Exercise A0.III.5.17

Definition (Rees algebra)

Let $R$ be a commutative ring and fix an ideal $I$ of $R$. The Rees algebra of $I$ is a ring structure on the infinite direct sum

$${\rm Rees}_R(I) = I^0 \oplus I^1 \oplus I^2 \ldots,$$

(where $I^0 = R$), along with a map $\sigma : R \to {\rm Rees}_R(I)$ sending $a \mapsto (a, 0, 0, \ldots).$

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Proposition (P1)

There is a ring structure on ${\rm Rees}_R(I)$, and $\sigma$ makes it into an $R$-algebra.

Proof

Note that $I^n$ is an abelian group. We already know that a direct sum of a countable set of abelian groups is an abelian group, so the only thing deserving our attention here is the multiplicative structure. The problem statement already gives a hint: since ${\rm Rees}_R(I)$ is supposed to be isomorphic to $R[x]$, we should expect multiplication in it to be the convolution

$$(a_1,a_2,a_3,\ldots)\cdot(b_1,b_2,b_3,\ldots) = (c_1,c_2,c_3,\ldots)$$

where

$$c_n = \sum_{p=0}^n a_p b_{n-p}.$$

Indeed, note that $a_p \in I^p$ and $b_{n-p} \in I^{n-p}$, thus

$$a_p = \sum_{k=0}^{m}i_{k,1}\ldots i_{k,p} \quad \text{where}\quad i_{k,l} \in I, m \in \mathbb{N},$$

and the picture is analogous for $b_{n-p}$. Using distributivity, we can see that $a_p b_{n-p}$ has to be a sum over $n$-element products of elements of $I$, hence $c_n \in I^n$.

Since we have already verified (1.13) that multiplication of polynomials is associative, we only need to observe that $c_n$ is simply the coefficient of $x^n$ in the product of two polynomials $\left(\sum_{k=0}^{m}a_k x^k\right)\left(\sum_{k=0}^{m}b_kx^k\right)$ where $m$ is some non-negative integer such that both sequences are identically zero on indices above $m$. Since both sequences have finite support, we know that such $m$ exists.

Clearly $(1,0,0,\ldots)$ is the multiplicative identity.

To verify right-distributivity, we check

$$\begin{align*} (a \cdot (b + c))_n &= \sum_{p=0}^n a_p \cdot (b_{n-p}+c_{n-p}) \\ &= \sum_{p=0}^n (a_p b_{n-p}+ a_p c_{n-p}) \\ &= \sum_{p=0}^n a_p b_{n-p}+ \sum_{p=0}^n a_{p}c_{n-p} \\ &= (a \cdot b)_n + (a \cdot c)_n, \end{align*}$$

and the same argument works for left-distributivity. Thus, ${\rm Rees}_R(I)$ is indeed a ring.

Note that, since $R$ is commutative,

$$(a \cdot b)_n = \sum_{p=0}^n a_p b_{n-p} = \sum_{p=0}^n a_{n-p} b_{p} = \sum_{p=0}^n b_{p} a_{n-p} = (b \cdot a)_n.$$

So, ${\rm Rees}_R(I)$ has to be commutative as well.

Observe that $\sigma$ is simply the injection $i_0 : I^0 \to \oplus_n I^n$, so we already know that it is a ring homomorphism. Since ${\rm Rees}_R(I)$ is commutative, $\sigma(R)$ has to lie in the center of ${\rm Rees}_R(I)$, making it into an $R$-algebra.

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Proposition (P2)

Let $a \in R$ such that $a$ is not a zero-divisor, and let $I = (a)$. Then, ${\rm Rees}_R(I)$ is isomorphic to $R[x]$ as an $R$-algebra.

Proof

Consider the elements of $I^n$. They have the form

$$\sum_{k=0}^{m} r_ka^n = a^n\sum_{k=0}^{m} r_k \quad \text{where} \quad r_k \in R,m\in\mathbb{N}.$$

Any element of $R$ can be written as $\sum_{k=0}^{m} r_k$, so we can replace this factor with $r \in R.$ This gives us a convenient picture: elements of $I^n$ are of the form $ra^n$.

Thus, the tuples that live in ${\rm Rees}_R(I)$ look like

$$(r_0a^0, r_1a^1, r_2a^2,\ldots),$$

and we’d like to have the map $\varphi : {\rm Rees}_R(I) \to R[x]$ given by the rule

$$(r_0a^0, r_1a^1, r_2a^2,\ldots) \mapsto r_0 + r_1x + r_2x^2 + \ldots$$

But is it well-defined? Well, if $r_pa^n = r_qa^n$ then $r_pa^n - r_qa^n = (r_p - r_q)a^n = 0$. Since $a$ is not a zero-divisor, neither is $a^n$ and hence $r_p = r_q$. And since the elements of ${\rm Rees}_R(I)$ have finite support, the resulting polynomials must have a finite degree.

It is clear that $\varphi$ preserves sums and additive inverses, since these are preserved pointwise. It is also clear that it sends $(0, \ldots)$ to $0$ and $(1, 0, \ldots)$ to $1$.

To see that the multiplication is preserved take $u,v \in {\rm Rees}_R(I)$ and observe that

$$\begin{align*} a^n{\rm coeff}_n \left(\varphi\left(u \cdot v\right)\right) & = \sum_{p=0}^n u_p v_{n-p} \\ &= \sum_{p=0}^{n}a^p{\rm coeff}_p(\varphi (u))a^{n-p}{\rm coeff}_{n-p}(\varphi(v)) \\ &= a^n\sum_{p=0}^{n}{\rm coeff}_p(\varphi (u)){\rm coeff}_{n-p}(\varphi(v)) \\ &= a^n{\rm coeff}_n \left(\varphi\left(u\right) \cdot \varphi\left(v\right)\right), \end{align*}$$

where ${\rm coeff}_n(\beta)$ denotes the coefficient of $x^n$ in $\beta$. Extensionality and cancellation of $a^n$ force

$$\varphi(u \cdot v) = \varphi(u) \cdot \varphi(v).$$

Finally, note that

$$\begin{align*} \varphi(r w) &= \varphi(rw_0,rw_1,rw_2,\ldots)\\ &= rw_0 + rw_1x + rw_2 x^2 + \ldots\\ &= r(w_0+w_1x+w_2x^2 + \ldots). \end{align*}$$

This demonstrates that $\varphi$ is indeed an $R$-algebra homomorphism.

Finally, we must show that $\varphi$ is a bijection. Consider the map $\psi : R[x] \to {\rm Rees}_R(I)$ given by the inverse rule

$$r_0 + r_1x + r_2x^2 + \ldots \mapsto (r_0a^0, r_1a^1, r_2a^2,\ldots).$$

It is clear that $\psi$ exists as a set-function (since ${\rm coeff}_n$ is well-defined), and an $R$-algebra homomorphism with an inverse function is an isomorphism.

$\blacksquare$