On quotient rings

Posted on June 5, 2020 by rohanjoshi

In this post I will talk about how to compute the product and tensor product of quotient rings $R/I$ and $R/J$ . This sort of thing is usually left as an exercise (especially the first Corollary) and not proved in full generality in algebra courses, although it is not hard.

In all that follows $R$ is a commutative ring with identity and $I$ and $J$ are ideals of $R$ .

Lemma: If $I \subseteq J$ , there is a natural map $R/I \to R/J$ .

Proposition: The natural map $R/I \otimes R/J \to R/I+J$ is an isomorphism of $R$ -algebras.

Proof: To see surjectivity, notice that $1$ generates $R/(I+J)$ as an $R$ -module, and $1 \mapsto 1$ , since the map is a ring homomorphism. To see injectivity, notice that every element

$\sum_{i=1}^n [a_i]\otimes[b_i] \in R/I \otimes R/J \ \ (a_i, b_i \in R)$

is equal to the pure tensor $[c]\otimes 1 = 1 \otimes [c]$ where $c =\sum a_ib_i$ . If $[c] \otimes 1 \mapsto 0$ , then $c \in I+J$ . So $c=i+j$ , $i\in I, j\in J$ . Then $[c]\otimes1=[i+j]\otimes1=[i]\otimes1+1\otimes[j]=0+0=0$ .

Corollary: $\mathbb{Z}/m\otimes\mathbb{Z}/n\cong\mathbb{Z}/(\gcd(m,n)).$

Proposition (Chinese Remainder Theorem): The natural map $R/(I\cap J) \to R/I\times R/J$ is injective. If $I+J=(1)$ , it is also surjective, and thus an isomorphism.

Proof: To see injectivity, notice that if $[c]\mapsto(0,0)$ , then $c\in I$ and $c\in J$ , so $c\in I\cap J$ so $[c]=0\in R/(I\cap J)$ . To see surjectivity, note that $I+J=(1)$ implies there exist $i\in I$ , $j\in J$ , such that $i-j=b-a$ , for any $a,b \in R$ . Consider $([a], [b]) \in R/I\times R/J$ , and set $i$ and $j$ . Then $a+j=b-j$ , so $a+i \mapsto ([a],[b])$ .

Corollary: If $\gcd(m,n)=1$ , $\mathbb{Z}/m\times \mathbb{Z}/n\cong\mathbb{Z}/mn$ . In particular, by applying this repeatedly we have $\mathbb{Z}/p_1\dots p_n = \mathbb{Z}/p_1 \times \cdots \times \mathbb{Z}/p_n$ .

Also, note the following fact:

Proposition: If $I + J = (1)$ , then $I \cap J = IJ$ .

Proof: Clearly $IJ \subseteq I \cap J$ . To go the other way, note that $I + J = (1)$ means that there exist $i \in I$ , $j \in J$ , and $m, n \in R$ such that $mi + nj = 1$ . So, consider an element $a \in I \cap J$ . Then we have $a = a(mi+nj) = m(ia) + n(aj)$ . Since $a \in J$ , $m(ia) \in IJ$ , and since $a \in I$ , $n(aj) \in IJ$ . So $a \in IJ$ .

Remark: One can also think of this in terms of Tor: $\text{Tor}_1(R/I, R/J) = (I\cap J)/IJ$ , and when $I+J = (1)$ this Tor group vanishes.

A special case of Krull’s intersection theorem

Posted on August 31, 2019 by rohanjoshi

In this post I will prove a special case of Krull’s intersection theorem, which can be proved without invoking the Artin-Rees lemma. This result is useful in the study of discrete valuation rings. The following proof is from Serre’s Local Fields.

Proposition: Let $R, \frak{m}$ be a noetherian local domain where $\frak{m} = (\pi).$ Then $\bigcap_{n \geq 0} \frak{m}^n = 0$ .

Proof: Suppose $y \in \bigcap_{n \geq 0} \frak{m}^n$ . Then $y = \pi^n x_n$ for all $n \geq 0$ . So for all $n$ $\pi^nx_n = \pi^{n+1}x_{n+1}$ . Since $R$ is a domain, $x_n = \pi x_{n+1}$ .

