11

Consider the ring of polynomials in two variables $\mathbb{C}[x,y]$. Show that the ideal $\langle xy^3, x^2y^2, x^3y\rangle$ cannot be generated by two elements.


Until now, I assumed by contradiction that $\langle xy^3, x^2y^2, x^3y\rangle=\langle g, h \rangle$ for some $g,h \in \mathbb{C}[x,y]$. Then, we have that:

$g=xy^3a + x^2y^2b + x^3yc$

$h=xy^3d + x^2y^2e + x^3yf$

for $a, b, c, d, e, f$ in the polynomial ring. However, I'm not sure if this is the correct approach as I'm not sure on how to proceed. I was thinking that maybe one can write an element in $\langle xy^3, x^2y^2, x^3y\rangle$ as the square of a sum. What would be the best approach to do this?

Thanks for the help.

user26857
  • 52,094
Guy
  • 989
  • 2
    Please read this answer and try to do the same for your ideal. – user26857 Jan 22 '16 at 07:49
  • I guess in this case we would take the quotient by $m=(x,y)$ right? The step I don't quite follow is that if $\mathbb{dim}(I/mI)=3$ (say if $I$ is the original ideal in this problem), then the minimal number of generators of $I$ would have to be 3. As well, I'm not quite sure on how to derive that the dimension of the quotient has to be $3$. Could you try to extend this as an answer? I think your argument is very elegant, but I'm not sure if I can quite follow it.

    Thanks a lot for the help!

    – Guy Jan 22 '16 at 08:03

1 Answers1

8

Set $I=\langle xy^3, x^2y^2, x^3y\rangle$ and $\mathfrak m=\langle x, y\rangle$.

$\dim I/\mathfrak mI=3$

Just show that the residue classes of the generators of $I$ are linearly independent over $K$, that is, if $axy^3+bx^2y^2+cxy^3\in mI$ with $a,b,c\in K$ then $a=b=c=0$. This is obvious since the generators of $\mathfrak mI$ are homogeneous of degree $5$.

If $\dim I/\mathfrak mI=3$ then the minimal number of generators of $I$ is $3$.

This is very well explained here.

user26857
  • 52,094
  • That makes sense. Is there an easier way however, to show that $\mathbb{dim}(I/mI)=3$ implies that the minimal number of generators is $3$? – Guy Jan 22 '16 at 08:56
  • @Guy Well, if $I$ is generated by less than three elements then $I/mI$ is generated by less than three elements, so $\dim I/mI<3$. – user26857 Jan 22 '16 at 12:59