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Let $k$ be a field, $R = k[x,y]$ and $I = (x,y)$, so that $\operatorname{Spec}(R) = \mathbb{A}^2_k$ and $V(I) = \{0\} \subset \mathbb{A}^2_k$. By definition, the blow-up of the plane at the origin is the projective scheme associated to the Rees algebra $S_\bullet = \bigoplus_{n \geq 0} I^n$. Therefore in our case $S_\bullet = R[X,Y]$ where $X,Y$ are the elements $x,y \in I$ lying in degree 1.

On the other hand, $\operatorname{Proj}(S_\bullet) = \mathbb{P}^1_R$ so we seem to have an isomorphism $\mathrm{Bl}_0 \mathbb{A}^2_k \cong \mathbb{P}^1_{\mathbb{A}^2_k}$. Unless I am missing something about $S_\bullet$ the difference between the two must be in their structure morphism to $\mathbb{A}^2_k$, since $\mathrm{Bl}_0 \mathbb{A}^2_k$ only has a $\mathbb{P}^1_k$ above the origin and a point $\operatorname{Spec}(k)$ elsewhere, whereas $\mathbb{P}^1_{\mathbb{A}^2_k} \cong \mathbb{P}^1_\mathbb{Z} \times \mathbb{A}^2_k$ has a copy of $\mathbb{P}^1_\mathbb{Z}$ over each point of the plane. If so, how can I describe the difference between the two structure morphisms?

Eduardo de Lorenzo
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    The component of degree $0$ of $S$ is $k[x,y]$, whereas the component of degree $0$ of $k[x,y]$ is $k$, so $S\not\simeq k[x,y]$ as a graded ring. – Damian Rössler Nov 14 '23 at 12:26
  • @DamianRössler I didn't claim that $\mathbb{P}^1_k \cong \mathbb{P}^1_R$, the rings I am comparing are the ones associated to $\mathrm{Bl}0 \mathbb{A}^2_k$ (which is $R[I]$) and to $\mathbb{P}^1{\mathbb{A}^2}$ (which is $R[X,Y]$). – Eduardo de Lorenzo Nov 14 '23 at 13:41

1 Answers1

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The rings of $\mathrm{Bl}_0 \mathbb{A}^2_k$ and $\mathbb{P}^1_R$ are not isomorphic, the issue is mainly notational. The Rees algebra $R[I] = R[X,Y]$ is not a polynomial ring in two variables. It remembers the relation $xY - yX = 0$, where the lower case $x,y$ are in degree $0$ and the upper case $X,Y$ are in degree 1.

One possible way to deal with this notational issue is to do as in the Wikipedia article on the Rees algebra and denote it by $R[It] = \bigoplus_{n \geq 0} I^n t^n \subset R[t]$. This way it is clear that the multiplication table of $R[It] = R[xt, yt]$ has something to do with the one of $R$ (and hence $x(yt) - y(xt) = (xy-yx)t = 0$) whereas in the polynomial ring $R[u, v]$ there are no relations, and in particular $xv - yu \neq 0$.

Eduardo de Lorenzo
  • 366