# Fiber product of singular varieties

Let $$f\colon X\to Y$$ be a morphism between irreducible $$\mathbb{C}$$-varieties, such that $$f$$ is 1-1 and surjective, but $$df$$ fails to be injective along a subvariety $$Z$$. (This forces $$Y$$ to be singular).

My question is, in this case, with the (possible non-reduced) scheme structure, What is the scheme structure of the fiber product $$Y\times_{Y\times Y} X\times X=X\times_Y X$$ ?

Thanks for the comment @abx. Probably I should be more precise, what I would like to ask is, what is the scheme structure of $$X\times_Y X$$? For example, the base set is just the diagonal $$\Delta_X$$ union with the double point set, but as a scheme there should be a non-reduced structure along the non-immersion locus, I suppose. My question was to ask this scheme structure. For example, consider $$X,Y\subset \mathbb{C}^4$$ parametrized by $$X=(x, -6x^2, 8x^3, w)$$, and $$Y=(3x^4, -6x^2, 8x^3, w)$$, with $$f\colon (x,y,z,w)\mapsto (x^4+yx^2+zx, y,z,w)$$. $$X$$ is a smooth surface, while $$Y$$ is not. $$f$$ is 1-1 and surjective, but not local immersion along $$(0,0,0,w)$$. In this case what is the scheme structure of $$X\times_Y X$$?

• The first one is $X$, and the second one $X\times_YX$. This has nothing to do with singularities, or with the special properties of $f$ — just compose the appropriate cartesian diagrams. The question would be a better fit for MSE.
– abx
Sep 18, 2019 at 9:23
• Thanks for the comment. Probably I should be more precise, what I would like to ask is, what is the scheme structure of $X\times_Y X$? For example, the base set is just the diagonal $\Delta_X$ union with the double point set, but as a scheme there should be a non-reduced structure along the non-immersion locus, I suppose. My question was to ask this scheme structure. Sep 19, 2019 at 1:15
• You will get the diagonal with an embedded component along $Z$; the precise structure depends very much on the particular morphism you choose. You might try first the case where $f$ is the normalization of the cuspidal cubic $y^2=x^3$ to see what happens. Your example is doable in the same way, though more complicated.
– abx
Sep 19, 2019 at 4:07
• @abx Dear abx: Thanks for your reply! that's exactly what I was expecting for, that the fiber product should have an embedded component along Z. (I did the cuspidal curve case, but I struggled in my example since I had trouble finding the kernel for the tensor product of $\mathbb{C}$ algebras. I might will try to hit Macaulay 2, lol.) Sep 19, 2019 at 4:37