I'm doing Exercise 1.13 on Gathmann's commutative algebra notes:
Show that the equation of ideals $$(x^3-x^2,x^2y-x^2,xy-y,y^2-y)=(x^2,y)\cap (x-1,y-1)$$ holds in the polynomial ring $\mathbb C[x,y]$. Is this a radical ideal? What is its zero locus in $\mathbb A_\mathbb C^2$?
Attempt: Let the four polynomials on the LHS be zero, and we can find that the only solutions are $x=0,y=0$ and $x=1,y=1$, so the zero locus are the two points $(0,0)$ and $(1,1)$. Since in $\mathbb C[x,y]$ we have 1-1 correspondence between subvarieties and radical ideals, and the radical ideal corresponds to the two points is $(x,y)\cap (x-1,y-1)$, the ideal in question is not radical.
Questions: (1) How to show the equation holds? (2) I think my approach for the second question may be wrong, or at least not rigorous since I don't know how to show the the ideal corresponds to the two points is $(x,y)\cap (x-1,y-1)$, what is the correct way of checking if an ideal is radical?
Edit: Using the hints from red whisker, since the two ideals on RHS are coprime, the intersection is the same as product $(x^2,y)\cdot (x-1,y-1)\subset (x^2(x-1),x^2(y-1),y(x-1),y(y-1))$, so RHS is contained in LHS. The LHS is contained in the RHS since all four terms in LHS are the product of one term in the first ideal and one term in the second ideal in RHS, so all four terms are in the intersection, so the LHS ideal is in the intersection. For the counterexample to the ideal being radical: $x(x-1)$ is not contained in the ideal but $x^2(x-1)^2$ is.
Best Answer
There isn't a 1-1 correspondence between subvarieties and ideals because the ideal of a subvariety is radical. What you have shown above by looking at the zero loci is that the radicals of the two ideals are the same.
To show that the two ideals are the same, you need to show that everything on the left lives in the right and vice versa. One of these is perhaps easy to see, for the other here's a hint: the two ideals on the right are comaximal (so the intersection is the product; alternatively, write out what elements in the intersection look like).
The ideal corresponding to two points is indeed the intersection because going from zero loci to ideals turns unions into intersections. Showing that an ideal is radical seems difficult in general, but showing that it is not radical is easier: find some element which lands in the ideal when raised to some power but is itself not in the ideal (for this particular case, think of functions that vanish on the zero locus).