# Abstract Nonsense

## Product of Rings (Pt. I)

Point of Post: In this post we discuss the product of rings including the classification for ideals of finite products and the universal property of product rings.

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Motivation

As for groups, topological spaces, representations, in any category it is always useful to define product structures. So, in this post we shall describe what it means to form the ‘product ring’ of the set of rings $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$. We will then show that similarly to quotient rings there is a universal mapping property which characterizes product rings. We will also discuss the situation of finding the ideals of a product of unital rings in terms of the ideals of the individual rings, and in fact characterize the difference infiinitely and finitely many factors in the product (i.e. whether we are taking the product of finitely many or infinitely many rings) based on the ideals of the product ring.

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Definitions and Characterizations

We begin by giving a certain universal mapping characterization of product rings. Namely, let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of rings. We say that a ring $R$ satisfies the is a product of $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ if there exists a set of epimorphisms $\left\{p_\alpha:R\to R_\alpha\right\}_{\alpha\in\mathcal{A}}$ such that for any ring $S$ and any set of morphisms $\left\{f_\alpha:S\to R_\alpha\right\}_{\alpha\in\mathcal{A}}$ there exists unique morphism $g:S\to R$ such that $p_\alpha\circ g=f_\alpha$. The important thing to note is:

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Theorem: Let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of rings. Then, any two products of $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ are isomorphic.

Proof: Let $P$ and $P'$ be two products of $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ with the associated sets of epimorphisms $\left\{p_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{p'_\alpha\right\}_{\alpha\in\mathcal{A}}$. We first note by definition that since $p'_\alpha:P'\to R_\alpha$ is a set of maps there exists a unique map $f:P'\to P$ such that $p_\alpha\circ f=p'_\alpha$ and similarly there exists a unique map $g:P\to P'$ such that $p'_\alpha\circ g=p_\alpha$. We thus find that $p_\alpha=p'_\alpha\circ g=p_\alpha\circ f\circ g$ and similarly $p'_\alpha=p'_\alpha\circ g\circ f$. Since both $p_\alpha$ and $p'_\alpha$ are both epic (epimorphism) we may conclude that from the first of these equalities that $f\circ g=\text{id}_P$ and from the second that $g\circ f=\text{id}_{P'}$. Thus, we may conclude that $f$ and $g$ are isomorphisms with $g=f^{-1}$. $\blacksquare$

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We thus see that if (and in general this is a big if) products exist then they are, up to isomorphism, unique. Thus, it suffices to show that there actually exist models of products of a set of rings. In particular, given a set of rings $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ we define the the product of $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ to be the set $\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha$ (with the usual function definition) with the addition $f+g$ of $f,g$ in $\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha$ to be $(f+g)(\alpha)=f(\alpha)+g(\alpha)$ and the product $fg$ to be equal to $(fg)(\alpha)=f(\alpha)g(\alpha)$. We define the canonical projections $\displaystyle \pi_\beta:\prod_{\alpha\in\mathcal{A}}R_\alpha\to R_\beta$ to be the map $\pi_\beta(f)=f(\beta)$. Depending upon the connotation this map may also be called the evaluation homomorphism and is denoted $\text{ev}_\beta$. What we now note is that the product is a product. Indeed, the canonical projections are epimorphisms and evidently given a set of morphisms $f_\alpha:R\to R_\alpha$ the function $\displaystyle f:R\to\prod_{\alpha\in\mathcal{A}}R_\alpha$ given by $(f(r))(\beta)=f_\beta(r)$ has the property that $\pi_\beta\circ f=f_\beta$ and evidently is the unique function to do this. Thus, when we think about products we shall be thinking about this model of a product for the infinite case. For the finite case there is a nicer way to think about the product of rings instead of the function definition. Namely, we identify for a finite set of rings $\left\{R_1,\cdots,R_n\right\}$ the product $\displaystyle \prod_{\alpha\in[n]}R_\alpha$ with the set $R_1\times \cdots\times R_n=\left\{(r_1,\cdots,r_n):r_k\in R_k\right\}$ with coordinate wise multiplication and addition. It’s important to note that we arbitrarily chose the order the tuples appeared in (i.e. we chose the first tuple to contain elements of $R_1$) but since evidently any reordering of the factors is canonically isomorphic (prove the obvious identification is an isomorphism if you’ve never seen it done before!). In particular, if $R_1=\cdots=R_n=R$ we denote $R_1\times\cdots\times R_n$ by $R^n$ when convenient.

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It’s important to note that if each ring is unital then the product is naturally unital with unit $1(\alpha)=1_\alpha$ or in the finite case $1=(1_1,\cdots,1_n)$.

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Product of Morphisms and the Ideals of a Product of Unital Rings

An obvious question now is, what are the ideals of a product? To answer this we first define an object which should be familiar to most people. Namely, suppose that have to sets of rings $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{S_\alpha\right\}_{\alpha\in\mathcal{A}}$ and a set of morphisms $\left\{f_\alpha:R_\alpha\to S_\alpha\right\}$ we can form the product morphism

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$\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha:\prod_{\alpha\in\mathcal{A}}R_\alpha\to\prod_{\alpha\in\mathcal{A}}S_\alpha$

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given by

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$\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)(g)\right)(\beta)=f_\beta(g(\beta))$

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In other words, the $\beta^{\text{th}}$ “coordinate” is acted on by $f_\beta$. To see precisely what this means it may be helpful to consider the finite case when given maps $f:R_1\to S_1$ and $g:R_2\to S_2$ we have that $(f\times g)(r_1,r_2)=(f(r_1),f(r_2))$. It’s evident that the product of morphisms is itself a morphism. It’s also evident that

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$\displaystyle \ker\left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)=\prod_{\alpha\in\mathcal{A}}\ker f_\alpha$

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With this in mind we are in good position to prove:

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Theorem: Let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of rings and $\mathfrak{a}_\alpha\in\mathfrak{L}(R_\alpha)$ for each $\alpha\in\mathcal{A}$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}\mathfrak{a}_\alpha$ is an ideal of $\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha$.

