1
$\begingroup$

What is the dimension of the union of fields $L_1$ and $L_2$ over $L_1 \cap L_2$ ($L_1$ and $L_2$ have dimensions $n_1$ and $n_2$ over $L_1 \cap L_2$)as a vector space? I think it should be $n_1n_2$ but I am not able to prove it for some (probably dumb) reason.

By union of fields, I mean the smallest field containing the union lying in some algebraic closure of $L_1$ and $L_2$.

$\endgroup$
1
$\begingroup$

What you are calling the union of fields $L_1$ and $L_2$ is called the compositum of $L_1$ and $L_2$, or $L_1 L_2$.

Let $F:=L_1\cap L_2$, and suppose that $L_1$ has basis $\{x_1,\ldots,x_{n_1}\}$ over $F$ and that $L_2$ has basis $\{y_1,\ldots,y_{n_2}\}$ over $F$. Then $L_1 L_2$ is spanned over $F$ by the $x_i y_j$'s, so it can have dimension at most $n_1 n_2$ over $F$. However, the dimension can be smaller. For example, let $\omega$ be a primitive cube root of 1, let $L_1:=\mathbb{Q}(\sqrt[3]{2})$, and let $L_2:=\mathbb{Q}(\omega \sqrt[3]{2})$. Then $L_1\cap L_2=\mathbb{Q}$ so $n_1=n_2=3$, but $L_1 L_2=\mathbb{Q}(\omega, \sqrt[3]{2})$ so $[L_1 L_2:L_1\cap L_2]=6<9=n_1 n_2$.

$\endgroup$
  • $\begingroup$ Thanks for your answer, does my question hold true if $L_1$ and $L_2$ are Galois? $\endgroup$ – Asvin Apr 5 '15 at 6:23
  • $\begingroup$ Yes, if at least one of $L_1$ and $L_2$ is Galois over $F$, then $[L_1 L_2:F]=n_1 n_2$. $\endgroup$ – Polichinelle Apr 7 '15 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.