## You are here

Homereduced direct product

## Primary tabs

# reduced direct product

Let $\{A_{i}\mid i\in I\}$ be a set of algebraic systems of the same type, indexed by $I$. Let $A$ be the direct product of the $A_{i}$’s. For any $a,b\in A$, set

$\operatorname{supp}(a,b):=\{k\in I\mid a(k)\neq b(k)\}.$ |

Consider a Boolean ideal $L$ of the Boolean algebra $P(I)$ of $I$. Define a binary relation $\Theta_{L}$ on $A$ as follows:

$(a,b)\in\Theta_{L}\quad\mbox{ iff }\quad\operatorname{supp}(a,b)\in L.$ |

###### Lemma 1.

$\Theta_{L}$ defined above is a congruence relation on $A$.

###### Proof.

Since $L$ is an ideal $\varnothing\in L$. Therefore, $(a,a)\in\Theta_{L}$, since $\{k\in I\mid a(k)\neq a(k)\}=\varnothing$. Clearly, $\Theta_{L}$ is symmetric. For transitivity, suppose $(a,b,(b,c)\in\Theta_{L}$. If $a(k)\neq c(k)$ for some $k\in I$, then either $a(k)\neq b(k)$ or $b(k)\neq c(k)$ (a contrapositive argument). So

$\operatorname{supp}(a,c)\subseteq\operatorname{supp}(a,b)\cup\operatorname{% supp}(b,c).$ |

Since $L$ is an ideal, $\operatorname{supp}(a,c)\in L$, so $(a,c)\in\Theta_{L}$, and $\Theta_{L}$ is an equivalence relation on $A$.

Next, let $\omega$ be an $n$-ary operator on $A$ and $a_{j}\equiv b_{j}\;\;(\mathop{{\rm mod}}\Theta_{L})$, where $j=1,\ldots,n$. We want to show that $\omega(a_{1},\ldots,a_{n})\equiv\omega(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod% }}\Theta_{L})$. Let $\omega_{i}$ be the associated $n$-ary operators on $A_{i}$. If $\omega(a_{1},\ldots,a_{n})(k)\neq\omega(b_{1},\ldots,b_{n})(k)$, then $\omega_{k}(a_{1}(k),\ldots,a_{n}(k))\neq\omega_{k}(b_{1}(k),\ldots,b_{n}(k))$, which implies that $a_{j}(k)\neq b_{j}(k)$ for some $j=1,\ldots,n$. This implies that

$\operatorname{supp}(\omega(a_{1},\ldots,a_{n}),\omega(b_{1},\ldots,b_{n}))% \subseteq\bigcup_{{j=1}}^{n}\operatorname{supp}(a_{j},b_{j}).$ |

Since $L$ is an ideal, and each $\operatorname{supp}(a_{j},b_{j})\in L$, we have that $\operatorname{supp}(\omega(a_{1},\ldots,a_{n}),\omega(b_{1},\ldots,b_{n}))\in L$ as well, this means that $\omega(a_{1},\ldots,a_{n})\equiv\omega(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod% }}\Theta_{L})$. ∎

Definition. Let $A=\prod\{A_{i}\mid i\in I\}$, $L$ be a Boolean ideal of $P(I)$ and $\Theta_{L}$ be defined as above. The quotient algebra $A/\Theta_{L}$ is called the $L$-*reduced direct product* of $A_{i}$. The $L$-reduced direct product of $A_{i}$ is denoted by $\prod_{L}\{A_{i}\mid i\in I\}$. Given any element $a\in A$, its image in the reduced direct product $\prod_{L}\{A_{i}\mid i\in I\}$ is given by $[a]\Theta_{L}$, or $[a]$ for short.

Example. Let $A=A_{1}\times\cdots\times A_{n}$, and let $L$ be the principal ideal generated by $1$. Then $L=\{\varnothing,\{1\}\}$. The congruence $\Theta_{L}$ is given by $(a_{1},\ldots,a_{n})\equiv(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod}}\Theta_{L})$ iff $\{i\mid a_{i}\neq b_{i}\}=\varnothing$ or $\{1\}$. This implies that $a_{i}=b_{i}$ for all $i=2,\ldots,n$. In other words, $\Theta_{L}$ is isomorphic to the direct product of $A_{2}\times\cdots\times A_{n}$. Therefore, the $L$-reduced direct product of $A_{i}$ is isomorphic to $A_{1}$.

