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.

Solution 1:

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 $\mathbb C$: $$ (a+ib)\cdot (c+id)= (ac-bd) + i(bc+ad) $$

Solution 2:

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

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

Solution 3:

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.