# If $R$ and $S$ are fields, either prove or disprove that $R\times S$ is a field

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.

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Can you find a multiplicative inverse to $(1,0)$? –  Jyrki Lahtonen Mar 21 '13 at 15:36

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?

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$R \times S$ 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.

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

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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?

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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.

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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.

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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 $\mathbb{R}^2$.

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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.

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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.

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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