1
$\begingroup$

I was asked this question and I thought I know how to approach it but I'm completely stuck. The question is as follows:

Let $$ A=\mathbb{C}[x_1,x_2,x_3,x_4]/(x_4x_3-x_2x_1, x_1^2x_3-x_4^3x_2) $$ Find a polynomial ring $B$ with a finite injective map into $A$.

So far, I tried using Noether normalisation, and found reduced the ring to 3 variables, $x_2',x_3,x_4$, with $x_2=x_2'-x_1$, but I can't get rid of $x_1$ in the polynomials I got after plugging in the new value for $x_2$.

The polynomial I got are: $$ x_{4}x_{3}-x_{1}x_{2}'+x_{1}^{2};\ x_{1}^{2}x_{3}-x_{4}^{3}\left(x_{2}'-x_{1}\right) $$

But I see no way to cancel out $x_1$ so I can continue, and I need that in order to find a polynomial which is in the intersection of I and the polynomial ring I got.

I'm mostly looking for a way to cancel out $x_1$, I don't want the complete answer just yet :)

$\endgroup$
0
$\begingroup$

Answering my own question, for future generation, the way to do this is to complete the square in the following manner:

$$\left(x_{1}-\frac{1}{2}x_{2}'\right)^{2}-\left(\frac{1}{4}x_{2}'^{2}-x_{3}x_{4}\right)$$

and then calculating until you get $x_1$ to be of degree 1: $$ x_{4}x_{3}-x_{1}x_{2}'+x_{1}^{2}\\\\x_{1}^{2}x_{3}-x_{4}^{3}\left(x_{2}'-x_{1}\right)\\\\x_{3}\left[x_{4}x_{3}-x_{1}x_{2}'+x_{1}^{2}\right]-x_{1}^{2}x_{3}+x_{4}^{3}\left(x_{2}'-x_{1}\right)\\\\x_{3}^{2}x_{4}-x_{1}x_{2}'x_{3}+x_{4}^{3}x_{2}'-x_{1}x_{4}^{3}\\\\x_{3}^{2}x_{4}+x_{4}^{3}x_{2}'-x_{1}\left(x_{2}'x_{3}+x_{4}^{3}\right) $$

and manipulating it algebraically to get $$ x_{1}\left(x_{2}'x_{3}+x_{4}^{3}\right)-\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)+\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\\\\\left(x_{1}-\frac{1}{2}x_{2}'\right)\left(x_{2}'x_{3}+x_{4}^{3}\right)+\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\\\\\left[\left(x_{1}-\frac{1}{2}x_{2}'\right)\left(x_{2}'x_{3}+x_{4}^{3}\right)+\left(\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\right)\right]\cdot\\\cdot\left[\left(x_{1}-\frac{1}{2}x_{2}'\right)\left(x_{2}'x_{3}+x_{4}^{3}\right)-\left(\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\right)\right]\\\\\left(x_{1}-\frac{1}{2}x_{2}'\right)^{2}\left(x_{2}'x_{3}+x_{4}^{3}\right)^{2}-\left(\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\right)^{2}\\$$ Then subtracting $$\\\left(x_{1}-\frac{1}{2}x_{2}'\right)^{2}\left(x_{2}'x_{3}+x_{4}^{3}\right)^{2}-\left(\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\right)^{2}-\\-\left[\left(x_{1}-\frac{1}{2}x_{2}'\right)^{2}-\left(\frac{1}{4}x_{2}'^{2}-x_{3}x_{4}\right)\right]\left(x_{2}'x_{3}+x_{4}^{3}\right)^{2}\\\\=-\left(\frac{1}{2}x_{2}'\left(x_{2}'x_{3}+x_{4}^{3}\right)-x_{3}^{2}x_{4}-x_{4}^{3}x_{2}'\right)^{2}+\left(\frac{1}{4}x_{2}'^{2}-x_{3}x_{4}\right)\left(x_{2}'x_{3}+x_{4}^{3}\right)^{2}\in I $$ And then I got a polynomial in $I$ which is also in the lower algebra $\Bbb C[x_2',x_3,x_4]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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