The Direct Product of Two Rings

# The Direct Product of Two Rings

Definition: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ both be rings. Define $T = R \times S = \{ (a, b) : a \in R, b \in S \}$. The Direct Product is the set $T$ paired with the operation of addition $+$ defined for each $(a, b), (c, d) \in T$ as $(a, b) + (c, d) = (a +_1 c, b +_2 d)$ and the operation of multiplication $*$ defined as $(a, b) * (c, d) = (a *_1 c, b *_2 d)$. |

For example, consider the ring of real number $(\mathbb{R}, +_1, *_1)$ where $+_1$ and $*_1$ denote standard addition and standard multiplication. Also consider the ring of $2 \times 2$ matrices with real coefficients $(M_{22}, +_2, *_2)$ where $+_2$ and $*_2$ denote standard matrix addition and standard matrix multiplication. Then the direct product between these two rings is the following Cartesian product:

(1)\begin{align} \quad T = \mathbb{R} \times M_{22} = \{ (a, B) : a \in \mathbb{R}, B \in M_{22} \} \end{align}

Alongside addition $+$ defined for all $(a, B), (c, D) \in T$ by:

(2)\begin{align} \quad (a, B) + (c, D) = (a +_1 c, B + D) = \left (a +_1 c, \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} +_2 \begin{bmatrix} d_{11} & d_{12}\\ d_{21} & d_{22} \end{bmatrix} \right ) \end{align}

And multiplication $*$ defined for all $(a, B), (c, D) \in T$ by:

(3)\begin{align} \quad (a, B) * (c, D) = (a *_1 c, B *_2 D) = \left ( a *_1 c, \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} *_2 \begin{bmatrix} d_{11} & d_{12}\\ d_{21} & d_{22} \end{bmatrix} \right ) \end{align}

In the following theorem we will show that the direct product of rings forms a ring itself.

Theorem 1: Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be rings. Then the direct product $T = R \times S$ is a ring. |

**Proof:**We will show that all of the ring axioms hold for the direct product of any two rings. Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ both be rings and let $T = R \times S$. Define $+$ and $*$ on $T$ as we have above, and let $(a, b), (c, d), (e, f) \in T$.

- Clearly $T$ is closed under $+$ since $(a +_1 c) \in R$ and $(b +_2 d) \in S$ so:

\begin{align} \quad (a, b) + (c, d) = (a +_1 c, b +_2 d) \in T \end{align}

- We also have $+$ is associative due to the associativity of $+_1$ on $R$ and $+_2$ on $S$:

\begin{align} \quad (a, b) + [(c, d) + (e, f)] = (a, b) + (c +_1 e, d +_2, f) = (a +_1 [c +_1 e], b +_2 [d +_2 f]) \\ = ([a +_1 c] +_1 e, [b +_2 d] +_2 f) = (a +_1 c, b +_2 d) + (e, f) = [(a, b) + (c, d)] + (e, f) \end{align}

- The identity on $+$ is $(0_1, 0_2)$ where $0_1$ is the identity of $+_1$ on $R$ and $0_2$ is the identity of $+_2$ on $S$:

\begin{align} \quad (a, b) + (0_1, 0_2) = (a +_1 0_1, b +_2 0_2) = (a, b) \end{align}

(7)
\begin{align} \quad (0_1, 0_2) + (a, b) = (0_1 +_1 a, 0_2 +_2 b) = (a, b) \end{align}

- For each $(a, b) \in T$ the inverse under $T$ is $(-a, -b)$ where $-a \in R$ is the inverse of $a \in R$ under $+_1$ and $-b \in S$ is the inverse of $b \in S$ under $+_2$:

\begin{align} \quad (a, b) + (-a, -b) = (a +_1 (-a), b +_2 (-b)) = (0_1, 0_2) \end{align}

(9)
\begin{align} \quad (-a, -b) + (a, b) = ((-a) +_1 a, (-b) +_2 b) = (0_1, 0_2) \end{align}

- The operation $+$ is commutative since:

\begin{align} \quad (a, b) + (c, d) = (a +_1 c, b +_2 d) = (c +_1 a, d +_2 b) = (c, d) + (a, b) \end{align}

- $T$ is closed under $*$ since $(a *_1 c) \in R$ and $(b *_2 d) \in S$ so:

\begin{align} \quad (a, b) * (c, d) = (a *_1 c, b *_2 d) \in T \end{align}

- $*$ is also associative due to the associativity of $*_1$ on $R$ and $*_2$ on $S$:

\begin{align} \quad (a, b) * [(c, d) * (e, f)] = (a, b) * (c *_1 e, d *_2, f) = (a *_1 [c *_1 e], b *_2 [d *_2 f]) \\ = ([a *_1 c] *_1 e, [b *_2 d] *_2 f) = (a *_1 c, b *_2 d) + (e, f) = [(a, b) * (c, d)] * (e, f) \end{align}

- The identity of $*$ on $T$ is $(1_1, 1_2)$ where $1_1$ is the identity of $*_1$ on $R$ and $1_2$ is the identity of $*_2$ on $S$:

\begin{align} \quad (a, b) * (1_1, 1_2) = (a *_1 1_1, b *_2 1_2) = (a, b) \end{align}

(14)
\begin{align} \quad (1_1, 1_2) * (a, b) = (1_1 *_1 a, 1_2 *_2 b) = (a, b) \end{align}

- Left distributivity also holds since:

\begin{align} \quad (a, b) * [(c, d) + (e, f)] = (a, b) * (c +_1 e, d +_2 f) = ([a *_1 c] +_1 [a *_1 e],[b *_2 d] +_2 [b *_2 f]) = (a, b) * (c, d) + (a, b) * (e, f) \end{align}

- And right distributivity as well:

\begin{align} \quad [(a, b) + (c, d)] * (e, f) = (a +_1 c, b +_2 d) * (e, f) = ([a *_1 e] +_1 [c *_1 e], [ b *_2 f] +_2 [d *_2 f]) = (a, b) * (e, f) + (c, d) * (e, f) \end{align}

- Therefore all of the ring axioms are satisfied, so the direct product $T = R \times S$ is a ring. $\blacksquare$