I found these examples in Eisenbud-Harris' Geometry of Schemes and I am rather stuck understanding the algebra machinery under them. Suppose $k$ is an algebraically closed field.

Let $X=\mathrm{Spec}(k[x,u]/(xu))$ be the union of two intersecting lines in the affine plane and let $\varphi:X\longrightarrow \mathrm{Spec}(k[t])$ be the morphism induced by $k[t]\longrightarrow k[x,u]/(xu),t\mapsto x+u+(xu)$.

I want to show that the fibre of $\varphi$ above $p=(t-a)\in \mathrm{Spec}(k[t])$ consists of two distinct points (i.e. $\mathrm{Spec}(K\times K)$) if $a\neq 0$ and a single double point if $a=0$, i.e. the scheme $\mathrm{Spec}(K[t]/(t^2)$).

I don't want to use the definition via fiber products, as in the book this is given later and at this stage the only definition it provides is "scheme-theoretic preimage" of a morphism $\varphi : \mathrm{Spec}(R)\longrightarrow \mathrm{Spec}(S)$ at a point $\mathfrak p\in \mathrm{Spec}(S)$ as the affine scheme $\mathrm{Spec}(R/e(\mathfrak p))$ where $e(\mathfrak p)$ is the extended ideal via the ring morphism $S\longrightarrow R$ associated with $\varphi$.

My attempt: Here we have $\mathfrak p =(t-a)$ and the first problem is calculating the extension. The generic element in $e(t-a)$ should be $$\sum_{i}(f_i+(xu))\cdot \varphi((t-a)\cdot g)=(x+u-a)\cdot g(x+u)\cdot \sum_i f_i +(xu)$$ where $g\in k[t],f_i\in k[x,u]$. If I am correct, then $e(t-a)=(x+u-a)$ in the quotient ring $k[x,u]/(xu)$. I'm not really convinced of that but I can't fault it.

It follows then that $$\varphi^{-1}(\mathfrak p)=\mathrm{Spec}\left(\frac{k[x,u]/(xu)}{(u+x-a)/(xu)}\right)$$ as sets. But then $$\frac{k[x,u]/(xu)}{(x+u-a)/(xu)}\simeq k[x,u]/(x+u-a)$$

I'm stuck from here and I am neither very convinced about the above. Could you please help me?


1 Answer 1


Your last two lines are incorrect: I think you're trying to use the Third Isomorphism Theorem, but that's only valid if $(xu)\subset (x+u-a)$, which is certainly false here.

The correct way to use the Third Isomorphism Theorem is to note that the ideal generated by $\overline{x+u-a}$ in $k[x,u]/(xu)$ is actually $(x+u-a,xu)/(xu)$. If this is not clear, try proving the inclusions in both directions.

Then $(k[x,u]/(xu))/((x+u-a,xu)/(xu)) \cong k[x,u]/(x+u-a,xu)\cong k[x] / (x(a-x))$, which will give you the answer you want.

  • $\begingroup$ Thanks, I suspected something like that. I have only a question: do you use the fact that $e(t-a)=(x+u-a,xu)$ here? And if yes, why is that true? $\endgroup$
    – Caligula
    Jun 21, 2017 at 15:52
  • $\begingroup$ No, this is not quite right. The extension of a principal ideal $(z) \subset R$ is always the principal ideal $(\varphi(z)) \subset S$. But we are also taking the quotient by the ideal $(ux)$. I will update the post with the correct use of the Third Isomorphism Theorem. $\endgroup$
    – Slade
    Jun 21, 2017 at 16:53
  • $\begingroup$ Thank you, now it is clear. $\endgroup$
    – Caligula
    Jun 22, 2017 at 9:20

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