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I recently had a homework problem asking to prove that the direct product of rings (or rings with identity) are still rings (with identity), and it seemed really silly to go through all the steps in order to conclude what was extremely trivial.

My question is, are there instances when this is not the case? Are there certain properties, or certain algebraic structures whereupon taking their direct products doesn't maintain the original algebraic structure/properties?

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    $\begingroup$ The direct product of two cyclic groups isn't (necessarily) cyclic. $\endgroup$ – Gerry Myerson Jan 11 '13 at 2:02
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    $\begingroup$ Also, a direct product of integral domains is never an integral domain (since $(1,0)\cdot(0,1)=0$). You can also consider PIDs. The direct product of PIDs is not necessarily PID, and you can conclude several results won't hold by considering other things, such as fields. $\endgroup$ – Clayton Jan 11 '13 at 2:04
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    $\begingroup$ The direct product of two simple groups is not simple, $\endgroup$ – Michael Joyce Jan 11 '13 at 4:07
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A direct product of integral domains is never an integral domain since $(1,0)\cdot(0,1)=0$. You can also consider PIDs; $\Bbb Z$ is a PID while $\Bbb Z\times\Bbb Z$ is no. The direct product of fields is not a field, say $\Bbb Q$ and $\Bbb Q\times\Bbb Q$.

There is also a problem of checking that if you don't use a canonical operation, checking that it still makes sense as an operation. For example, we can identify $\Bbb C$ with $\Bbb R^2$ by defining an operation $$(a,b)\cdot(c,d)=(ac-bd,ad+bc).$$ It is not immediately obvious that this is a well-defined operation that satisfies the axioms for a ring.

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  • $\begingroup$ Isn it that the product of two (unital say) principal rings is a principal ring? I.e. I have the impression that the product of two PIDs is not a PID because the product of two integral domains is not an integral domain, so that the principalitity does not add anything. $\endgroup$ – M. Luethi Aug 31 '16 at 11:22
  • $\begingroup$ @M.Luethi: That is correct; see here for the details. $\endgroup$ – Clayton Sep 1 '16 at 19:04
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The question requires care when interpreting what "taking direct products mean". One way to interpret this is: given two algebraic structures $A,B$ of the same kind, where $A,B$ are sets with extra structure, when will the same structure not be defined naturally on the product $A\times B$ as sets". This interpretation is open to some debate due to the use of the word 'naturally'.

Another interpretation uses the categorical framework and leads to two possible precise interpretations. Once you fix your favourite algebraic structure you can consider the category of all those algebraic structures together with their structure preserving functions (almost always will this indeed yield a category). Call this category $C$. Assuming that a typical algebraic structure of interest consist of a single set together with extra structure, there is a forgetful functor $U:C\to Set$. The question now becomes: if $C$ admits binary categorical products does the forgetful functor commute with binary products. I think this is a precise way to interpret what you had in mind. One instance where the answer is yes is when this forgetful functor has a left adjoint. That means that whenever free structures exist the answer to your question is: products are always constructed naturally on the cartesian product of the underlying sets.

This happens, for instance, for groups (since free groups exist), for abelian groups (since free abelian groups exists) etc. but not for fields (free fields do not exist). Non-algebraic cases fit into this setting as well, including examples from topology and analysis.

Another possible interpretation is to wonder whether the category $C$ at all admits products. For instance, the category of all cyclic groups doesn't have all binary products.

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  • $\begingroup$ Thanks a lot for your answer. I am deeply interested in this topic - the direct product of two algebraic structures. Are there any literatures discussing your expplanation in much further details? I would like to dive into it. Thanks. $\endgroup$ – scaaahu Jan 11 '13 at 3:23
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    $\begingroup$ If you read through the first four chapters of Mac Lanes "Categories for the working mathematician" you will have all the CT you need to understand these things and you can then roam around to find more. There are rather elementary considerations of CT though, so don't expect anything too deep. $\endgroup$ – Ittay Weiss Jan 11 '13 at 3:26
  • $\begingroup$ Your point is well taken. However, I happen to know some not-so-well-known structures not fit well into CT. The direct products are even more weird. So, it's still worh doing it. I'll keep your words in my mind, though. Thanks again. $\endgroup$ – scaaahu Jan 11 '13 at 3:44
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Such properties are sometimes called " productive". For example, a product of compact topological spaces is compact (Tychonov). Not all properties meet this [stringent] criterion.

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    $\begingroup$ This idea spans several mathematical disciplines. $\endgroup$ – ncmathsadist Jan 11 '13 at 2:06
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Some properties by definition do not necessarily "maintain the original algebraic structure" for the direct product, such as, say, being "irreducible". Perhaps the question gets more focus if we stay with groups (or rings, modules and algebras).
The direct product of simple groups need not be simple, the direct product of groups of a square-free order is not of square-free order, the direct product of cyclic groups need not be cyclic, and the direct product of complete groups need not be complete.

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