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Let $R = R_{1} \oplus R_{2} \oplus \cdots \oplus R_{n}$ be a direct sum of rings $r_{i}$. I need to determine conditions on the direct summands $R_{i}$ in order for the ring $R$ to have the identity and conditions on the $R_{i}$s in order for the ring $R$ to be commutative.

Since elements of $R$ are $n$-tuples $(x_{1},x_{2},\cdots,x_{n})$ where each $x_{i} \in R_{i}$ for each $i$, consider two $n$-tuples $x = (x_{1},x_{2},\cdots, x_{n})$ and $y = (y_{1},y_{2},\cdots, y_{n}) \in R$, where $x_{i},y_{i}\in R_{i}$. Then, multiplication in $R$ is defined componentwise, so $xy = (x_{1}y_{1},x_{2}y_{2},\cdots , x_{n}y_{n})$.

Now, the only way for $xy = x$ is for $y = 1_{R}$, the multiplicative identity in $R$. Likewise, the only way for $xy=y$ is for $x = 1_{R}$, the multiplicative identity in $R$. So, let's say that $xy = x$, Then, $(x_{1}y_{1},x_{2}y_{2},\cdots , x_{n}y_{n}) = x$, which implies that $x_{i}y_{i} = x_{i}$ $\forall i$. Therefore, $y_{i} = 1_{R_{i}}$ $\forall i$.

My hypothesis, therefore, is that $R$ has the multiplicative identity $1_{R}$ if and only if each $R_{i}$ contains $1_{R_{i}}$. Is this hypothesis correct? I believe that showing the opposite direction, that each $R_{i}$ possesses the identity implies that $R$ posseses the identity is much, much easier to show.

Moreover, I surmise that $R$ is commutative if and only if each $R_{i}$ is commutative.

This is so, since, suppose that each $R_{i}$ is commutative, then $xy = (x_{1},x_{2},\cdots , x_{n})(y_{1},y_{2},\cdots , y_{n}) = (x_{1}y_{1}, x_{2}y_{2}, \cdots , x_{n}y_{n}) (y_{1}x_{1}, y_{2}x_{2}, \cdots , y_{n}x_{n}) = (y_{1},y_{2}, \cdots , y_{n})(x_{1},x_{2}, \cdots , x_{n}) = yx$.

In the other direction, suppose that $R$ is commutative. Then $xy=yx$, and so $(x_{1},x_{2},\cdots , x_{n})(y_{1},y_{2},\cdots, y_{n}) = (y_{1},y_{2},\cdots, y_{n})(x_{1},x_{2},\cdots , x_{n})$ Multiplication is defined componentwise, so this becomes $(x_{1}y_{1}, x_{2}y_{2}, \cdots, x_{n}y_{n}) = (y_{1}x_{1}, y_{2}x_{2}, \cdots y_{n}x_{n})$. And, by equality of tuples, we must have that $x_{i}y_{i} = y_{i}x_{i}$ $\forall i$, which means that each $R_{i}$ is commutative.

I wanted to check that I and my line of reasoning were correct.

Thank you

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    Commutative proof is good. For the identity proof, you really need to show that if one of the $R_{i}$ does not have a multiplicative identity, then the product can't. – 2017-02-27
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    Scratch that: if $R$ has an identity element, then you should be able to show that the $ith$ coordinate of that identity element is the identity element of $R_{i}$. – 2017-02-27
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    @User0112358 in which direction? In the direction where $R$ has identity $\implies$ $R_{i}$ has identity? Or the direction where each $R_{i}$ has identity $\implies$ $R$ has identity? – 2017-02-27
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    I think you have done the second. You have to do the first. If $R$ has an identity, then denote it $e:=(e_{1}, \dots , e_{i}, \dots e_{n})$. Can you show $e_{i}$ is the identity for $R_{i}$? – 2017-02-27
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    @User0112358 I thought I had done the first, and needed to do the second... – 2017-02-27
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    Of course! Sorry. Well, the other direction is easy! Right? If each ring has an identity, then the tuple of all of them will be the identity. – 2017-02-27

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