Defining $R\times R$ as a ring? 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?

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 $\mathbb R[x]/(x^2)$.
