Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

3 Answers 3

up vote 3 down vote accepted

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

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

share|improve this answer

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.

share|improve this answer

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 $R$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.