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I am working through section IV.2 about blow-ups of the book Geometry of schemes by Eisenbud and Harris. I am having some trouble understanding the details of the following example, which they discuss in this section. I have very little expercience with schemes and I am sure my problem has to do with this.

Pick some algebraically closed field $K$ and define the affine plane $\mathbb{A}_K^2=\mathrm{Spec}(K[x,y])$. Set $U'=\mathrm{Spec}(K[x',y'])$ and $U''=\mathrm{Spec}(K[x'',y''])$. Consider the morphisms of schemes $$ \phi':U'\rightarrow \mathbb{A}_K^2 \quad\mbox{and}\quad \phi'':U''\rightarrow \mathbb{A}_K^2 $$ induced by the ring morphisms $$ \begin{matrix} (\phi')^{\#}:K[x,y]\rightarrow K[x',y']\\ x \mapsto x'\\ y \mapsto x'y' \end{matrix} \quad\mbox{and}\quad \begin{matrix} (\phi'')^{\#}:K[x,y]\rightarrow K[x'',y'']\\ x \mapsto x''y''\\ y \mapsto y''. \end{matrix} $$

Observe that $(\phi')^{\#}\vert_{K[x,y,x^{-1}]}:K[x,y,x^{-1}]\rightarrow K[x',y',{x'}^{-1}]$ is an isomorphism with inverse $$ \begin{matrix} K[x',y',{x'}^{-1}]\rightarrow K[x,y,x^{-1}]\\ x'\mapsto x\\ y'\mapsto yx^{-1}. \end{matrix} $$ Similarly we see that $(\phi'')^{\#}\vert_{K[x,y,x^{-1}]}:K[x,y,x^{-1}]\rightarrow K[x'',y'',{y''}^{-1}]$ is an isomorphism with inverse $$ \begin{matrix} K[x'',y'',{y''}^{-1}]\rightarrow K[x,y,x^{-1}]\\ x''\mapsto xy^{-1}\\ y''\mapsto y. \end{matrix} $$ Therefore, we have the dual isomorphisms $$ \phi'\vert_{U'_x}:U'_x=\mathrm{Spec}(K[x',y',x'^{-1}])\rightarrow U_x=\mathrm{Spec}(K[x,y,x^{-1}]) $$ and $$ \phi''\vert_{U''_y}:U''_y=\mathrm{Spec}(K[x'',y'',y''^{-1}])\rightarrow U_y=\mathrm{Spec}(K[x,y,y^{-1}]). $$ We write $U_{xy}=U_x\cap U_y=\mathrm{Spec}(K[x,y,x^{-1},y^{-1}])$ and observe that $U_{xy},U'_{xy}$ and $U''_{xy}$ can be identified via these isomorphisms.

Let's compute the morphisms of schemes $\psi=(\phi')^{-1}\circ\phi'':U_{xy}''\rightarrow U_{xy}'.$ This map is induced by the ring morphism $$ \begin{matrix} K[x',y',x'^{-1},y'^{-1}] & \rightarrow & K[x,y,x^{-1},y^{-1}] & \rightarrow & K[x'',y'',x''^{-1},y''^{-1}]\\ x' & \mapsto & x & \mapsto & x''y''\\ y' & \mapsto & yx^{-1} & \mapsto & {x''}^{-1}. \end{matrix} $$ Define $Z= \frac{U'\cup U''}{U_{xy}'\simeq U_{xy}''}$ and let $\phi:Z\rightarrow \mathbb{A}_K^2$ be the `gluing' of $\phi'$ and $\phi^{''}$.

The idea is that this scheme $Z$, together with the map $\phi$, is the blow-up of $\mathrm{A}_K^2$ in the origin. In particular $\phi^{-1}(x,y)$ should be isomorphic to $\mathbb{P}_K^1$ and $\phi$ should be an isomorphism everywhere else.

I am having trouble verifying this statement. This is what I tried in order to compute $\phi^{-1}(x,y)$:

First I wanted to consider $U'$ and $U''$ individually, expecting that $\phi'^{-1}(x,y)$ and $\phi''^{-1}(x,y)$ would both be isomorphic to an affine line. Then by gluing I expected this to become the projective line when considering $\phi^{-1}(x,y)$.

In order to compute $\phi'^{-1}(x,y)$, recall that $\phi'$ sends $p \in \mathrm{Spec}(K[x',y'])$ to the preimage of $pK[x',y']_{p}$, under the composite map $$ \begin{matrix} \alpha: K[x,y] & \rightarrow & K[x',y'] & \rightarrow & K[x',y']_{p}\\ x & \mapsto & x' & \mapsto & \frac{x'}{1}\\ y & \mapsto & x'y' & \mapsto & \frac{x'y'}{1}. \end{matrix} $$ Hence we are interested in finding $p \in \mathrm{Spec}(K[x',y'])$ such that $\phi'(p)=(x,y)$, i.e. $\alpha^{-1}(p)$ should equal $(x,y)$. So we want $p=\alpha(x,y)$. Compute $\alpha(x,y)=(x',x'y')=(x')$. Therefore, $\phi'^{-1}(x,y)$ consists just of one point, the ideal $(x')$. This is clearly not as expected.

Does anyone know what mistakes I am making? Thank you in advance.

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I am not sure why you are complicating things with $\alpha$ (but it could be me it is pretty late here as I am writing this and maybe I forgot I need to localize).

By definition, $\phi'(p) = (\phi'^\#)^{-1}(p)$ (where $^{-1}$ means the inverse image of a set and not an inverse map).

So you want to find the prime ideals $p$ such that $(\phi'^\#)^{-1}(p) = (x,y)$, which means $\{g \in K[x,y] \mid \phi'^\#(g) \in p\} = (x,y)$

$\phi'^\#$ is injective but not surjective, so this means that $p \cap Im \phi'^\# = \phi'^\#((x,y))$, and thus
$p \cap K[x',x'y'] = (x',x'y')$.

In fact you can show that (because $(x',x'y')$ is a maximal ideal of $K[x',x'y']$)
this is equivalent to $x' \in p$ and $1 \notin p$.

And there are many nontrivial prime ideals of $K[x',y']$ containing $x'$, such as all the $(x',y'-k)$ for $k \in K$

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