Let $K \subseteq L_1,L_2 \subseteq L$ be fields. I want to show the following: $$[L_1L_2:K] \leq [L_1:K][L_2:K]$$

My attempt: The definition of $$L_1L_2:=L_1(L_2):=\bigcap \{F:F\text{ is a subfield of }L\text{ and }L_1 \subseteq F\text{ and }L_2 \subseteq F\}$$

$$\begin{split} [L_1L_2:K]&=\dim_K L_1L_2\\ [L_1:K]&=\dim_K L_1\\ [L_2:K]&=\dim_K L_2\\ \end{split}$$

I know that the $K$-vector space $L_1L_2$ is a linear subspace of the $K$-vector space $L$. The same is true for the $K$-vector spaces $L_1,L_2$.

From linear algebra, I know the following theorem: $$\dim(U_1+U_2)=\dim U_1 +\dim U_2 - \dim (U_1 \cap U_2)$$

If I can show that $L_1L_2= U_1+U_2$, the result should follow. Now I have to show that $L_1L_2=U_1+U_2$

Rewriting the definition of $$L_1L_2:=L_1(L_2):=\bigcap \{F:F\text{ is subfield of }L\text{ and }L_1 \cup L_2 \subseteq F\}$$

This means that $L_1 \cup L_2 \in L_1L_2$. From linear algebra, I know that $\operatorname{span}(U_1 \cup U_2)=U_1+U_2$. Thus, $L_1L_2=U_1+U_2$ and thus $[L_1L_2:K] \leq [L_1:K][L_2:K]$ holds.

My Question: Is my answer correct?

  • 2
    $\begingroup$ Suppose $e_1, \ldots, e_n$ is a basis of $L_1$ and $f_1, \ldots, f_m$ is a basis of $L_2$. If you can show that the $e_i f_j$ generate $L_1 L_2$, then the result follows. $\endgroup$ Commented Mar 22, 2023 at 2:46
  • $\begingroup$ @CharlesHudgins I am a little bit lost right now, how could I show that? $\endgroup$
    – wanymose
    Commented Mar 22, 2023 at 3:12
  • $\begingroup$ @CharlesHudgins Since $E:=\{e_1,...,e_n\} \subset L_1$ and $F:=\{f_1,...,F_n\} \subset L_2$, $E \cup F \subset L_1L_2$ but how do I show that it generates $L_1L_2$. Could you give me a hint? $\endgroup$
    – wanymose
    Commented Mar 22, 2023 at 3:49
  • $\begingroup$ What does a generic element of $L_1 L_2$ look like? Expand it in terms of the bases for $L_1$ and $L_2$. Combine like terms. Each term should now look like $k e_i f_j$ for some $k \in K$. $\endgroup$ Commented Mar 22, 2023 at 3:50
  • $\begingroup$ In general, a good place to start to understand some construction is to try to explicitly write down a generic element. $\endgroup$ Commented Mar 22, 2023 at 3:51


You must log in to answer this question.