Htykuut

Given a field-extension $Ksubset M$. Let $x,y in $ M be algebraic,
$ fin K[X,Y]$ and $z=f(x,y) in M$. Show that $z$ is algebraic.

So far I got this obvious step:

Since $x$ and $y$ are algebraic so we can find a $gin K[X,Y]$ with $g(x,y)=0$

I can't see a connection, can someone give me a hint?
I tried seeing it with a example, but i can't generalize it.

Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).

Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.

Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.

Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.