# If R and S are rings then $R \times S$ is never a field

This question is from the book "Abstract Algebra" by Dummit and Foote.(Exercise: 7.6.4):

Problem: Prove that if $R$ and $S$ are any two non-zero rings then $R \times S$ is never a field.

My problem with this question is that it doesn't feel right. If $R$ and $S$ are rings then $R \times S$ is also so. (The addition and multiplication on $R \times S$ is defined as $(r_1,s_1)+(r_2,s_2)=(r_1+r_2,s_1+s_2)$ and $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$ respectively where $(r_1,s_1),(r_2,s_2) \in R \times S$.) So proving that $R \times S$ is a field reduces to proving commutativity of multiplication, existence of multiplicative indentity and existence of multiplicative inverse for each element of it. Now, if the two rings are fields themselves then multiplication is commutative, the identy is $(1_R,1_S)$ and inverse of an element $(r,s)$ is $(r^{-1},s^{-1})$. So $R \times S$ is also a field. Am I right? Please help.

Update: The question comes from a section about the Chinese Remainder Theorem. Is it possible to prove this fact using Chinese Remainder Theorem?

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In your example $(r,s)$ what happens if $r$ or $s$ is the zero element of $R$ or $S$? –  JSchlather Jun 29 '11 at 5:45
I can't think of a way to use the Chinese Remainder Theorem; and I cannot imagine any way in which you wouldn't, at some point, simply realize that you have zero divisors. –  Arturo Magidin Jun 29 '11 at 6:20
@Arturo: Yeah it's a stupid question. The zero just didn't kick my head at that point of time. But thanks anyway. –  Sayantan Jun 29 '11 at 6:29
Of course this doesn't work when you think rings need not have 1. Then for $R=\{0\}$ and $S$ a field... –  GEdgar Jun 29 '11 at 13:04
@GEdgar: huh? The problem states that the rings are nonzero. (And by the way, the zero ring has a $1$: it just happens to also be called $0$.) –  Pete L. Clark Jun 30 '11 at 9:42

Since $R$ is nonzero, there is an $r\in R$, $r\neq 0$. Since $S$ is nonzero, there exists $s\in S$, $s\neq 0$.

Since $r\neq 0$, then $(r,0_S)\in R\times S$ is not zero; and since $s\neq 0$, then $(0_R,s)\in R\times S$ is not zero.

What is the product of $(r,0_S)$ and $(0_R,s)$?

Can that happen if nonzero elements are cancellable?

Are nonzero elements cancellable in a division ring?

(This shows a slightly stronger statement: a direct product of nontrivial rings always has zero divisors.)

Alternatively. Prove that $R\times\{0\}$ is a proper ideal of $R\times S$ (here you use that $R\neq 0$ and $S\neq 0$). Then remember that fields are simple.

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Hint: What is the inverse of $(1,0)$?

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Let us prove that $R\times S$ is not a field if $R$ and $S$ are any two non-zero rings. The following steps lead to a solution:

(1) Note that $(1_R,0_S)\in R\times S$ where $1_R$ is the multiplicative identity of $R$ and $0_S$ is the additive identity of $S$.

(2) If $T$ is a ring, if $0_T$ is the additive identity of $T$, and if $t\in T$, prove that $t0_T=0_T=0_Tt$. (Hint: use the distributive law and write $0_T=0_T+0_T$.)

(3) Prove that $(1_R,1_S)$ is the multiplicative identity of $R\times S$ where $1_R$ and $1_S$ are the multiplicative identities of $R$ and $S$, respectively.

(4) If $(r,s)\in R\times S$, prove that $(r,s)\cdot (1_R,0_S)=(r,0)$.

(5) Finally, prove that there does not exist $(r,s)\in R\times S$ such that $(r,s)\cdot (1_R,0_S)=(1_R,1_S)$. Deduce that the element $(1_R,0_S)\in R\times S$ has no multiplicative inverse. (Hint: note carefully where you use the fact that $R$ and $S$ are non-zero rings.)

I hope this helps!

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HINT $\$ Factorizations correspond to idempotents. Nontrivial product rings contain idempotents $\rm\:e = e^2\:$ that are nontrivial $\rm\: e\ne 0,1\:,\:$ e.g. $\rm\ e = (0,1)\in R\times S\:.\:$ But a field or domain has only trivial idempotents since $\rm\ e\: (1-e) = 0\ \Rightarrow\ e = 0,1\:.\:$ Note that the nontrivial idempotent $(0,1)$ may be viewed as constructed by diagonalizing out of the set of trivial idempotents $(0,0),\: (1,1)\:.$

Conversely a nontrivial idempotent $\rm\:e\:$ yields a nontrivial factorization $\rm\:R \cong e\:R \times (1-e)\:R\:$ (the Peirce decomposition). This decomposition extends to any finite set of idempotents with sum $1$ and mutually orthogonal $\rm\:e_i\ e_j = 0\ \ if\ \ i\ne j\:.$ For example, the Chineses Remainder Theorem may be viewed in this manner.

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I think that should be 'Peirce' instead of 'Pierce': en.wikipedia.org/wiki/Benjamin_Peirce –  wildildildlife Jun 29 '11 at 13:44
Yes, curiously it's misspelled "Pierce" in many places, including textbooks. –  Bill Dubuque Jun 29 '11 at 13:59