Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

...and if it's important, do those ideas have any generalization to more "exotic" number systems?

The motivation for my question comes from reading some of the excellent answers posted to other questions, such as a recent one asking whether $\sqrt{1 + 24n}$ always yields primes. In particular, I've been struck by commenter Bill Dubuque's repeated use of what he terms "modular reduction", casting a problem in $\mathbb{Z}_n$ to make it much easier to solve in the general case for $\mathbb{Z}$.

What I don't quite grok is why this works; why can we do this? Is there a deep reason? At first glance, to me, there doesn't seem anything inherent in the axioms of a given $\mathbb{Z}_k$ that necessarily ties it intimately with $\mathbb{Z}$; all we care is that is has $k$ elements and it is closed under the binary operator of addition. It doesn't seem to encode information about $\mathbb{Z}$'s other elements.

Now, in saying that, I'm not sure I'm on the right foot here at all, so I'll analogize to something I know a little better: One sees, in some textbook definitions, an identification of $\mathbb{C}$ with an ordered pair $(a,b)$, $(c,d)$ with rules for addition and multiplication that go to $(a + c, b + d)$ and $(ac - bd, ad + bc)$, respectively, with no immediate hint about the importance of $\mathbb{C}$ in that is the algebraic closure of $\mathbb{R}$, which is a highly nontrivial theorem that needs to be proved through the FTA. Is there a similar relationship between $\mathbb{Z}_n$ and $\mathbb{Z}$, and is that extensible to other systems?

I'd appreciate any answers and references tailored to someone who's taken up to the middling undergraduate math level. (e.g. Linear Algebra, elementary abstract Algebra, undergraduate Complex Analysis...)

share|improve this question
5  
They are the quotients of $\mathbb{Z}$, a.k.a. homomorphic images. en.wikipedia.org/wiki/Quotient_ring#Examples –  Jonas Meyer Feb 13 '11 at 21:17

1 Answer 1

up vote 15 down vote accepted

Well, the first thing to say is that the ring $\mathbb{Z}/n\mathbb{Z}$ is the quotient of the ring $\mathbb{Z}$ by the ideal $n \mathbb{Z}$.

Is this already familiar to you? If not, you should take a course and/or read a book on basic abstract algebra -- and, in the meantime, check out this wikipedia article. If so, could you clarify your question: what more than this do you want to know?

Added: well, I certainly understand and appreciate that you're looking for insight. In this case I'm not sure exactly how to provide it. But I'll try...

Here is a number theorist's trick: there is in fact a unique homomorphism $f$ from $\mathbb{Z}$ to any ring $R$. From this it follows that if $P = P(x_1,\ldots,x_n) = 0$ is any polynomial equation with $\mathbb{Z}$-coefficients, then any solution $(a_1,\ldots,a_n) \in \mathbb{Z}^n$ induces a solution $(f(a_1),\ldots,f(a_n)) \in R^n$. Thinking contrapositively, if you can find a ring -- any ring -- for which the equation $P = 0$ has no solutions in $R^n$, then you know right away that it has no solutions in $\mathbb{Z}$. The most classical -- and also, not coincidentally, effective -- choices of rings $R$ to choose are $\mathbb{R}$ (the real numbers) and $\mathbb{Z}/n\mathbb{Z}$. For instance:

Since $P(x,y) = x^2 + y^2 + 3 = 0$ has no solutions over $\mathbb{R}$, it has no solutions over $\mathbb{Z}$.

Since $P(x,y) = x^2 + y^2 - 3 = 0$ has no solutions over $\mathbb{Z}/4\mathbb{Z}$, it has no solutions over $\mathbb{Z}$.

As for your more foundational question as to whether the finite rings $\mathbb{Z}/n\mathbb{Z}$ are defined in reference to $\mathbb{Z}$: by coincidence you have hit upon one of my standard pedagogical rants. The answer is a resounding yes. If you don't think of the elements of $\mathbb{Z}/n\mathbb{Z}$ as equivalence classes of integers, how do you know that the addition and multiplication operations are commutative and associative, and that multiplication distributes over addition? To show this, you do not have to say "quotient map" if you really don't want to, but you have to use it, i.e., that the operations $+,\cdot$ in $\mathbb{Z}/n\mathbb{Z}$ are defined in terms of those in $\mathbb{Z}$.

share|improve this answer
    
I was looking for insight into things: Why can we go back and forth between $\mathbb{Z_n}$ and $\mathbb{Z}$ to solve equations, starting from the "binary operation" definition (or do you need $\mathbb{Z}$ to define the $\mathbb{Z_k}$?), and secondly in what sense does $Z$ fulfill a missing attribute that $Z_k$ lacks? –  Uticensis Feb 13 '11 at 21:28
6  
@Billare: We can't go "back-and-forth", but any solution in $\mathbb{Z}$ to a certain type of equation will yield a solution in $\mathbb{Z}_n$, and solutions in $\mathbb{Z}_n$ will yield information about possible solutions in $\mathbb{Z}$. This is very different from the identification of $\mathbb{C}$ as a set of ordered pairs. $\mathbb{Z}_n$ is not a description of $\mathbb{Z}$, it's a "rough approximation" of $\mathbb{Z}$. –  Arturo Magidin Feb 13 '11 at 21:34
1  
@Billare: as a former (now mostly inactive) wikipedian, your description of the math articles fills me with pride: that's exactly how it should be. Or, well, not exactly: it would be nice if you could learn new subjects from wikipedia articles, but that's kind of a bonus. Speaking as a post-PhD mathematician, I have often learned about new math topics from wikipedia articles...but not always: some of them are a bit impenetrable to me too. What I want to see when I look at a math wikipedia article is some reasonable overview and a good list of references to primary sources.... –  Pete L. Clark Feb 13 '11 at 22:11
1  
@Pete L. Clark What you've added made things much clearer, thank you. This fact, "there is in fact a unique homomorphism f from $\mathbb{Z}$ to any ring $R$", does it have an easy proof, amenable to being assigned as an exercise? –  Uticensis Feb 13 '11 at 22:19
1  
@Billare: yep. According to standard conventions, rings have a multiplicative identity $1$ and any homomorphism of rings sends the $1$ in the source ring to the one in the target ring. In the case of $\mathbb{Z}$, this forces $2$ to map to $1_R + 1_R$, and so forth... –  Pete L. Clark Feb 13 '11 at 22:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.