Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

That's the question from my homework. I am thinking $R\times S$ is not a field, but I'm not sure. I understand the definition of a field, but I am not sure how to proceed.

share|cite|improve this question
Can you find a multiplicative inverse to $(1,0)$? – Jyrki Lahtonen Mar 21 '13 at 15:36
up vote 18 down vote accepted

Hint: Every non-zero element in a field has a multiplicative inverse. Can you find a non-zero element in the product of two fields that has no inverse?

share|cite|improve this answer

$R \times S$ is never a domain, even if $R$ and $S$ are domains because $(1,0)\cdot(0,1)=(0,0)$ shows that $R \times S$ has zero divisors.

share|cite|improve this answer

Hint: Use the characterization of a field in terms of its ideals. What are the ideals of $R \times S$?

share|cite|improve this answer

If $R^*$ is the set of units of a ring, then $(R \times S)^* = R^* \times S^*$. A commutative ring $R$ with unit is a field if and only if $R^* = R - \{0\}$. If $R \times S$ is a field, we would have the following equality. $$(R \times S)^* = R^* \times S^* = (R \times S) - \{(0,0)\}$$

Is this possible?

share|cite|improve this answer

For any $x\in (F,+,\cdot)$ we require that if $x\neq 0_F$ then there exists an element $x^{-1}$ so that $x\cdot x^{-1}=x^{-1}\cdot x=1_F$. Consider $(r,0_S)\in R\times S$. Supposing that there is an element $(r,0_S)^{-1}=(a,b)\in R\times S$ then we would have $(r,0_S)\cdot(a,b)=(1_R,1_S)=(1,1)_{R\times S}$ (it is easy to check the last equality). This would force $0_S\cdot b=1_S$. This is a contradiction since $S$ is a field and we must have $s\cdot 0_S=0_S$ for all $s\in S$. So $(r,0)$ has no inverse and $R\times S$ cannot be a field.

share|cite|improve this answer

Hint $\, $ Factorizations $\rm\: R\times S\:$ have nontrivial idempotents [e.g. $\rm\,e = (0,1)\,$], i.e $\rm\:e(e\!-\!1)=0,\ e\neq 0,1,\:$ so $\rm\:e\:$ is a zero-divisor $\ne 0.$ Conversely, a nontrivial idemptotent yields a factorization (see the Peirce decomposition).

Here's a concrete example of this intimate correspondence between idempotents and factorizations. In $\rm\:\Bbb Z/n = \:$ integers mod $\rm\,n,\,$ idempotents correspond to factorizations of $\rm\:n\:$ into coprime factors. Namely, if $\rm\:e^2 = e\in\Bbb Z/n\:$ then $\rm\:n\:|\:e(e\!-\!1)\:$ so $\rm\:n = jk,\,\ j\:|\:e,\,\ k\:|\:e\!-\!1,\:$ so $\rm\:(j,k)= 1\:$ by $\rm\:(e,e\!-\!1) = 1.\:$ Conversely if $\rm\:n = jk\:$ for $\rm\:(j,k)= 1,\:$ then CRT $\rm\,\Rightarrow\, \Bbb Z/n\cong \Bbb Z/j\times \Bbb Z/k,\:$ which has nontrivial idempotents $\rm\:(0,1),\,(1,0).\:$ It is easy to explicitly work out the details of the correspondence. Some integer factorization algorithms can be viewed as searching for nontrivial idempotents.

share|cite|improve this answer

I assume you define addition on $R\times S$ by $(r_1,s_1)+(r_2,s_2):=(r_1+r_2,s_1+s_2)$ and multiplication by $(r_1,s_1)\cdot(r_2,s_2):=(r_1r_2,s_1s_2)$, then the multiplicative unit is $(1,1)$ and
$$ (1,0)\cdot(0,1)=(0,0). $$ So none of these nonzero elements is invertible.

But if, for instance, $R=S$ are commutative and if $r^2+s^2$ is invertible for every $(r,s)\neq (0,0)$, then the product $$(r_1,s_1)\cdot(r_2,s_2):=(r_1r_2-s_1s_2, r_1s_2+r_2s_1)$$ puts a field structure on $R\times S$ where the inverse is given by $$ (r,s)^{-1}=(r(r^2+s^2)^{-1},-s(r^2+s^2)^{-1}). $$ E.g. $R=S=\mathbb{R}$, this is the $\mathbb{C}$ field structure on $\mathbb{R}^2$.

share|cite|improve this answer

Let $F_1$ and $F_2$ be two finite fields. Then $F_1 \times F_2$ has cardinality $|F_1| \cdot |F_2|$. But a finite field has cardinality a prime power, so it is sufficient to suppose that $F_1$ and $F_2$ have different characteristics to conclude that $F_1 \times F_2$ is not a field.

share|cite|improve this answer

If you dont consider the product field should have the operation inherited from the original fields it can be a little bit harder question, but you can give counterexamples even in this case.

Consider $\mathbb{R} \times \mathbb{R}^2 \cong \mathbb{R}^3$, you can show $\mathbb{R}^3$ can't have a field structure (not easy), an easier example is to consider finite fields of different characteristic $\mathbb{F}_p$ and $\mathbb{F}_q$ and ask what is the characteristic of the supposed field $\mathbb{F}_p \times \mathbb{F}_q$ and get a contradiction.

share|cite|improve this answer
If you are allowed to redefine the operations of the (now only set-theoretic) product, then $\mathbb R^3$ can be a field (since $\left|\mathbb R\right| = \left|\mathbb R^3\right|$). – azimut Apr 19 '13 at 5:19

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.