Confusion regarding proof of a proposition in Field Theory (Dumb Question)

Question is to prove that :

For finite extensions $E/k$ and $F/k$ Prove that $[EF:k]\leq [E:k][F:k]$

where $EF$ is the smallest field extension which contains both $E$ and $F$

Supposing $E=k(x_1,x_2,\cdots,x_n)$ and $F=k(y_1,y_2,\cdots,y_m)$ we can see that $EF=k(x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_m)$

So, $[EF:k]\leq m+n$ which is definitely less than $mn=[E:k][F:k]$

So, $[EF:k]\leq m+n\leq mn=[E:k][F:k]$

When we know a strong result that $[EF:k]\leq m+n$ why are we looking for a less strong result that $[EF:k]\leq m+n\leq mn=[E:k][F:k]$

This Question was asked by my friend and i was unable to convince him that this does not make sense....

To be frank, at this instant I do not see anything wrong here...

• The dimension of a field extension is generally not the number of elements which you've adjoined. For instance if you have $\mathbb{Q}(\sqrt{2})$, you've only adjoined one element, but $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, not $1$. – Ben West Aug 4 '14 at 6:33
In general the number of generators of a finite extension has nothing to do with the degree. Of course in your argument you could chose the $x_i$ and $y_i$ to be elements of a basis of the respective extension. But then the union of the two sets generates $EF$ as a field (or ring) over $k$, but not as a vector space. To get a set of generators as a vector space you must consider the products $x_iy_j$ too. Which leads to the bound $\leq mn$.
• I guess i am following you but i am not (I do not know)... I understand first two lines.. I would take same $x_i,y_j$ as in generators of $E$ and $F$.. Then i do not get your point.... you are saying $EF=k(x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_m)$ is not a vector space over $k$... ?? – user87543 Aug 4 '14 at 7:10
• @PraphullaKoushik: No, he is saying that $\{x_1, x_2, \dots, x_n, y_1, y_2, \dots, y_m\}$ is not a generating set of this vector space. – Dune Aug 4 '14 at 7:47