# Finding a polynomial ring embedded in a fg algebra over complex numbers

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 :)

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 $$\Bbb C[x_2',x_3,x_4]$$.