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Prime ideals 
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#1
May1212, 03:38 PM

P: 31

This is a basic abstract algebra question.
Q1. Is this (x1, x2) a prime ideal in C[x1, x2, x3, x4] ? Q2. What about this: (x_{1} x_{4}x_{2} x_{3}, x_{1} x_{3}x_{2}^{2})? Q3. Is this a prime ideal (this is the twisted cubic in projective 3space): (x_{1} x_{4}x_{2} x_{3}, x_{1} x_{3}x_{2}^{2}, x_{2} x_{4}x_{3}^{2})? Thanks everyone. 


#2
May1212, 03:49 PM

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P: 18,061

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
May1212, 04:05 PM

P: 6

A1. Yes
A2. Yes A3. No? Not sure. x2 (x1*x4x2*x3)+x3 (x1*x3x2^2)+x1 (x2*x4x3^2)=0 (*) and x3 (x1*x4x2*x3)+x4 (x1*x3x2^2)+x2 (x2*x4x3^2)=0 (**) Are (*) and (**) relevant at all? 


#4
May1212, 04:09 PM

Mentor
P: 18,061

Prime ideals



#5
May1212, 05:54 PM

P: 31

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 x4plane). Q2. What about this: (x1 x4x2 x3, x1 x3x2^2)? No since x2 (x1 x4x2 x3) x3 (x1 x3x2^2) = x1(x2 x4x3^2) is in the ideal but neither x1 nor (x2 x4x3^2) is in the ideal. Q3. Is this a prime ideal (this is the twisted cubic in projective 3space): (x1 x4x2 x3, x1 x3x22, x2 x4x32)? Yes, because this is the twisted cubic, which is irreducible. 


#6
May1212, 05:59 PM

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P: 18,061




#7
May1212, 06:28 PM

P: 31

> Yes, but strictly speaking you'll need to show that neither x1 nor (x2 x4x3^2) is in the ideal.
Thanks. > This is hardly a proof. Yes, but it seems quite messy to do it directly, using the polynomials. 


#8
May1212, 06:39 PM

P: 31

> 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
May1312, 07:03 PM

Sci Advisor
HW Helper
P: 9,453

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
May1312, 09:27 PM

P: 31

Oh I see. Thank you mathwonk!



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