Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I tried to prove some property of fields, but I could not, and I hope someone can help me with that. I have a question about fields and roots.

If I have an arbitrary family (each of them is a set of roots $(x,y)$ of a polynomial in two variables over an algebraically closed field), is the intersection of all of them the set of roots of another polynomial?

The set $X$ is an algebraically closed set, the set of roots in this 2 variables polynomial is clearly of the form $(x,y)$.

Do I need to know some special property of algebra to prove this, or only the definition of algebraically closed set?

share|improve this question
    
Interesting problem! I suspect that the answer is easy for anyone who knows any classical algebraic geometry. If you replace the complex numbers by the reals, the answer is "not necessarily." In that case, I can produce a family of zero sets of polynomials whose intersection is not the zero set of any polynomial. –  André Nicolas Aug 5 '11 at 18:41
add comment

2 Answers 2

up vote 3 down vote accepted

I mentioned in an earlier, less relevant answer that the intersection $\{(0, 0)\}$ of the zero sets of $x$ and $y$ was not the zero set of a single polynomial. One way to see this is via the following lemma.

Let $k$ be an algebraically closed field, and let $f \in k[x, y]$ be a non-constant polynomial. Then the zero set of $f$ is infinite.

Here are some hints toward proving this. Since $f$ is not a constant, some variable — which might as well be $x$ — appears in it. Then we can write $$ f(x, y) = \sum_{i = 0}^n g_i(y)x^i $$ where each $g_i \in k[y]$, and $n > 0$ is such that $g_n$ is not the zero polynomial. The key is that $g_n$ has only finitely many zeros in $k$. Can you take it from there?

share|improve this answer
add comment

Your set X can be given as the roots of a finite set of polynomials, but possibly not a single one (for example if X is the empty set).

There's a bit of abstract algebra that deals specifically with this kind of question. The key point is that if $F$ is an algebraically closed field, then $F[x,y]$ is "Noetherian", which means that any ideal is finitely generated.

What does this have to do with your question? Suppose you have two polynomials $f$ and $g$. Notice that a point $(a,b)$ is a root of polynomials $f$ and $g$ if and only if it is a root of the polynomial $p(x,y) f(x,y) + q(x,y)g(x,y)$, for any polynomials $p$ and $q$. In other words, it needs to be a root of all the polynomials in the ideal generated by $f$ and $g$.

Therefore instead of considering your family of polynomials, it's enough to consider the ideal generated by that family. Since $F[x,y]$ is Noetherian, that ideal is finitely generated, and those generators are the polynomials you're looking for. You should be able to read about this in any textbook on abstract algebra/ring theory.

share|improve this answer
4  
The empty set is the set of roots of the polynomial $1$. –  André Nicolas Aug 5 '11 at 17:01
    
where i can find a proof about this nice property of polynomials over algebraically closed fields? –  Daniel Aug 5 '11 at 19:28
    
@Daniel: you can probably find a proof in any book on commutative algebra. –  Qiaochu Yuan Aug 5 '11 at 19:38
    
thanks for all man, only a last question, is necessary the hypothesis of being algebraically closed? i can´t found a counterexample, that´s not happen if i don´t have this nice property sorry for ask )= –  Daniel Aug 5 '11 at 19:52
1  
@Daniel: please do not use answers to make comments. If you are having trouble logging into your account it is because you are unregistered. Registration will fix this. –  Qiaochu Yuan Aug 5 '11 at 20:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.