# Why is the compositum of finite Galois extension a Galois extension?

Could anyone help me with that?

Let say that $K_1$ and $K_2$ are finite Galois extensions of $K$, I want to show that $K_1K_2$ is Galois over $K$.

Because the extensions are galois I can find separable polynomials $f_1,f_2$ in $K[X]$ such that $K_i$ is the splitting field of $f_i$.

The proof that I read affirms that $K_1K_2$ is the splitting field of $f_1f_2$ and concludes that $K_1K_2$ is Galois over $K$. But my problem is that $f_1f_2$ need not to be separable as $f_1$ and $f_2$ could share a root. But then this root would be in $K_1 \cap K_2$ which would allow us to cancel the extra term in $K_1 \cap K_2$ giving us only $K_1K_2/K_1 \cap K_2$ to be Galois...

What am I missing?

• Can't you assume that $f_1,f_2$ are irreducible over $K$? That would rule out common roots (other than the trivial $K_1=K_2$). – Greg Martin Oct 19 '16 at 22:16
• yes this is easier! – WrabbitW Oct 19 '16 at 22:52

## 1 Answer

You're right that this proof isn't quite correct if $f_1$ and $f_2$ have common roots. To fix it, you can let $f$ be the least common multiple of $f_1$ and $f_2$ in $K[X]$, and then show that $K_1K_2$ is the splitting field of $f$. Since $f_1$ and $f_2$ have no repeated roots, neither can $f$.

Alternatively, by the primitive element theorem, you can choose $f_1$ and $f_2$ to be irreducible (namely, the minimal polynomial of a primitive element). Then $f_1$ and $f_2$ have no common roots unless $K_1=K_2$, but in that case $K_1K_2=K_1$ is obviously Galois.