Another Galois theory question: Is this proof correct?

Question

Theorem: If L is algebraic over M, and M is algebraic over K, then L is algebraic over K
Proof: Pick x in L. Then since L is algebraic over M, there exist m_0, m_1,...,m_n in M such that
m_n*x^n+m_(n-1)*x^(n-1)+
...+m_1*x+m_0=0.
Since this is a polynomial in K(m_0, m_1,...,m_n), immediately we have that x is algebraic over K(m_0, m_1,...,m_n). Thus
[K(m_0, m_1,...,m_n,x):K(m_0, m_1,...,m_n)] [K(m_0, m_1,...,m_n):K] Hence
[K(m_0, m_1,...,m_n,x):K]=
[K(m_0, m_1,...,m_n,x):K(m_0, m_1,...,m_n)]
*[K(m_0, m_1,...,m_n):K] So K(m_0, m_1,...,m_n,x) is a finite extension of K, hence K(m_0, m_1,...,m_n,x) is algebraic over K. This implies x is algebraic over K. Since x was an arbitrary element of L, it follows that L is algebraic over K.

Answer 1

This proof appears correct to me.

If this is your proof, I have one comment...in the part where you say

"...immediately we have that x is algebraic over K(m_0, m_1,...,m_n). Thus
[K(m_0, m_1,...,m_n,x):K(m_0, m_1,...,m_n)]
I would highlight that this is true because K(m_0, m_1, ..., m_n, x) : K(m_0, m_1, ..., m_n) is a *simple* algebraic extension. I had to re-read this part several times before I realized what was going on. (Algebraic extensions in general, of course, need not be finite.)

(Then again, it might just be that I'm being slow to follow things today; I just thought I'd offer the suggestion. Use your own judgment.)

Answer 2

Well my friend it looks correct to me... Bravo!