quotient ring | Arithmetic variety

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 $a \in IJ$ .