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Suppose I have an ideal like $\mathrm{rad}(\langle y - x^2, z - x^3 \rangle)$. How do I go about finding generators for the ideal?

If I could show that $\langle y - x^2, z - x^3 \rangle$ is a prime ideal then obviously $y-x^2$ and $z-x^3$ would generate $\mathrm{rad}(\langle y - x^2, z - x^3 \rangle)$, but I don't see how to do this.

Thanks.

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3  
There is a map $k[x, y, z] \to k[x]$ having this ideal as its kernel. Can you guess what that is? –  Dylan Moreland Feb 9 '12 at 19:28
    
Ah, this is a very good comment, Dylan. –  Georges Elencwajg Feb 9 '12 at 20:05

2 Answers 2

up vote 6 down vote accepted

The trick is to replace $y$ by $x^2$ and $z$ by $x^3$ in the quotient ring $k[x,y,z]/\langle y-x^2, z-x^3\rangle$ so that the quotient becomes: $k[x,y,z]/\langle y-x^2, z-x^3\rangle =k[x,x^2,x^3]=k[x]$.
Hence, since the quotient is a domain, your wish has come true : the ideal $\langle y-x^2, z-x^3\rangle$ is prime.

A generalization
If you have a quotient of a polynomial ring $k[x_0,x_1,..., x_n]$ of the special form $$A=k[x_0,x_1,\dots, x_n]/\langle x_0-g(x_1,\dots,x_n),f_1(x_0,\dots, x_n),\dots,f_i(x_0,\dots, x_n),\dots\rangle$$ you can apply the same trick of replacing every occurrence of $x_0$ by $g(x_1,\dots,x_n)$ thus obtaining $$A=k[x_1,\dots, x_n]/\langle f_1(g(x_1,\dots,x_n),\dots, x_n),\dots,f_i(g(x_1,\dots,x_n),\dots, x_n),\dots\rangle.$$ In the example you submitted I implicitly used that twice (but only wrote the end result), exploiting the rule for successive quotients $$k[x,y,z]/\langle f,g\rangle=(k[x,y,z]/\langle f\rangle)/\langle \bar g\rangle.$$ Edit: a geometric interpretation
We have an embedding $\mathbb A^1 \to \mathbb A^3: t\mapsto (t,t^2,t^3)$ with image an irreducible afine variety $ V$ known as the affine twisted cubic curve.
The calculation above says that this variety is the ideal-theoretic (or scheme-theoretic) intersection of the affine quadrics $y=x^2$ and $z=x^3$.
It is interesting to notice that the closure $\bar V$ of $V$ in $\mathbb P^3$ is no longer a scheme-theoretic complete intersection even though it is set-theoretically the intersection of a quadric and a cubic surface.

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Thanks. Funnily enough, the question I'm trying to do asks me to determine that ideal, and then to show that $k[x,y,z]/\mathrm{rad}(\langle y-x^2,z-x^3 \rangle) $ is a polynomial algebra in one variable. Is there a way to determine the ideal without thinking about the quotient ring? –  Matt Feb 9 '12 at 19:40
1  
Dear @Matt, it is a difficult problem to decide whether an ideal is reduced and if it isn't to determine its radical.This is the realm of Gröbner bases and computer algebra systems: MAPLE, REDUCE, AXIOM, MATHEMATICA, MACSYMA, MACAULAY,... I wouldn't try to avoid quotient rings in doing bare-handed computations. –  Georges Elencwajg Feb 9 '12 at 20:24
    
+1 @GeorgesElencwajg Nice trick!! –  user38268 Feb 10 '12 at 1:36

You could start by showing the ideal $I=(y-x^2,z-x^3)$ is in fact radical so that finding generators for its radical becomes easy!

So let $f\in k[x,y,z]$ be such that there is an $n\geq0$ such that $f^n\in I$. It is easy to see that there exist $g$, $h\in k[x,y,z]$ and $r\in k[x]$ such that $$(\clubsuit)\qquad\qquad f(x,y,z)=g(x,y,z)\cdot(y-x^2)+h(x,y,z)\cdot(z-x^3)+r(x),$$

and using Newton's formula for powers of trinomials— we see immediately from this that $$f(x,t,z)^n\equiv r(x)^n\mod I,$$ so that in fact $r(x)^n\in I$. Suppose for a second that we manage to prove that

$(\star)\qquad\qquad$ we have $k[x]\cap I=0.$

Then, since $r(x)^n$ is in that intersection, we have $r(x)^n=0$ and therefore $r(x)=0$. Looking back at $(\clubsuit)$ we see now that $f\in I$, and we are happy.

Can you prove $(\star)$?

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(Behind this is, of course, the idea of Groebner bases which Georges mentions in a comment!) –  Mariano Suárez-Alvarez Feb 9 '12 at 20:28

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