1
$\begingroup$

Why is this a ring homomorphism?

$$\phi:R\to R/I$$ where $I$ is an ideal, given by $\phi:r\mapsto r+I$.


To be a ring homomorphism it needs to be a homomorphism of addition and multiplication, i.e:

1) $\phi(ab)=\phi(a)\phi(b)$

2) $\phi(a+b)=\phi(a)+\phi(b)$

where $a,b\in R$


Here we have:

1) $\phi(ab)=ab+I$ and $\phi(a)\phi(b)=ab+aI+bI+I^2=ab+I^2$ and I imagine then that $I^2=I$?

2) $\phi(a+b)=a+b+I$ and $\phi(a)+\phi(b)=a+b+2I=a+b+I$.

So this holds I suppose if $I^2=I$, does this just come because $aI=I$ and let $a=I$ and then $II=I$?

$\endgroup$
  • $\begingroup$ Notice that the quotient ring operations are defined as: $(a+I) + (b+I) = (a+b) + I$ and $(a+I)\cdot (b+I) = (ab) + I$. $\endgroup$ – jiyanez Jun 21 '15 at 5:01
1
$\begingroup$

For Qn 1. The elements of $aI, bI, I^2$ are all in $I$, from the definition of ideal of a ring. So every element of the product formed by picking one each from $(a+I)$ and $(b+I)$ lies in the coset $ab+I$. Now you can see how things fall into place.

$\endgroup$
  • $\begingroup$ Oh yes thank you. I wasn't thinking right. The ideal $I$ generates everything via addition and multiplication puhhh. $\endgroup$ – homomorphism Jun 21 '15 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.