# Prime ideals

1. May 12, 2012

### naturemath

This is a basic abstract algebra question.

Q1. Is this (x1, x2) a prime ideal in C[x1, x2, x3, x4] ?

Q3. Is this a prime ideal (this is the twisted cubic in projective 3-space):
(x1 x4-x2 x3, x1 x3-x22, x2 x4-x32)?

Thanks everyone.

2. May 12, 2012

### micromass

Staff Emeritus
What are your thoughts??

Hint: I always like to check if something is a prime ideal by checking if the quotient ring is an integral domain.

3. May 12, 2012

### math2012

A1. Yes
A2. Yes
A3. No? Not sure.

-x2 (x1*x4-x2*x3)+x3 (x1*x3-x2^2)+x1 (x2*x4-x3^2)=0 (*)

and

-x3 (x1*x4-x2*x3)+x4 (x1*x3-x2^2)+x2 (x2*x4-x3^2)=0 (**)

Are (*) and (**) relevant at all?

4. May 12, 2012

### micromass

Staff Emeritus
Why yes to both??

I don't see how (*) and (**) are relevant to this.

5. May 12, 2012

### naturemath

So I'm guessing the following.

Q1. Is this (x1, x2) a prime ideal in C[x1, x2, x3, x4] ?

Yes since the quotient is an integral domain (an irredu variety-- it's the x3 x4-plane).

No since
x2 (x1 x4-x2 x3) -x3 (x1 x3-x2^2) = x1(x2 x4-x3^2) is in the ideal but neither x1 nor (x2 x4-x3^2) is in the ideal.

Q3. Is this a prime ideal (this is the twisted cubic in projective 3-space):
(x1 x4-x2 x3, x1 x3-x22, x2 x4-x32)?

Yes, because this is the twisted cubic, which is irreducible.

6. May 12, 2012

### micromass

Staff Emeritus
Yes, but strictly speaking you'll need to show that neither x1 nor (x2 x4-x3^2) is in the ideal.

This is hardly a proof.

7. May 12, 2012

### naturemath

> Yes, but strictly speaking you'll need to show that neither x1 nor (x2 x4-x3^2) is in the ideal.

Thanks.

> This is hardly a proof.

Yes, but it seems quite messy to do it directly, using the polynomials.

8. May 12, 2012

### naturemath

> This is hardly a proof.

I'm thinking of writing the ideal (or any ideal) as a primary decomposition, but for even that, is there a systematic way to decompose an ideal in such a way? Or do you recommend other (more) feasible options?

9. May 13, 2012

### mathwonk

the tricky part is that many ideals have the same zero locus but at most one of those ideals is prime. in the case of the twisted cubic, if an ideal I has the twisted cubic as its zero locus, the irreducibility of the cubic implies the radical of I is prime, but not necessarily I itself. so you have to do the algebra.

10. May 13, 2012

### naturemath

Oh I see. Thank you mathwonk!