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?
Best Answer
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 $xv - yu \neq 0$.