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I feel a bit stupid, but I know that the normal definition of $R\times R$ as $R \times S = \{(r, s) : r \in R, s \in S\}$, under $(r, s) + (r', s')=(r+r',s+s')$ and $(r, s) \cdot (r', s')=(rr', ss')$ is a ring.

But, can you define $R \times R$ otherwise as a ring?

I'm trying to decide whether $R \times R$ has any non-zero nilpotent elements. Obviously it does not under the normal definition, but can you define $R \times R$ as a ring otherwise such that there are non-zero nilpotent elements?

3 Answers 3

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Hint: You can define multiplication in $R \times R$ by thinking about the complex numbers.

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Presuming that R refers to real numbers: "but can you define $\mathbb R \times \mathbb R$ as a ring otherwise such that there are non-zero nilpotent elements?"

Sure. Let $(r, s) \cdot (r', s') = (rr', r's+rs')$, for which $(0, 1)$ is a nilpotent.

For those comfortable enough with quotients of polynomial rings, this is the formula for multiplication in $\mathbb R[x]/(x^2)$.

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    R does indeed refer to the real numbers, sorry for my poor formatting. Thank you, that was rather helpful. I was unable to thinking of any multiplication definition that was associative and commutative, but you have helpfully provided one! – 2017-02-11
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    I don't even see why it matters we specify the real numbers. As far as I know all of this is ture for an arbitrary ring $R$. – 2017-02-11
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    @rschwieb You're right, it doesn't matter. – 2017-02-12
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Yes, you can, but it is not necessarily useful. There is a very precise meaning to the symbol $\times$: it is the product in the category you are considering. It is, in some sense, the smaller object that contains the whole structure of both the object you are taking the product of (more precisely, it satisfies a certain universal property that you can find here). If you define the ring structure in another way, you lose this property.

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    I don't see why it wouldn't be useful in my case? If I can define R x R (R here meaning real numbers, not sure how to do the actual symbol) as a ring as Dustan suggests, then R x R does indeed have non-zero nilpotent elements; if I can't, then it does not. I'm just not sure if I can do this! – 2017-02-11
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    @Atlas the symbol is given by \mathbb{R} . Sure, you can define many different ring structures on the *set* $\mathbb{R}^2$, and some of them are interesting. But my point still stands: the product in the category of rings is defined that way for a very precise reason. – 2017-02-11