The direct product and its properties
Definition (The product of two rings)
Let \(R_{1}, R_{2}\) be rings, with structures \((R_{1}, +_{R_{1}}, \cdot_{R_{1}}, 0_{R_{1}}, 1_{R_{1}})\) and \((R_{2}, +_{R_{2}}, \cdot_{R_{2}}, 0_{R_{2}}, 1_{R_{2}})\). Define
\[R_{1} \times R_{2} := \{ (x_{1}, x_{2}) :\ x_{1} \in R_{1},\ x_{2} \in R_{2} \},\]with the componentwise operations:
\[(x_{1}, x_{2}) + (y_{1}, y_{2}) = (x_{1} +_{R_{1}} y_{1},\ x_{2} +_{R_{2}} y_{2}), \qquad (x_{1}, x_{2}) \cdot (y_{1}, y_{2}) = (x_{1} y_{1},\ x_{2} y_{2}).\]The zero in \(R_{1} \times R_{2}\) is \(\vec{0} = (0_{R_{1}}, 0_{R_{2}})\) and the one is \(\vec{1} = (1_{R_{1}}, 1_{R_{2}})\). Then \((R_{1} \times R_{2}, +, \cdot, \vec{0}, \vec{1})\) is a ring.
Definition (The direct product of a family of rings)
In general, if \(\{R_{i}\}_{i \in I}\) is a set of rings, we can construct the ring
\[\prod_{i \in I} R_{i}\]in an analogous way, with the operations defined componentwise.
Example (\(\mathbb{R}^{2}\) has zero divisors)
\(\mathbb{R}^{2} = \mathbb{R} \times \mathbb{R}\) is a ring, and
\[\underbrace{(1, 0)}_{\neq (0,0)} \cdot \underbrace{(0, 1)}_{\neq (0,0)} = (0, 0).\]So, even though \(\mathbb{R}\) has no zero divisors, \(\mathbb{R}^{2}\) does.
Note (The units of the product)
Let \(\{R_{i} :\ i \in I\}\) be a set of rings and \(R = \prod_{i \in I} R_{i}\). Considering the units \(R^{\times}\), we have that
\[R^{\times} = \prod_{i \in I} R_{i}^{\times}.\]Theorem (Subrings and ideals of a product)
Let \(\{R_{i} :\ i \in I\}\) be a set of rings and \(R = \prod_{i \in I} R_{i}\). Then:
- if for each \(i \in I\), \(S_{i}\) is a subring of \(R_{i}\), then \(\prod_{i \in I} S_{i}\) is a subring of \(R\);
- if for each \(i \in I\), \(A_{i} \subseteq R_{i}\) is an \(R_{i}\)-ideal, then \(\prod_{i \in I} A_{i}\) is an \(R\)-ideal;
- if \(K\) is an ideal of \(R \times S\), then there exist \(I\), an \(R\)-ideal, and \(J\), an \(S\)-ideal, such that \(K = I \times J\).
Proof: For (1): we use the subring criterion. Since \(1_{R_{i}} \in S_{i}\) for each \(i\), we have that \(\vec{1} = (1_{R_{i}})_{i} \in \prod_{i} S_{i}\); and if \(a = (a_{i})_{i}\) and \(b = (b_{i})_{i}\) lie in \(\prod_{i} S_{i}\), then \(a - b = (a_{i} - b_{i})_{i}\) and \(a b = (a_{i} b_{i})_{i}\) have all their components in the \(S_{i}\), since each \(S_{i}\) is a subring.
For (2): \(\prod_{i} A_{i} \neq \emptyset\) since it contains \(\vec{0}\); it is closed under componentwise sums; and if \(r = (r_{i})_{i} \in R\) and \(x = (x_{i})_{i} \in \prod_{i} A_{i}\), then \(r x = (r_{i} x_{i})_{i}\) has each component in \(A_{i}\) by the absorption of each \(A_{i}\).
For (3): let \(K\) be an ideal of \(R \times S\). If \((r, s) \in K\), absorption with \((1, 0)\) and \((0, 1)\) gives
\[(r, 0) = (1, 0) \cdot (r, s) \in K, \qquad (0, s) = (0, 1) \cdot (r, s) \in K.\]Define
\[I = \{ r \in R :\ (r, 0) \in K \}, \qquad J = \{ s \in S :\ (0, s) \in K \}.\]Then \(I\) is an \(R\)-ideal: \(0 \in I\); if \(r, r' \in I\), then \((r + r', 0) = (r, 0) + (r', 0) \in K\); and if \(a \in R\), then \((a r, 0) = (a, 0) \cdot (r, 0) \in K\) by the absorption of \(K\). Analogously \(J\) is an \(S\)-ideal. Let us see that \(K = I \times J\). “\(\subseteq\)”: if \((r, s) \in K\), by the above \((r, 0), (0, s) \in K\), that is, \(r \in I\) and \(s \in J\). “\(\supseteq\)”: if \(r \in I\) and \(s \in J\), then
\[(r, s) = (r, 0) + (0, s) \in K,\]by the closure of \(K\) under sums.