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 $II=I$?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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