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 want to show that given a ring homomorphism $\phi:R\to S,$ if $\phi$ is injective, then $\ker(\phi)=\{0\}.$

Given $\phi(r)=\phi(r'),$ by the definition of an injection, we have $r=r'.$ There is a fact that $\phi(r)=\phi(r')$ if and only if $r-r'\in\ker(\phi).$ Using this fact, given $r-r'\in \ker(\phi),$ we have $r=r'$ i.e. $r-r'=0.$

My question is, given any $a\in \ker(\phi),$ do there always exist $r,r'\in R$ such that $a=r-r'$? If not, then my proof will not work. Further what does the structure of $\ker (\phi)$ look like?

Thanks!

share|improve this question
4  
Sure, $r=a$, $r'=0$. –  André Nicolas Apr 10 '12 at 15:22
    
The Ker($\phi$) is an ideal. That is to say it is an abelian group and $ri\in Ker(\phi),ir\in Ker(\phi)$ for any $r\in R, i\in Ker(\phi)$. –  wxu Apr 10 '12 at 15:27
    
You may find of interest my post here on the relationship between congruences, ideals, and subalgebras of the square. –  Bill Dubuque Apr 10 '12 at 16:12

2 Answers 2

up vote 3 down vote accepted

The answer to your specific question is certainly, just take $r=a$ and $r'=0$.

Your answer is correct, but there is a simpler version of it that would be I think clearer to you. Suppose that $a$ is in the kernel of $\phi$. We want to show that $a=0$.

Since $a\in\ker(\phi)$, we have $\phi(a)=0$. But (always) $\phi(0)=0$. Thus, by injectivity, $a=0$.

share|improve this answer

What you have written above is by and large correct, however this is a general piece of advice, I think it would be very instructive for you to really try to write out every detail of the definitions that you use above. The reason why this is instructive is that the proposition that you mention above is nothing more really than playing around with the involved definitions.

For example ask yourself if you can:

  • State a definition of a function
  • State a definition of an injective function
  • State a definition of the kernel of a function
  • State a definition of a homeomorphism
  • State a definition of a ring

It is important that you understand every single word of the definitions. Make sure that you understand all quantifiers appearing in the definitions. If anything is unclear and does not "click", try to identify the source. This will save you a lot of time later. Also remember that the definitions are often useless without good examples of objects satisfying the definitions. After all we are trying to understand a complicated reality out there. Non examples are also good, i.e. examples that "almost" satisfy the definition, in the sense that everything but a single hypothesis in the definition is satisfied.

Lastly, try to have fun. Play around with the definitions and ask yourself a lot of dumb questions!

share|improve this answer
    
+1, well stated! –  wxu Apr 10 '12 at 16:28

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.