Abstract Nonsense

Crushing one theorem at a time

Quotient Rings (Pt. I)


Point of Post: In this post we give the construction of quotient rings by ideals as well as the canonical projection R\twoheadrightarrow R/\mathfrak{a}.

Motivation

Intuitively what we’d like to do is make a construction similar to that of a quotient group but for rings. Namely, for some subset A of a ring R we’d like to define a ring structure on the set R/A of equivalence classes x\sim y if and only if x-y\in A, which ‘collapses’ A down to a point in such a way that the canonical map \pi:R\to R/A:r\mapsto [r] is a ring morphism. We know from group theory that since \pi:R\to R/A is a group morphism we must have that A\unlhd R, but since R is abelian this is equivalent to A\leqslant R. So, the obvious question then is what multiplicative conditions must be imposed on A? What we shall see is that A must be an ideal of R for this construction to work. A little experimentation actually makes this obvious since 0\in A and so \pi(0)=[0]=0_{R/A} and thus for any a\in A and r\in R we have [ra]=[r][a]=[r]0_{R/A}=0_{R/A}=[a] and so ra\in A and similarly ar\in A.

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Quotient Ring

Let R be a ring and \mathfrak{a}\in\mathfrak{L}(R). Since \mathfrak{a}\leqslant R and R is abelian (so that \mathfrak{a}\unlhd R) we may form the quotient group R/\mathfrak{a}. Recall that the elements of R/\mathfrak{a} are the sets \left\{r+a:a\in\mathfrak{a}\right\} written r+\mathfrak{a} and their group structure is defined by (r+\mathfrak{a})+(s+\mathfrak{a}))=(r+s)+\mathfrak{a}. We know from basic group theory that since R is abelian so is R/\mathfrak{a}.

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What we now claim is that the ‘multiplication’ (r+\mathfrak{a})(s+\mathfrak{a})=rs+\mathfrak{a} is well-defined on R/\mathfrak{a}. Indeed, if r+\mathfrak{a}=r'+\mathfrak{a} and s+\mathfrak{a}=s'+\mathfrak{a} then r'=r+a for a\in\mathfrak{a} and s'=s+b for b\in\mathfrak{a}. We see then that

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(r'+\mathfrak{a})(s'+\mathfrak{a})=(r+a)(s+b)+\mathfrak{a}=rs+rb+as+ab+\mathfrak{a}=rs+\mathfrak{a}

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the last conclusion made since rb,as,ab\in\mathfrak{a} (since \mathfrak{a} is an ideal and a,b\in\mathfrak{a}) and so rb+as+ab\in\mathfrak{a} and so the last claim follows by definition. What we now claim is that with this multiplication and addition R/\mathfrak{a} becomes a ring. Indeed:

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Theorem: Let R be a ring and \mathfrak{a}\in\mathfrak{L}(R), then R/\mathfrak{a} with the addition (r+\mathfrak{a})+(s+\mathfrak{a})=(r+s)+\mathfrak{a} and multiplication (r+\mathfrak{a})(s+\mathfrak{a})=rs+\mathfrak{a} is a ring.

Proof: We already know from group theory that \left(R/\mathfrak{a},+\right) is a group and we know from the above discussion that the multiplication is a binary operation, so it suffices to show that it’s associative and left and right distributive. The first of these is clear since

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\left((r+\mathfrak{a})(s+\mathfrak{a})\right)(t+\mathfrak{a})=(rs)t+\mathfrak{a}=r(st)+\mathfrak{a}=(r+\mathfrak{a})((s+\mathfrak{a})(t+\mathfrak{a}))

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To see left distributivity we note that

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\begin{aligned}(r+\mathfrak{a})\left((s+\mathfrak{a})+(t+\mathfrak{a})\right) &=(r+\mathfrak{a})\left((s+t)\mathfrak{a}\right)\\ &=r(s+t)\mathfrak{a}\\ &=(rs+rt)+\mathfrak{a}\\ &= \left(rs+\mathfrak{a}\right)+\left(rt+\mathfrak{a}\right)\\ &=\left(r+\mathfrak{a}\right)(s+\mathfrak{a})+(r+\mathfrak{a})(t+\mathfrak{a})\end{aligned}

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and right distributivity is done similarly. \blacksquare

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With this structure we call the ring R/\mathfrak{a} a quotient ring or when we want to be more precise the quotient ring of R by \mathfrak{a}. We note that this structure (by construction) satisfies our desired result that:

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Theorem: Let R be a ring and \mathfrak{a}\in\mathfrak{L}(R) then the map \pi:R\twoheadrightarrow R/\mathfrak{a}:r\mapsto r+\mathfrak{a} is an epimorphism with \ker\pi=\mathfrak{a}.

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This map is called the canonical projection of R onto R/\mathfrak{a} or canonical projection for short. From the existence of the canonical projection we are able to conclude that R/\mathfrak{a} inherits many of the properties of R. For example if R is commutative or unital then so is R/\mathfrak{a} since these are both properties preserved under epimorphism. But, we are also able to gleam from this theorem the converse of a previously noted theorem. In particular, we can now legitimately say:

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Theorem: Let R be a ring, then J\subseteq R is an ideal if and only if there exists a ring S and morphism f:R\to S for which J=\ker f.

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We note next that there is a certain ‘mapping property’ that the canonical projection possess, namely:

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Theorem: Let R and S be rings, \mathfrak{a}\in\mathfrak{L}(R) and f:R\to S a ring morphism with \mathfrak{a}\subseteq\ker f. Then, there exists a unique morphism \eta:R/\mathfrak{a}\to S for which \eta\circ \pi=f.

Proof: Consider the mapping \eta:R/\mathfrak{a}\to S defined by \eta\left(r+\mathfrak{a}\right)=f(r). This is well-defined since if r+\mathfrak{a}=r'+\mathfrak{a} then r'=r+a for a\in\mathfrak{a} and so \eta\left(r'+\mathfrak{a}\right)=f(r')=f(r+a)=f(r)+f(a)=f(r)=\eta\left(r+\mathfrak{a}\right). This is a morphism since

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\eta\left((r+\mathfrak{a})+(s+\mathfrak{a})\right)=f(r+s)=f(r)+f(s)=\eta(r+\mathfrak{a})+\eta(s+\mathfrak{a})

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and evidently \eta(\pi(r))=\eta(r+\mathfrak{a})=f(r) for all r\in R and so f=\eta\circ \pi.

and

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\eta\left((r+\mathfrak{a})(s+\mathfrak{a})\right)=\eta\left(rs+\mathfrak{a}\right)=f(rs)=f(r)f(s)=\eta(r+\mathfrak{a})\eta(s+\mathfrak{a})

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Since uniqueness is clear the conclusion follows. \blacksquare

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

 

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June 25, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , ,

9 Comments »

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