L/K$ into a compositum of cyclic extensions over $K$?

My guess: We know that $L = L^{C_1}L^{C_2}...L^{C_r}$ is the compositum of all the fixed fields by these cyclic groups. However, the extension $L^{C_i}/K$ is not cyclic (actually $L/L^{C_i}$ is) and so this doesn't work. I am considering now the following decomposition:

$$L = L^{\{id\}\times C_2\times...\times C_r}L^{C_1\times \{id\}\times...\times C_r}\cdots L^{C_1\times C_2\times...\times C_{r-1}\times \{id\}}$$

It is clear then that each of these fields in the decomposition are cyclic extensions of $K$ with Galois group $C_i$. Is this decomposition true? Namely, does this last equality really hold? And if not, how can i construct a decomposition of $L$ into cyclic extensions of $K$?

Thanks in advance!

Answer 1

Let $G_i = C_1 \times \ldots \times C_{i-1} \times \{ id \} \times C_{i+1} \ldots \times C_n$.

Then the composition $L^{G_1} L^{G_2} \ldots L^{G_n}$ is the smallest subextension of $L$ containing $L^{G_1},L^{G_2},\ldots,L^{G_n}$.

By the fundamental theorem of Galois theory, it is the field corresponding to the largest subgroup of $Gal(L/K)$ contained in $G_1,G_2,\ldots,G_n$, which is $ G_1 \cap G_2 \cap \ldots \cap G_n$

But $G_1 \cap G_2 \cap \ldots \cap G_n = \{id_G\}$, so $L^{G_1} L^{G_2} \ldots L^{G_n}= L^{\{id_G\}} = L$

Answer 2

To be a bit more general, we have the following theorem:

Let $L/F$ be a finite Galois extension whose Galois group $G$ may be written as $G= G_1\times G_2\times\cdots \times G_n$, and let $L_i$ denote the fixed field of $G_1\times\cdots\times \{1\}\times\cdots \times G_n$, where the $\{1\}$ occurs in the i-th position of the direct product. Then each $L_i/F$ is Galois, $L_{i+1}\cap (L_1\cdots L_i)=F,$ and $L=L_1\cdots L_n$.

Mercio already gave a proof to the last assertion. Note that the original assumption that $C_i$ are cyclic wasn't used, so there are only notation changes, namely, $G_i$ becomes $C_i, L_i$ becomes $L^{G_i}$, and $F$ becomes $K$ in Mercio's answer.

For the first assertion, notice that $(G_1\times\cdots\times \{1\}\times\cdots \times G_n )\lhd (G_1\times\cdots\times G_n)$ regardless of the position of $\{1\}$ and therefore by the fundamental theorem of Galois theory, $L_i/F$ is Galois for each $i$.

Similarly to Mercio's answer we have that $L_1\cdots L_i$ corresponds to the intersection $\{1\}\times\cdots\times\{1\}\times G_{i+1}\times\cdots\times G_n$ , and $L_{i+1}\cap (L_1\cdots L_i)$ corresponds to the smallest subgroup generated by $\{1\}\times\cdots\times\{1\}\times G_{i+1}\times\cdots\times G_n$ and $G_1\times G_2\times\cdots\times\{1\}\times G_{i+2}\cdots \times G_n$, which is obviously $G_1\times G_2\times\cdots \times G_n=G$. Then since the fixed field of $G$ is $F$, we get $L_{i+1}\cap (L_1\cdots L_i)=F$, which proves our second assertion.