Exercise A0.III.5.18

Proposition

Let $R$ be a commutative ring, $I$ an ideal of $R$, and $a \in R$ such that $a$ is not a zero divisor. Then ${\rm Rees}_R(I) \cong {\rm Rees}_R(aI).$

Proof

First, recall that $(aI)^n = a^n I^n.$ This holds in any commutative ring.

We will construct the isomorphism $\varphi : {\rm Rees}_R(I) \to {\rm Rees}_R(aI)$ explicitly.

First, let’s define a family of maps $\varphi_n : I^n \to a^nI^n$ as $\varphi_n : x \mapsto a^n x$. Clearly, this map is an $R$-module homomorphism. We also define $\psi_n : a^n I^n \to I^n$ as $\psi_n : a^n x \to x$. We can do this, since multiplication by $a$ is injective and all elements of $a^n I^n$ are multiples of $a^n$.

These maps are inverses of each other and an $R$-module homomorphism with an inverse (set-) function is an isomorphism. Now, let $\varphi$ be the direct sum of all $\varphi_n.$ Similarly, let $\psi$ be the direct sum of all $\psi_n.$

Is it possible that $\varphi$ fails to be the inverse of $\psi$? For every $n$, $\varphi_n \circ \psi_n = {\rm id}_{a^nI^n}$ and similarly $\psi_n \circ \varphi_n = {\rm id}_{I^n}$. So $\varphi \circ \psi$ and $\psi \circ \varphi$ are both direct sums of identity arrows and thus must themselves be identities.

Now we need to check that $\varphi$ is an $R$-algebra homomorphism. Recall (5.17) that $(1,0,...)$ is the multiplicative identity of any Rees algebra. Since all components of $\varphi$ preserve $0$, and since its zeroth component is the identity arrow on $R$, $\varphi$ preserves the multiplicative identity.

To verify that $\psi$ preserves multiplication, let $u, v \in {\rm Rees}_R(aI)$ and observe that

$$\begin{align*}(u \cdot v)_n &= \sum_{p=0}^n (a^pu'_p) (a^{n-p}v'_{n-p}) \\&= a^n \sum_{p=0}^n u'_p v'_{n-p},\end{align*}$$

where $u'_m$ is just the $m$th component of $u$ with the $a^m$ factor dropped. In other words, $u'_m = \psi_m(u_m).$

So,

$$\begin{align*}(\psi(u \cdot v))_n &= \psi_n((u\cdot v)_n) \\&= \psi_n \left(a^n \sum_{p=0}^n u'_p v'_{n-p} \right) \\&= \sum_{p=0}^n u'_p v'_{n-p} \\&= (\psi(u)\cdot\psi(v))_n,\end{align*}$$

and this forces $\varphi$ to be a ring homomorphism as well.

Thus, $\varphi$ is an isomorphism of $R$-algebras.

$\blacksquare$