The example above can be generalized: if $J\subseteq I$, then

$\prod{}_{{P(J)}}\{A_{i}\mid i\in I\}\cong\prod\{A_{i}\mid i\in I-J\}.$ |

For $a\in A=\prod\{A_{i}\mid i\in I\}$, write $a=(a_{i})_{{i\in I}}$. It is not hard to see that the map $f:\prod_{{P(J)}}\{A_{i}\mid i\in I\}\to\prod\{A_{i}\mid i\in I-J\}$ given by $f([a])=(a_{i})_{{i\in I-J}}$ is the required isomorphism.

Remark. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filter $F$. All there is to do is to replace $\operatorname{supp}(a,b)$ by its complement: $\operatorname{supp}(a,b)^{c}:=\{k\in I\mid a(k)=b(k)\}$. The congruence relation is now $\Theta_{{F^{{\prime}}}}$, where $F^{{\prime}}=\{I-J\mid J\in F\}$ is the ideal complement of $F$. When $F$ is prime, the $F^{{\prime}}$-reduced direct product is called a *prime product*, or an *ultraproduct*, since any prime filter is also called an ultrafilter. Ultraproducts can be more generally defined over arbitrary structures.

# References

- 1 G. Grätzer: Universal Algebra, 2nd Edition, Springer, New York (1978).

## Mathematics Subject Classification

08B25*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## Ultraproduct and cardinality

Is it possible using these model-theoretic tools to show that if I is an infinite set and \prod_I K_i is a product of fields and if M is a nonprincipal maximal ideal of this ring (thus corresponding to a nonprincipal ultrafilter) then \prod_I K_i/M has cardinality greater than that of K_i for any i in I?

## Re: Ultraproduct and cardinality

Not my area but I saw talk recently by Kazhdan where he talked of this type of result. He mumbled something about the implicit use of the continuum hypothesis -- so probably not entirely provable but otherwise the techinques are still model theoretic. That would be my only contribution to your question. I'll check back soon and see if someone smarter answers with a better insight.

Good luck,

James

## Re: Ultraproduct and cardinality

I think I have a counterexample to my question.

Take any product of fields R:=\prod_I K_i, with I infinite (it is proven btw that the factor fields of this ring are uncountable and non-archimidean, but more is not known.. as far as I know), take a nonprinicipal maximail ideal of this ring (which corresponds to a nonprincipal ultrafilter of I). Let K=R/M be the factor field wrt M,

then, R':=Kx\prod_I K_i has a factor field from a non-principal maximal ideal (which corresponds to a non-principal ultrafilter of I) which is K, K is also the factor field of the principal maximal ideal (0,1)R' .. (R' can be considered as a tuple here). So the two "ultraproducts" have the same cardinality.

When and where did you saw the talk of Kazhdan? What was the title of his talk? I would be interested in it. Although, this is also not my area.

JosÃ©

## Re: Ultraproduct and cardinality

in the first sencetence of my statement of the counterexampl, I meant factor fields over non-principal maximal ideals are non-archimidean and uncountable.

## Re: Ultraproduct and cardinality

Another counterexample is as follows: let I be infinite, and each K_i be Z_2, then R = \prod_I K_i is a Boolean ring. Let I be the ideal generated by functions f_i whose ith value is 1 and 0 everywhere else. Then the constant function whose value is 1 is not in I. Extend I to a prime ideal M excluding f. This is possible because R is Boolean. By the same reason, M is maximal. Furthermore, M is non-principal, for if g\in R generates M, then g is 0 at some j. But then f_j\in I does not lie in M, a contradiction. Finally, R/M has order 2 because M is a maximal ideal in a Boolean ring.

## Re: Ultraproduct and cardinality

You are telling me that you have an ultraproduct from a nonprincipal ultrafilter by which has order 2, which is the same as the order of the fields in your product, in fact in your example you are saying R/M=Z_2 (although I doubt this!)... Why has R/M order 2? You argue that this is because R is a boolean ring.. but how does this show R/M has order 2 if M is nonprincipal. In fact I do believe in this special case R/M has order >2 for M nonprincipal. By "order" you surely mean the cardinality, right? not the characterization of the field R/M. I know this to be a counterexample too, maybe you meant order >2 :) .. in fact I believe one can say more for this special case..

