# Dimension of union of fields over intersection of fields.

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 $L_2$.

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 $[L_1 L_2:L_1\cap L_2]=6<9=n_1 n_2$.
• Thanks for your answer, does my question hold true if $L_1$ and $L_2$ are Galois? – Asvin Apr 5 '15 at 6:23
• Yes, if at least one of $L_1$ and $L_2$ is Galois over $F$, then $[L_1 L_2:F]=n_1 n_2$. – Polichinelle Apr 7 '15 at 1:09