# Blow-up an irreducible component

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.

$$\newcommand{\Spec}{\operatorname{Spec}}$$
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 $$\Spec(R)\leftarrow \Spec(R/I)$$.