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I have to prove that there are exactly 2 congruences over $\mathbb {R}$ seen as a model/structure $\tau = (\varnothing, {+,*}, \varnothing, \operatorname{arity}(+) = \operatorname{arity}(*)=2)$ where $+$ and $*$ are the usual operations.

I know that these two congruences are the trivials, namely $\Theta = \{(a,b) \mid a,b\in \mathbb {R}\}$ and $\Theta = \{(a,a) \mid a\in \mathbb {R}\}$.

But I don't know how to prove this are the only congruences, i.e., the only equivalence relationships that satisfy: $(a1,b1)\in \Theta \bigwedge (a2,b2)\in \Theta \Rightarrow (a1+a2, b1+b2)\in \Theta \bigwedge (a1*a2, b1*b2)\in \Theta $

I've thought of trying to prove by reductio ad absurdum but when I tried to start the proof couldn't manage to figure out anything.

Any helps\hints are welcome.

Srivatsan
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leandrodemarco
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2 Answers2

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Let $\Theta$ be a congruence on $\mathbb{R}$, and suppose that $(a,b)\in\Theta$ with $a\ne b$. You know that $(-a,-a)\in\Theta$, so $(0,b-a)\in\Theta$, where $b-a\ne 0$. Now see if you can carry out the following steps:

  1. Use a similar trick to show that $(0,1)\in\Theta$.
  2. Then show that $(0,r)\in\Theta$ for all $r\in\mathbb{R}$.
  3. Use the first trick to show that $(x,y)\in\Theta$ for all $x,y\in\mathbb{R}$.
  4. Conclude that if $\Theta$ isn’t equality, it must be the trivial congruence.
Brian M. Scott
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HINT $\ $ Congruences biject with ideals. But $\mathbb R\:$ is a field so it has only two ideals, $(0)$ and $(1)$.

See also my post here which discusses the following more general result

THEOREM $\ $ The following are equivalent for a ring $\rm\:R\:$ and set $\rm\ S\subset R\times R$

$\rm(1)\quad S\ $ is a congruence on $\rm\:R\:$

$\rm(2)\quad S\ $ is a subalgebra of $\rm\:R\times R\:$ and $\rm\:S \supset (1,1)\: R$

$\rm(3)\quad I\: :=\: \{\: r\in R\: :\ (r,0)\in S \:\}\ $ is an ideal in $\rm\:R\:$

Community
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Bill Dubuque
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