Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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.
share|improve this answer
add comment

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 $\rm\:R\:$

share|improve this answer
add comment

Your Answer

 
discard

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.