# Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field [duplicate]

Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field and that why any external direct sum of two fields cannot be a field. I believe it has much to do with the lack of every non-zero element having an inverse, however I am having difficulty seeing it.

-

## marked as duplicate by rschwiebDec 4 '14 at 18:08

$(1, 0)\cdot(0,1) = ?$ –  Lierre Nov 5 '12 at 17:38

Because $(x,0)$ and $(0,x)$ lack multiplicative inverses even when $x\ne0$.

(This is true of fields in general; not only of $\mathbb{R}$.)

-

As Lierre points out, there will always be zero divisors in such a ring, even if both summands (or, factors, depending on your viewpoint) are fields. Such is life.

-

Use the fact that for two rings $R$ and $S$ we have $(R \times S)^* = R^* \times S^*$, where $R^*$ means the set of units of $R$.

-