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Let $k$ be any field, and let $f,g\in k[x,y]$ be two irreducible polynomials such that $g$ is not divisible by $f$. Prove that $V(f,g)\subseteq A_k^2$ is finite.

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1 Answers1

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No need to consider $f,g$ irreducible polynomials, it's enough to assume $\gcd(f,g)=1$. Then $\gcd(f,g)=1$ in $K(X)[Y]$, and therefore there exist $u,v\in K(X)[Y]$ such that $1=uf+vg$. In order to clear the denominators we can find a polynomial $w\in K[X]$ such that $wu,wv\in K[X,Y]$. From $w=(wu)f+(wv)g$ it's obvious that $f(a,b)=g(a,b)=0$ implies $w(a)=0$ and since $w$ has only finitely many roots in $K$ we are done.

  • @user26857 Hi. I can see why the $x$-coordinate has a finite number of roots in $(f,g)$ of $\mathbb{A}^2$, but how are you able to conclude that the $y$-coordinate is also finite? – math123456789 Feb 15 '17 at 03:37
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    @jm.byrnes, you do the same thing with the variables interchanged, of course... – Mariano Suárez-Álvarez Jul 07 '17 at 22:35
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    How did you prove that "$\gcd(f,g)=1$. Then $\gcd(f,g)=1$ in $K(X)[Y]$"? – No One Jul 12 '17 at 19:58
  • Use Gauss' lemma (Theorem 2.1 from https://math.stanford.edu/~conrad/210APage/handouts/gausslemma.pdf). Basically, any positive-degree factor of f in K(X)[Y] leads to a positive-degree factor of f in K[X, Y]. – Fred Akalin Dec 27 '23 at 06:43