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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.

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marked as duplicate by rschwieb abstract-algebra Dec 4 '14 at 18:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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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.

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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 $R$.

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