Let \(R\) be a ring with identity. Suppose \(I_1,\ldots,I_n\) are left ideals such that, as additive groups, \[ R=I_1\oplus I_2\oplus\cdots\oplus I_n. \] Prove that there are elements \(u_i\in I_i\) such that for every \(a_i\in I_i\), \[ a_i u_i=a_i, \qquad a_i u_j=0\quad(j\ne i). \]
Solution
1. There are unique elements \(u_i\in I_i\) such that\[ 1=u_1+\cdots+u_n. \]
The hypothesis \[ R=I_1\oplus\cdots\oplus I_n \] means that every element of the additive group of \(R\) has a unique expression as a sum of elements from the respective \(I_i\). Apply this to the identity element \(1\in R\).
2. If \(a_i\in I_i\), then\[ a_i=a_i u_1+\cdots+a_i u_n, \]and each \(a_i u_j\) belongs to \(I_j\).
Multiplying the equality in step 1 on the left by \(a_i\) gives \[ a_i=a_i1=a_i u_1+\cdots+a_i u_n. \] Since \(u_j\in I_j\) and \(I_j\) is a left ideal, multiplication on the left by \(a_i\in R\) gives \(a_i u_j\in I_j\).
3. For every \(a_i\in I_i\),\[ a_i u_i=a_i, \qquad a_i u_j=0\quad(j\neq i). \]
The element \(a_i\) already has the direct-sum decomposition \[ a_i=0+\cdots+0+a_i+0+\cdots+0, \] with \(a_i\) in the \(i\)th summand. Step 2 gives another decomposition of \(a_i\), whose \(j\)th component is \(a_i u_j\in I_j\). Uniqueness of the direct-sum decomposition therefore gives \(a_i u_i=a_i\) and \(a_i u_j=0\) for \(j\neq i\).
4. Q.E.D.
The elements \(u_i\) constructed in step 1 satisfy the required identities by step 3.