# Abstract Nonsense

## 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.

June 25, 2011 -

## 9 Comments »

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