# Direct Sums

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

$R = R_1 \oplus ... \oplus R_m = \oplus_{i=1}^n R_i = \sum_{i=1}^n R_i = {\{ { (x_1,...,x_m)|x_i \in A_i }\}}$

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

Let $$R$$ be a ring and $$J_1,...,J_n\triangleleft R$$. Define $$\phi:R\rightarrow \oplus_{i=1}^n R_i$$ by

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

, so $$\phi$$ is a ring homomorphism with $$ker\phi = { \cap_{i=1}^n { J_i} }$$.

Proposition:

1. $$\phi$$ is injective if and only if $$\cap_{i=1}^n J_i = \{0\}$$

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

(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 $$n \gt 2$$ 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

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

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 $$J_1,...,J_n\triangleleft R$$ 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 $$J\triangleleft R$$ 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.

Ben Lynn blynn@cs.stanford.edu 💡