Ideal of External Direct Sum of Rings

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Theorem

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.

Let

$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$

be their direct product.


For each $k \in \closedint 1 n$, let:

${R_k}' = \set {\tuple {x_1, \ldots, x_n} \in R: \forall j \ne k: x_j = 0}$


Then:

$\forall k \in \closedint 1 n: {R_k}'$ is an ideal of $R$.


Proof

Let $y = \tuple {y_1, y_2, \ldots, y_n} \in R$.

Let $x = \tuple {x_1, x_2, \ldots, x_n} \in R_k$

By definition of direct product, we have:

$x \circ y = \tuple {x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}$
$y \circ x = \tuple {y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}$

But we have:

$\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$

because $x_j = 0$.

So it follows that $x \circ y \in R_k$ and $y \circ x \in R_k$.

Hence ${R_k}'$ is an ideal of $R$.

$\blacksquare$


Sources

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