## 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.