I believe I have read something in these lines:

If R/M is an ultrapower instead of an ultraproduct, i.e. the K_i's in R are all isomorphic to say K, then R/M for M non-principal has cardinality strictly greater than K

## Re: Ultraproduct and cardinality

> You are telling me that you have an ultraproduct from a

> nonprincipal ultrafilter by which has order 2, which is the

> same as the order of the fields in your product, in fact in

> your example you are saying R/M=Z_2 (although I doubt

> this!)... Why has R/M order 2? You argue that this is

> because R is a boolean ring.. but how does this show R/M has

> order 2 if M is nonprincipal. In fact I do believe in this

> special case R/M has order >2 for M nonprincipal. By "order"

> you surely mean the cardinality, right? not the

> characterization of the field R/M. I know this to be a

> counterexample too, maybe you meant order >2 :) .. in fact I

> believe one can say more for this special case..

Fact: if R is Boolean, and M any maximal ideal in R (principal or not), then R/M is isomorphic to Z_2:

if x is an element of R not in M, then R=(x,M), so that 1=x+m for some m in M. As a result, 1+x=1-x=m is in M. This shows that x+M is either 0+M or 1+M, depending on whether x is in M or not. In other words, R/M is isomorphic to {0,1}.

For more info, please see, for example, Halmo's book "Lectures on Boolean Algebras" (1974, Springer)... great book, I wish they reprint this!

> I believe I have read something in these lines:

> If R/M is an ultrapower instead of an ultraproduct, i.e. the

> K_i's in R are all isomorphic to say K, then R/M for M

> non-principal has cardinality strictly greater than K

Do you remember what your source is?

I found, in "Models and Ultraproducts, an Introduction" by Bell and Slomson (1969, North-Holland), the following:

Suppose I is countably infinite, and {s_i : i in I} is a set such that each s_i is a finite cardinal. Let F be an ultrafilter on I, and P the ultraproduct \prod s_i/F. Then

1. if the set {i in I : s_i=n} is in F for some finite cardinal n, then the card(P)=n,

2. otherwise, the card(P)=2^{\alelph_0}.

However, I did not use this fact in the proof...

## Re: Ultraproduct and cardinality

My reference was a Fundamentae Mathematica paper written by Dana Scott, Morel and Frayne. The paper is very old (also in the 60's) and I had a bad copy of it.. it basically states the fact you just wrote in the end of your post. except that they used "{i in I : s_i=n}\not\in F" in the second part of what you stated, and I read "\in" instead of "\not\in" as the print was very bad.

Yes thanks, having R as Boolean does makes it easier as all other elements are orthogonal to each other.

## Re: Ultraproduct and cardinality

There is something that is bothering me the whole time. You do know that (x,M)=Rx+M, how can you make sure that x+m =1, I can see that xr+m=1 for some r in R and m in M, but not x+m=1. Or is x+M already R?

Although from the fact you mentioned, R/M must indeed be Z_2 as

{i in I : |Z_2|=2} = I contains the ultrafilter making our maximal ideal.

## Re: Ultraproduct and cardinality

jocaps writes:

> My reference was a Fundamentae Mathematica paper written

> by Dana Scott, Morel and Frayne. The paper is very old

> (also in the 60's) and I had a bad copy of it..

By the way, you can get PDFs of old Fundamenta Mathematicae papers from the ICM site. The text is usually clear.

http://matwbn.icm.edu.pl/spis.php?wyd=1&jez=en

## Re: Ultraproduct and cardinality

Yes, I recently discovered that myself :) .. the text is indeed clear. I am loving open-access and I really hope that most professional mathematicians turn into publishing in open-access journals.

## Re: Ultraproduct and cardinality

Sorry, that was my poor first attempt. Here's my second attempt: if 1=rx+m, then 1+rx=rx+rx+m=m\in M. So x+rx = x+rx^2 = x(1+rx)\in M. Therefore, 1+x = (1+rx)+(x+rx)\in M as well.

Another way: if R is any commutative ring with 1, and M is any maximal ideal, then R/M is simple. Any commutative simple ring must be a field. So, if R is Boolean, M is maximal in R, then R/M is simple, and a field. But R has characteristic 2, R/M must be Z_2.