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Consider the map of affine schemes $$ \operatorname{Spec}\left( \dfrac{\mathbb{C}[x,y]}{\langle xy\rangle }\right)\stackrel{f}{\rightarrow} \operatorname{Spec}\mathbb{C}[t] $$ whose corresponding map of rings is $$ \mathbb{C}[t]\stackrel{f^*}{\rightarrow} \dfrac{\mathbb{C}[x,y]}{\langle xy\rangle } \hspace{10 mm}(\star) $$ where $t\mapsto x+y$.

It is clear geometrically that $f^{-1}(t-c)$ is two points for $c\not=0$ while for $c=0$, it is the double point $\mathbb{C}[x]/\langle x^2\rangle$.

But how does one deduce this only from the map of rings ($\star$)?

Thank you.

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up vote 4 down vote accepted

To compute $f^{-1}(c)=f^{-1}(\langle t-c\rangle),$ you should compute the pullback of $f$ to the spectrum of the residue field of the point $t=c.$ The residue field of $t=c$ is $\mathbb C,$ so we want to compute the fibre product $$\operatorname{Spec}\mathbb C \times_{\operatorname{Spec}\mathbb C[t]}\operatorname{Spec} \left( \dfrac{\mathbb C[x,y]}{\langle xy\rangle} \right).$$ In other words, we want to compute $\operatorname{Spec}\left( \mathbb C\otimes_{\mathbb C[t]}\dfrac{\mathbb C[x,y]}{\langle xy\rangle} \right).$ Simplifying the tensor product, we get $\dfrac{\mathbb C[x,y]}{\langle xy,x+y-c\rangle} \cong \dfrac{\mathbb C[x]}{\langle x(c-x)\rangle},$ and we recognise $\operatorname{Spec}\left( \dfrac{\mathbb C[x]}{\langle x(c-x)\rangle}\right)$ as one (with multiplicity $2$) or two points (with multiplicity $1$) depending on whether $c=0$ or $c\neq 0.$

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Thank you Andrew! – math-visitor Jul 23 '12 at 6:58
You're welcome :) – Andrew Jul 23 '12 at 7:04

I find it hard to see how one can avoid the geometry, since it's equivalent data, but I'll try. Working algebraically, in general one will recover the definition of the scheme-theoretic fibre, but things are particularly simple here.

Since $(t - c)$ is maximal, you're just looking for prime ideals of $\mathbb C[x, y]/(xy)$ which contain the extension $(\bar x + \bar y - c)$ of this ideal. What are the primes $\mathfrak p$ containing $(\bar x + \bar y - c)$? Well, $\mathfrak p$ has to contain $\bar x$ or $\bar y$. If, for example, $\bar x \in \mathfrak p$ then $\bar y - c \in \mathfrak p$, and $(\bar x, \bar y - c)$ is maximal. Similar things happen when $\bar y \in \mathfrak p$, and the two ideals we find coincide when $c = 0$.

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