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If one blows-up for example $V(y)$ in $V(xy)$ (ie the center is $V(y)$) the Rees-Algebra is $K[x,y,yt]/(xy)\simeq K[x,y,u]/(xy,xu)$ ie as grading ring $R[u]/(xu)$ with $R=K[x,y]/(xy)$. The Blow-up is then $\widetilde{X}=\text{Proj}(R[u]/(xu))=\text{Spec}(R/(x))=\text{Spec}(K[y])$ and that's good because in $\widetilde{X}$ the center $V(y)$ become a Cartier divisor ie $y$ is no more a zero divisor. Here we see that $\widetilde{X}$ is the other irreducible component.

Question: in general can we describe simply the blow-up of $V(g)$ in $\text{Spec}(R)$ with $g$ a zero divisor? The problem is that I can't get an isomorphisme like $R[yt]\simeq R[u]/(xu)$ here. In this document page 20 example 4.31 is written that the Rees algebra is $R[v]/(hv)$ with $hg=0$, but I can't get it and furthermore I can't see why $g$ should be regular in this ring.

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$\newcommand{\Spec}{\operatorname{Spec}}$

Let $R$ be a ring, $g\in R$ and set $I=\{f\in R: fg^k=0,\ \mbox{for some } k\geq 1\}$. I claim that $\Spec(R)\leftarrow\Spec(R/I)$ is the blow-up of $\Spec(R)$ along the principal ideal $(g)$. To show this I use the universal property of blow-ups. Firstly, it is clear that $g$ becomes regular at $R/I$. Secondly, if $\Spec(R)\leftarrow X$ is any morphism such that the image of $g\in \Gamma(X,\mathcal{O}_X)$ defines an invertible ideal, then in particular $g$ is regular in $\Gamma(X,\mathcal{O}_X)$ and therefore the image of $I$ in $\Gamma(X,\mathcal{O}_X)$ must be zero. This means that $R\to \Gamma(X,\mathcal{O}_X)$ factorizes uniquely through $R\to R/I$, whence $\Spec(R)\leftarrow X$ factorizes uniquely through $\Spec(R)\leftarrow \Spec(R/I)$.

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  • $\begingroup$ Thanks, all is clear. I found to this in the first answer $\endgroup$ Feb 28, 2020 at 19:57

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