Proof: Let $\pi_\alpha:R\to R/\mathfrak{a}_\alpha$ be the usual morphism and consider the morphism $\displaystyle \prod_{\alpha\in\mathcal{A}}\pi_\alpha$. By previous observation then we have that the kernel of this map is equal to $\displaystyle \prod_{\alpha\in\mathcal{A}}\ker\pi_\alpha$ which is equal to $\displaystyle \prod_{\alpha\in\mathcal{A}}\mathfrak{a}_\alpha$. Since we know the kernel of ring morphisms are ideals the conclusion follows. $\blacksquare$

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What we now show is that the converse is a characterization of whether we are taking the product of infinitely many or finitely many unital rings. Namely:

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Theorem: Let $\left\{R_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of unital rings. Then, all the ideals of $\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha$ are of the form $\displaystyle \prod_{\alpha\in\mathcal{A}}\mathfrak{a}_\alpha$ for $\mathfrak{a}_\alpha$ an ideal of $R_\alpha$ if and only if $\#(\mathcal{A})<\infty$.

Proof: The reason why all the ideals of say $R_1\times\cdots\times R_n$ are of the form $\mathfrak{a}_1\times\cdots\times\mathfrak{a}_n$ is as follows. We first note that if $\mathfrak{a}$ is an ideal then $\pi_j(\mathfrak{a})\overset{\text{def.}}{=}\mathfrak{a}_j$ is an ideal since we know ideals are preserved under epimorphisms. We thus have that $\mathfrak{a}\subseteq\mathfrak{a}_1\times\cdots\times\mathfrak{a}_n$. To prove the reverse inclusion we note that if $r_j\in\mathfrak{a}_j$ then there exists tuples $(r_1,\cdots,\ast),(\ast,r_2,\cdots,\ast),\cdots,(\ast,\cdots,r_n)\in\mathfrak{a}$ (where $\ast$ just denotes some arbitrary element–i.e. we know there exists some tuple in $\mathfrak{a}$ which has $r_j$ in the $j^{\text{th}}$ coordinate and it’s irrelevant what it is). Now, since $\mathfrak{a}$ is an ideal we have that $(0,\cdots,r_j,\cdots,0)=(\ast,\cdots,r_j,\cdots,\ast)(0,\cdots,1_j,\cdots,0)\in\mathfrak{a}$. Thus, $(r_1,\cdots,r_n)=(r_1,\cdots,0)+\cdots+(0,\cdots,r_n)\in\mathfrak{a}$ and so the opposite inclusion holds. This half of the theorem then follows.

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Conversely, suppose that $\mathfrak{A}$ is infinite and let $\mathfrak{a}$ be equal to all elements of $\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha$ with finite support (only finitely many non-zero coordinates). This is evidently an ideal but evidently not the product of ideals since it clearly isn’t even the product of sets. $\blacksquare$

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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.

July 11, 2011 -

## 11 Comments »

1. […] Point of Post: This is a continuation of this post. […]

Pingback by Product of Rings (Pt. II) « Abstract Nonsense | July 11, 2011 | Reply

2. […] if is). Moreover, is a morphism, and since it’s evidently bijective we find that (where we recall that is the -fold product of […]

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3. […] the form ______ is an ideal of .” We had a similar goal when we were trying to classify the ideals of product rings where we saw, at least for the finite product of unital rings, that the ideals of were precisely […]

Pingback by Matrix Rings (Pt. II) « Abstract Nonsense | July 13, 2011 | Reply

4. […] given rings and constructing new rings out of them, such as forming the matrix ring  and the -fold product ring . In this post we discuss yet another way of forming a ring in a ‘functorial’, namely […]

Pingback by Polynomial Rings (Pt. I) « Abstract Nonsense | July 19, 2011 | Reply

5. […] as the case for the product of rings and matrix rings there is a certain ‘functorial’ property for matrix rings. In […]

Pingback by Polynomial Rings (Pt. III) « Abstract Nonsense | July 19, 2011 | Reply

6. […] fact, it’s fairly easy to see that as rings is isomorphic to the product ring  (i.e. ‘-copies’ of ). Thus, the distinction between the two is purely connotative, […]

Pingback by Rings of Functions (Pt. I) « Abstract Nonsense | August 1, 2011 | Reply

7. […] that when each is secretly , and is finite we’ve discussed that a sort of converse to the first part of this theorem where every ideal of the product of […]

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8. […] existence of morphisms fitting the universal characterization is just a consequence of the way the coproduct is […]

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9. […] we claim is that if we take the trivial inverse system on a set of rings we get the product ring as a result. To see this we note that we have natural maps (where is the product ring) given by […]

Pingback by Inverse Limit of Rings (Pt. II) « Abstract Nonsense | December 26, 2011 | Reply

10. […] be a preordered set and an inverse system of rings over . Then, if denotes the subring of the product  ring over the defined by . Then, along with the natural projections are an inverse limit of the […]

Pingback by Inverse Limit of Rings (Pt. III) « Abstract Nonsense | December 26, 2011 | Reply

11. […] of unital rings with unital ring maps there are products, as we have already proven that the usual product of rings with the canonical projections form a […]

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