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We have a factor ring R/I for a ring R and an ideal I. I understand R/I = {r + I , r e R} and that each element is a coset of I. I also understand that R/I is a ring under the defined operations +, x on R/I (which i wont write out here)

Now, what I dont understand is, for example the ring of integers and the ideal generated by 5 is why:

Z/<5> = Z 5 (integers modulo 5),

Why does this factor ring equal that? The factor ring contains all the cosets of the ideal generated by 5, i.e all the cosets of the set {5a , a e Z}. so why does the factor ring equal intgers modulo 5?? thanks

user141673
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3 Answers3

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This is a special case of a general correspondence between congruences and ideals in rings. Namely, given any congruence one easily checks that the elements $\equiv 0\,$ form an ideal. Conversely, given an ideal $\,I,\,$ one can define an equivalence relation by $\,a\equiv_I b\iff a-b\in I\,$ and the ideal properties of $\,I\,$ translate into congruence properties of the equivalence relation, i.e. it is compatible with the ring operations, i.e. $\,A\equiv a,\ B\equiv b\,\Rightarrow\, A+B\equiv a+b,\,$ and similarly for other operations.

For further discussion see this answer. It includes another characterization of congruences, namely $\,R$-subalgebras of the square $\,S\subseteq R^2,\,$ e.g. the above compatibility condition becomes $\,(A,a),\,(B,b)\in S\,\Rightarrow\, (A,a)+(B,b) = (A+B,a+b)\in S,\,$ i.e. $\,S\,$ is closed under addition.

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Bill Dubuque
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When is $a + 5\mathbb Z = b + 5\mathbb Z$? It is the case when $a \in b + 5\mathbb Z$, i.e. when there is some $c\in\mathbb Z$ with $a=b+5c$. But this is the case if and only if $a\equiv b \pmod 5$.

Jesko Hüttenhain
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The cosets are exactly $\{5\mathbb{Z},1+\mathbb{Z},2+\mathbb{Z},3+\mathbb{Z},4+\mathbb{Z}\}$. If you understand how to apply the ring operations to these elements of the factor ring, then you can check directly that the operations coincide with those in the integers modulo 5. For instance, $(2+\mathbb{Z})(2+\mathbb{Z})=(4+\mathbb{Z}),(4+\mathbb{Z})+(3+\mathbb{Z})=2+\mathbb{Z}$ etc.

The point is that there's an isomorphism between the factor ring and the integers modulo 5 given by sending the coset $n+\mathbb{Z}$ to $n\pmod 5$, and similarly for the inverse. So even though the elements of the factor ring are infinite sets, as far as the ring structure is concerned they're entirely interchangeable with whatever objects you're used to playing the role of the elements of the ring of integers modulo 5.

Kevin Carlson
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