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I have a question, and this is it in entirety:

Let R be a non-zero ring(that is, R contains at least one element other than the zero element). Prove that $R \times R$ is NOT an integral domain.

Now I haven't done a problem like this before, and it seems to be(from the Wikipedia definition) that this ring $R$ needs to be commutative to be an integral domain. Which it isn't stated to be here. What exactly am I meant to do? I haven't been given anything else than the area in grey.

Exam in 5 days
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  • Commutativity doesn't matter. Even if it is commutative, it won't be an integral domain, because $(x,0)\times (0,y) = (0,0)$ – Prahlad Vaidyanathan Jun 23 '14 at 07:21
  • @PrahladVaidyanathan Oh my, wow, I need to take a break, I have asked two trivial questions in less than thirty minutes... Thank you, that solves it... – Exam in 5 days Jun 23 '14 at 07:23

1 Answers1

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Hint:

Elements of $R \times R$ will look like $(x, y)$ such that $x, y \in R$. Further, multiplication is defined as follows:

$$(a, b)(x, y) = (a \cdot x, b \cdot y)$$

Where $a \cdot x$ and $b \cdot y$ are computed in the original ring $R$. Addition is defined similarly.

Next, the additive identity (traditionally called "zero") in $R \times R$ will be the element $(0, 0)$. This is easy to see since $(x, y) + (0, 0) = (x+0, y+0) = (x, y)$.

Finally, to show that $R \times R$ is not an integral domain for any ring $R$, it is your task to think of at least two nonzero elements $(a, b)$ and $(c, d)$ in $R \times R$ such that $(a, b)(c, d) = (0, 0)$.

Some clarification: It is true that all integral domains are commutative. From this, if $R$ is not commutative, then $R \times R$ is not commutative (why?) and won't be an integral domain. However, even if $R$ is a commutative ring, $R \times R$ is never an integral domain. This follows from the work above, since the direct product will always have zero divisors.

Kaj Hansen
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