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given the ring $ \mathbb{Z}/n\mathbb{Z} $ is always true that $ \mathbb{Z}/n\mathbb{Z}=[\text{zero divisor}]+[\text{units}] $

how can evaluate the zero divisor and units ??

I believe that $ a x=0 \pmod n $ for zero divisor

and $ ax=1 \pmod n $ for units

I know how to solve a congruence but what is $a$ ?? thanks.

also what does it mean $ \mathbb{Z}= \prod_{p \in \mathbb{P}} \mathbb{Z}/p\mathbb{Z} $ for every prime $p$ ?

what are the generators of the gorup $ \mathbb{Z}/n\mathbb{Z} $ under the addition '+' and product '$\times$'

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You tagged this question [group-theory], but the very talk about units and zero divisors places it outside the scope of group theory and squarely into [ring-theory]. – Arturo Magidin Jan 11 '12 at 17:08
@Arturo, maybe elementary number theory would be an even better tag. The vocabulary is the vocabulary of ring theory, but the question is a question of number theory. – Gerry Myerson Apr 29 '12 at 13:32
@GerryMyerson: We can compromise and put both [ring-theory] and [elementary-number-theory] tags on it. Done. – Arturo Magidin Apr 29 '12 at 19:17

HINT $\ $ If $\rm\:a\in$ finite ring $\rm\:R\:$ then $\;\rm x\to a\:x\;$ is onto iff 1-1, so $\rm\:a\:$ is a unit iff $\rm\:a\:$ is not a zero-divisor.

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The general rule is that $a$ is a unit modulo $n$ is and only if it is prime to $n$, i.e. $\gcd(a,n)=1$. This is proven in every basic algebra/number theory textbook (for example, Dummit and Foot's "Abstract Algebra") but is so easy you can prove it on your own (use the fact that if $\gcd(a,b)=d$ then there exists integers $x,y$ such that $ax+by=d$).

As for the other question - $\mathbb{Z}/n\mathbb{Z}$ is not a "group" if you talk about two operations, but a ring, and "generators" is not very clear. Still, if I understand your basic question, there's a nice discussion of the structure of $\mathbb{Z}/n\mathbb{Z}$ for all values of $n$ in (for example) Ireland and Rosen's "Classical introduction to modern number theory" (and I think Dummit and Foote cover it as well). As before, this is a very standard result, given almost in every book on the subject.

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After reading the question , I wondered which rings $A$ have the property that every element in $A$ is a unit or is a zero-divisor.

1) If the zero ideal has a primary decomposition (which is always the case for a noetherian ring), then the set of zero divisors is the union of the associated primes of the ring: $zdiv(A)=\bigcup _{{\mathfrak p}\in Ass(A)}\mathfrak p$ so that the condition for the property is that every maximal ideal of $A$ be an assassin.

This holds for all artinian rings (e.g. the ring $\mathbb Z/n\mathbb Z$ of the question ), but also for higher dimensional rings, like the one-dimensional localization $A=R_{\mathfrak m}$ of the notorious ring $R=k[X,Y]/(X^2, XY)=k[x,y]\quad (k \text { a field})$ at the maximal ideal $\mathfrak m=(x,y)$.
Notice that a primary decomposition of $(0)$ is $(0)=(x)\cap {\mathfrak m}^2$ and that $Spec (A)=Ass(A)=\lbrace\mathfrak m,(x)\rbrace$, so that $zdiv(A)=\mathfrak m$ .

2) Some non-noetherian rings also have the property that every non-unit is a zero-divisor.
Class of examples: rings with only one prime ideal, like the non-noetherian $$A=k[X_1, X_2,...,X_n,...]/(X_1^2,..., X_n^2,...) $$

WARNING Needless to say, this post is absurdly highbrow with respect to the actual question.
As Gadi remarks, the answer to Jose's question is to be found in fairly elementary algebra or number theory books.
I just wanted to address a more general question which might (or might not!) be of interest to more advanced users of our site.

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Such rings are sometimes called quasi-regular, as I mention in my post here on embedding rings into quasi-regular rings. – Bill Dubuque Jan 11 '12 at 15:41
Ah, that's very interesting: thanks a lot for the word and the link, @Bill . I was a 100% sure that such an elementary property must have been investigated, but I had no idea what keyword to use to search the literature. – Georges Elencwajg Jan 11 '12 at 17:08
I've asked the exact same question in MO:… – lhf Jan 11 '12 at 17:15
Thanks a lot to you too, @lhf: the link is very rich and interesting. It is a little ironic that I thought I was explaining results to others and it turns out that I'm the one who is learning from users! Ironic but very pleasant! – Georges Elencwajg Jan 11 '12 at 18:15
@GeorgesElencwajg, yes, I've experienced the same surprise. I've learned a lot from all those generous and knowledgeable people here. – lhf Jan 11 '12 at 18:25

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