Why would anyone think that $[E_1 E_2 : E_2]=[E_1:F]$?

Question

I'm reading Lang's Undergraduate Algebra, here:

I'm a bit confused: Why would anyone think that $[E_1 E_2 : E_2]=[E_1:F]$?

Answer 1

The natural way to write $E_1E_2$ is by setting each element equal to a sum of quantities in the form $ab$ with $a\in E_1$ and $b\in E_2$, and then do the algebra with all the relations you can get. In fact, this is what typically children do at school. From a formal standpoint, we are saying that $E_1E_2$, as a $F$-algebra, is a quotient of the tensor product $E_1\otimes_F E_2$. It is clear that the easiest case one can face is when no quotient occurs, i.e. when $E_1E_2\cong E_1\otimes_F E_2$. Now, the following are always true:

• $\dim_F(E_1\otimes_FE_2)=\dim_FE_1\dim_FE_2$;
• $\dim_F(E_1E_2)=[E_1E_2:F]=[E_2E_1:E_1][E_1:F]=\dim_{E_1}(E_2E_1)\dim_F(E_1)$

And it is now clear that $E_1E_2\cong E_1\otimes_F E_2$ implies $$[E_2E_1:E_1]=\frac{\dim_F(E_1E_2)}{[E_1:F]}=\frac{\dim_F(E_1\otimes_FE_2)}{\dim_F(E_1)}=\dim_F(E_2)=[E_2:F]$$