We have $$I=\left(x^2+2y^2-3,y(x-y),y(y+1)(y-1)\right)\subset\mathbb{C}[x,y]$$ and I would like to decompose it as intersection of simpler ideals. How could I proceed?
For example, in this case, could I write $$I=(x^2+2y^2−3,y,y(y+1)(y−1))∩(x^2+2y^2−3,x−y,y(y+1)(y−1))\ ?$$ Then could I make other similar simplifications? Is it correct?
On
Here's another hint.
After lots of work (at least for me), you find that you don't really need the last generator of the ideal, because $$ \frac{y}{3}\left(x^2+2y^2-3\right) + \left(\frac{y+x}3 \right)(y^2-yx) = y^3-y=y(y-1)(y+1). $$
Hence your ideal has two generators $$ I = (x^2+2y^2-3,y^2-yx). $$ The first factor is irreducible, but the second factor is not. Then geometrically, this correspond to the intersection of the circle (or ellipse?) given by $x^2+2y^2-3$ and the union of the lines $y=0$ and $y=x$. These intersection points are easily computed to be $(\pm \sqrt{3},0)$ and $(\pm 1, \pm 1)$. Hence your ideal should decompose as $$ (y,x-\sqrt{3})\cap (y,x+\sqrt 3) \cap (x-1,y-1) \cap (x+1,y+1). $$
This must be checked algebraically of course, since this kind of geometric reasoning is only valid up to radicals.
NB: It is a small lie that I found the relation above by myself. I asked Macaulay2 to give a set of minimal generators for the ideal. Then once I knew there was only two generators, it was not too hard to find the relation by experimentation.
Here's a tool that may be helpful for this kind of stuff.
Proposition: Let $I,J,K$ be ideals of a ring $R$ with $J + K = R$. Then $$I + JK = (I + J)(I + K) = (I + J) \cap (I + K).$$
Proof: $(I+J)(I+K) = I^2 + I(J+K) + JK = I + JK$.
Edit: I probably should mention that the last equality is simply a special case of the fact that $IJ = I \cap J$ if $I + J = R$. You can check this using the previously proven rule for products: $IJ = ((I \cap J) + I)((I \cap J) + J) = (I \cap J) + IJ = I \cap J$.