Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field 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.
 A: We have to define a product in $k\times k$. If we define the natural product:
$$
(a,b) \cdot (c,d) = (ac, bd)
$$
then $k\times k$ is NOT a field because it has non trivial divisors of zero:
$$
(1,0)\cdot (0,1) = (0,0)
$$

But sometimes is possible to define a different product on $k\times k$ and this set become a field. For example let's consider $\mathbb R \times \mathbb R$ with the product:
$$
(a,b)\cdot (c,d) = (ac-bd,bc+ad)
$$
This is a well defined product and with this structure $\mathbb R\times \mathbb R$ is a field: infact is isomorphic to the field $\mathbb C$:
$$
(a+ib)\cdot (c+id)= (ac-bd) + i(bc+ad)
$$
A: Because $(x,0)$ and $(0,x)$ lack multiplicative inverses even when $x\ne0$.
(This is true of fields in general; not only of $\mathbb{R}$.)
A: 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.
A: 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 $R$.
A: No.  It isn't an integral domain.  Viz: $(1,0)\cdot(0,1)=(0,0)$.
