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If $f$ and $g$ are irreducible polynomials in $K[X,Y]$ that are not associates, show that the zero set $Z(f,g)$ is either empty or finite.

Here is what I been told to do: If $(f,g)≠K[X,Y]$, show $(f,g)$ contains a nonzero polynomial in $K[X]$ and similarly a nonzero polynomial in $K[Y]$.

and what i should do before that is let $R=K[X]$ and $F=K(X)$, and apply Gauss’s Lemma to show $f$ and $g$ are relatively prime in $F[Y]$.

Can any one help please, thanks a lot.

user138017
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2 Answers2

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The easiest way to see this might be using commutative algebra: One can assume that $K$ is algebraically closed. By assumptions there is no polynomial $h$ with $(h)$ prime and $(f,g) \subset (h)$. This shows that $K[X,Y]/(f,g)$ is zero-dimensional, because there is no prime of height 1 lying over $(f,g)$. Hence it is Artinian and has only finitely many maximal ideals.

But you are supposed to do it this way: $F[Y]$ is a principal ideal domain and by Gauss's Lemma $f,g$ are irreducible in this domain, hence we have $1=af+bg$ in $K(X)[Y]$ with $a,b \in K(X)$. Clearing the denominators of $a$ and $b$ will give us an element of $K[X]$ lying in $(f,g)$. Similarly we get an element of $K[Y]$ lying in $(f,g)$.

MooS
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  • thanks a lot for answering. can i ask how can I use these to conclude the zero set Z(f,g) is either empty or finite? – user138017 Feb 02 '15 at 09:12
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    This is now an immediate consequence of the fact that a polynomial in one variable does only admit finitely many zeros. – MooS Feb 02 '15 at 09:19
  • i see, by the Fundamental theorem of algebra. Thanks a lot. – user138017 Feb 02 '15 at 09:27
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    This is not what people usually call the Fundamental theorem of algebra. A polynomial in one variable of degree $n$ over a field has at most $n$ zeros. This does not require any assumptions on the field, it is just a consequence of the existence of division with remainder. – MooS Feb 02 '15 at 09:30
  • can i ask more in detail that how the denominators of a and b will give us an element of K[X] lying in (f,g)? and when you say denominator of a polynomial, are you meant the common denominator of all the coefficients of the polynomial? – user138017 Feb 02 '15 at 09:44
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    No, you should read more carefully. $a$ and $b$ are elements of $K(X)$, hence the quotient of two polynomials. With denominator I really mean the denominator of this fraction. – MooS Feb 02 '15 at 09:51
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The old style method is to take the resultant $R(x)$ of the two polynomials. Then we have $$R(x)=A(x,y)f(x,y)+B(x,y)g(x,y).$$

Since any common zeros will make the resultant zero there are a finite number of possible $x$ coordinates, it then follows that the number of points is finite.

user26857
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Rene Schipperus
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