Commutative Algebra

Let $R_1,...,R_m$ be rings. Then the direct sum of $R_1,...,R_m$ is the ring

$$R=R_1\oplus ...\oplus R_m={\oplus }_{i=1}^n{R}_i=\sum _{i=1}^n{R}_i=\{(x_1,...,x_m)\mid x_i\in A_i\}$$

with coordinatewise addition and multiplication. The projection mapping $p_i:R\to R_i$ for each $i$ is an onto ring homomorphism defined by $p(x_1,...,x_m)=x_i$.

Let $R$ be a ring and $J_1,...,J_n◃R$. Define $\varphi :R\to {\oplus }_{i=1}^n{R}_i$ by

$$\varphi :x\mapsto (J_1+x,...,J_n+x)$$

, so $\varphi$ is a ring homomorphism with $\ker \varphi ={\cap }_{i=1}^n{J}_i$.

Proposition:

1. $\varphi$ is injective if and only if ${\cap }_{i=1}^n{J}_i=\{0\}$

2. $\varphi$ is surjective if and only if $J_i,J_k$ are coprime for $i\ne k$

3. If $J_i,J_k$ are coprime for $i\ne k$ then ${\prod }_{i=1}^n{J}_i={\cap }_{i=1}^n{J}_i$.

Proof: (1) is clear.

For (2), suppose $\varphi$ is surjective. Then for some $x\in R$ we have $(J_1+x,...,J_n+x)=(J_1+1,J_2,...,J_n)$, thus $x\in (J_1+1)\cap J_2$, hence $1=(1-x)+x\in J_1+J_2$, so $J_1,J_2$ are coprime. Similarly, $J_i,J_k$ are coprime whenever $i\ne k$.

Conversely, suppose $J_i,J_k$ are coprime for $i\ne k$. Then for all $k\ge 2$ there exists some $u_k\in J_1,v_k\in J_k$ with $u_k+v_k=1$. Let $a\in R$. Set $x=a{v}_2...v_n$. Then $x\in J_k$ for $k\ge 2$, and

$$x=a(1-u_2)...(1-u_n)=a+(\text{terms containing}{u}_i)\in J_1+a$$

So $\varphi (x)=(J_1+a,J_2,...,J_n)$. Similarly for any $i$, we can show $(J_1,...,J_i+a,,...,J_n)\in \mathrm{im}\varphi$. Hence for any $a_1,...,a_n\in R$ we have $(J_1+a_1,...,J_n+a_n)={\sum }_{i=1}^n(J_1,...,J_i+a_i,...,J_n)\in \mathrm{im}\varphi$.

(3) Suppose $J_i,J_k$ are coprime for $i\ne k$. For $n=2$ we have $J_1\cap J_2=J_1{J}_2$ since $J_1,J_2$ are coprime. Now suppose $n>2$ and ${\prod }_{i=1}^{n-1}{J}_i={\cap }_{i=1}^{n-1}{J}_i$. Write $K={\prod }_{i=1}^{n-1}{J}_i$. Since for all $i=1,...,n-1$ there exists $x_i\in J_i,y_i\in J_n$ with $x_i+y_i=1$,we have

$$\begin{array}{ccc}1& =& 1-(x_1...x_{n-1})+(x_1...x_{n-1})\\ & =& 1-((1-y_1)...(1-y_{n-1}))+(x_1...x_{n-1})\\ & =& 1-(1+y)+x\end{array}$$

for some $y\in J_n,x\in K$. Thus $1\in J_n+K$. Hence

$$\prod _{i=1}^n{J}_i=K{J}_n=K\cap J_n$$

since $J_n,K$ are coprime.

Theorem: Let $R$ be a ring. Then:

1. Let $J_1,...,J_n◃R$ and $P$ be a prime ideal of $R$ with $P\supset {\cap }_{i=1}^n{J}_i$. Then $P\supset J_k$ for some $k$. Futhermore if $P={\cap }_{i=1}^n{J}_i$ then $P=J_k$ for some $k$.

2. Let $P_1,...,P_n$ be prime ideals of $R$ and suppose $J◃R$ satisfies $J\subset {\cup }_{i=1}^n{P}_i$. Then $J\subset P_i$ for some $i$.

Proof: (1) Suppose $P$ does not contain any of the $J_i$. Then for all $i$, choose some $x_i\in J_i\setminus P$. Set $y=x_1...x_n$. Then $y\in {\cap }_{i=1}^n\subset P$. But since $P$ is prime, for some $k$ we have $x_k\in P$, a contradiction. Hence $P\supset J_i$.

Next suppose $P={\cap }_{k=1}^n{J}_k$. Then $J_i\subset P\subset J_i$.

(2) If $n=1$ then $J\subset P_1$. This is the basis of an induction. We shall show that for some $j=1,...,n$, we have $J\subset {\cup }_{i\ne j}{P}_i$, from which the result follows.

Suppose not. Then for all $j=1,..,n$ choose $x_j\in J\setminus {\cup }_{i\ne j}{P}_i$. Since $J\subset {\cup }_{i=1}^n{P}_i$ we must have $x_j\in P_j$ for each $j=1,...,n$. Set

$$y=\sum _{j=1}^n{x}_1...x_{j-1}{x}_{j+1}...x_n$$

Then $y\in J={\cup }_{i=1}^n{P}_i$ so $y\in P_k$ for some $k=1,...,n$. But this means

$$x_1...x_{k-1}{x}_{k+1}...x_n=y-\sum _{j\ne k}{x}_1...x_{j-1}{x}_{j+1}...x_n\in P_k,$$

and since $P_k$ is prime, we have $x_j\in P_k$ for some $j\ne k$, a contradiction.