Multiplicative inverses of $\Bbb Z \times \Bbb Z$

Question

This is more of a clarification.. I know the only multiplicative inverses of $\Bbb Z$ are $\{-1, 1\}$. I want to say that by the same principle, the only multiplicative inverses of $\Bbb Z \times \Bbb Z$ are $\{(-1, -1), (1, 1)\}$.

I can't seem to convince myself that $\{(-1, 1), (1, -1)\}$ should not be included here.. should they be?

TIA

Yes, they should be. If $R$ and $S$ are rings with identity and $R^$ denotes the group of units, then $(R \times S)^ = R^* \times S^*$. This is a good exercise, though not hard. And, it answers your question. — Randall, Jul 06 '18 at 01:11
@randall of course, because right as joints preserve limits! :) — Andres Mejia, Jul 06 '18 at 01:21

score 0 · Answer 1 · answered Jul 06 '18 at 01:25

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Note that they are actually their own inverses: $(-1,1)\cdot (-1,1)=(1,1)$ which is the identity.

answered Jul 06 '18 at 01:25

Andres Mejia

20,977

1

And presumably OP really was looking for units. – Lubin Jul 06 '18 at 01:37
@Lubin yes, that’s the way I understood it. – Andres Mejia Jul 06 '18 at 01:38

score 0 · Answer 2 · answered Jul 06 '18 at 16:05

The unit group of a direct product is the direct product go its unit groups, see here:

Group of units of direct sum of rings is isomorphic to direct sum of the groups of units

So $$U(\Bbb{Z}\times \Bbb{Z})=U(\Bbb{Z})\times U(\Bbb{Z})=\{(\pm 1,\pm 1)\}.$$

Multiplicative inverses of $\Bbb Z \times \Bbb Z$

2 Answers2