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 \(ker\phi = { \cap_{i=1}^n { J_i} }\).


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 💡