Therefore consider the ascending chain $(x_0) \subseteq (x_1) \subseteq \cdots$ . This eventually stabilizes for high enough $n$ since $R$ is noetherian, so for some $n$ , $x_{n+1} = cx_n$ . Thus $x_{n+1} = c\pi x_{n+1}$ , so $(1-c\pi)x_{n+1} = 0$ . But $1-c\pi$ is a unit, so $x_{n+1} = 0$ , so $y = 0$ . $\square$

This theorem holds more generally even if $R$ is not assumed to be a domain, but the proof is more complicated (but still among the same lines).

Proposition: Let $R, \frak{m}$ be a noetherian local ring where $\frak{m} = (\pi)$ and $\pi$ is not nilpotent. Then $\bigcap_{n \geq 0} \frak{m}^n = 0$ .

Proof: Let $\frak{u}$ be the ideal of elements that kill some power of $\pi$ . We will use variables $u_1, u_2, \dots$ to refer to elements of $\frak{u}$ . Since $R$ is noetherian, $\frak{u}$ must be finitely generated, so all elements of $\frak{u}$ kill $\pi^N$ for some fixed $N$ .

Now suppose $y \in \bigcap_{n \geq 0} \frak{m}^n$ . $\pi^nx_n = \pi^{n+1}x_{n+1}$ , so $\pi^n(x_n - \pi x_{n+1})$ . Thus $x_n - \pi x_{n+1} \in \frak{u}$ .

Consider the ascending chain $\frak{u} + (x_0) \subseteq \frak{u} + (x_1) \subseteq \cdots$ . Since $R$ is noetherian it must eventually stablize, so for some $n$ , $x_{n+1}$ can be written as $u_1 + cx_n$ . But recall that $x_n = u_2 + \pi x_{n+1}$ . So $x_{n+1} = u_1 + c(u_2 + \pi x_{n+1}) = u_3 + c\pi x_{n+1}$ so $(1-c\pi)x_{n+1} = u_3$ . $1-c\pi$ is a unit since $\frak{m} = (\pi)$ , and $R$ is local, so $x_{n+1} \in \frak{u}$ . If we force $n$ to be large enough to surpass $N$ , then $\pi^{n+1}x_{n+1} = 0$ , so $y = 0$ . $\square$

Hilbert’s Basis Theorem

Posted on July 10, 2019 by rohanjoshi

Here is a proof of Hilbert’s Basis Theorem I thought of last night.

Let $R$ be a noetherian ring. Consider an ideal $J$ in $R[X]$ . Let $I_i$ be the ideal in $R$ generated by the leading coefficients of the polynomials of degree $i$ in $I$ . Notice that $I_i \subseteq I_{i+1}$ , since if $P \in I_i$ , $xP \in I_{i+1}$ , and it has the same leading coefficient. Thus we have an ascending chain $\dots \subseteq I_{i-1} \subseteq I_i \subseteq I_{i+1} \subseteq \dots$ , which must terminate, since $R$ is noetherian. Suppose it terminates at $i = n$ , so $I_n = I_{n+1} = \dots$ .

Now for each $I_i$ choose a finite set of generators (which we can do since $R$ is noetherian). For each generator, choose a representative polynomial in $J$ with that leading coefficient. This gives us a finite collection polynomials: define $J_i$ to be the ideal of $R[x]$ generated by these polynomials.

Let $J' = J_0 + J_1 + \dots + J_n$ . I claim $J = J'$ . Assume for the sake of contradiction that there is a polynomial $P$ of minimal degree (say $i$ ) which is in $J$ but $J'$ . If $i \leq n$ , there is an element of $P' \in J'$ with the same leading coefficient, so $P - P'$ is not in $J'$ but has degree smaller than $i$ : contradiction. If $P$ is of degree $i > n$ , then there is an element of $P'$ of $J_n$ which has the same leading coefficient as $P$ . Thus $P - x^{i-n}P'$ is of degree smaller than $i$ but is not in $J'$ : contradiction.