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Assume $k$ is algebraically closed, consider $V(f_{1},f_{2})\subset \mathbb{A}^{2}$ where $f_{1}$ is irreducible, and does not divide $f_{2}$. Is there a quick way to show $$V(f_{1},f_{2})$$ must consist of finitely many points?

One way of attaching this is to consider $f_{1},f_{2}$ as elements of $k[x][y]$. Then since $k[x][y]$ is a UFD because $k[x]$ is a PID, $f_{1}$ can be decomposed into $$\prod^{n}_{i=1} a[y-g^{1}_{i}[x]]$$ where $n$ is the degree of $f_{1}$ as a polynomial of $y$. So we can write $V(f_{1})=\{y:y=g^{1}_{i}[x]\}$. Then the two decompositions for $f_{1},f_{2}$ must be different. And the set $V_{ij}:g^{1}_{i}[x]-g^{2}_{j}[x]=0$ must be a finite set as $g^{1}_{i},g^{2}_{j}$ has finite degree. So we conclude their intersection must be a finite set as well.

Is the above proof correct? I am not that confident so I venture to ask. It is not intended for a test or homework. Zhen Lin pointed out my 'proof' did not use the fact $f_{1}$ is irreducible at all. So I must be wrong somewhere.

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The claim as stated is false: let $f_1 = f^2$ and $f_2 = f^3$; then $V(f_1, f_2) = V(f)$. You should probably assume $f_1$ and $f_2$ are radical. –  Zhen Lin Nov 17 '12 at 19:09
    
Oh, sorry. The lecture notes did assume $f_{1}$ is irreducible. –  Bombyx mori Nov 17 '12 at 19:11
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Irreducibility of $f_1$ is neither necessary nor sufficient (take $f_1=f_2$ !): you must assume that $f_1$ and $f_2$ have no common irreducible factor. –  Georges Elencwajg Nov 17 '12 at 19:12
    
@GeorgesElencwajg: Is $f_{1}$ irreducible and $f_{2}$ is not divisible by $f_{1}$ enough? Your condition is strong enough for my proof to go through (now I see where the problem is). –  Bombyx mori Nov 17 '12 at 19:14
    
Yes, it is enough . –  Georges Elencwajg Nov 17 '12 at 19:16
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