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What is the difference between ring $\mathbb{Z}_n$ and congruence to modulo $n$.

Are they same thing or what are significant differences between them except we can use integers greater than $n$ for modulo operations. They look like the same.

cingoz recai
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    Congruence modulo $n$ is a relation, so it cannot be a ring. They are entirely different types of things; this alone should suffice as an answer. – Marc van Leeuwen Nov 02 '12 at 13:58
  • @Marc That's not true generally - see my comment to your answer. – Bill Dubuque Nov 02 '12 at 14:25
  • @BillDubuque: just for clarity: by "That" you apparently mean "Congruence is a relation". You claim congruence can be defined as a subalgebra (of $\Bbb Z\times\Bbb Z$ in this case). That's fine with me, although it's not likely that OP considers it this way. And even then, this subalgebra is rather different from $\Bbb Z_n$. – Marc van Leeuwen Nov 02 '12 at 14:57
  • @Marc You stated that "a congruence cannot be a ring". That is false. In fact that a ring congruence is a ring is an essential property of a congruence. One cannot argue such semantic points so simply. – Bill Dubuque Nov 02 '12 at 15:07
  • @BillDubuque: As I said, it's fine with me; have a nice day. But I actually meant to say a relation cannot be a ring; maybe that is not even quite true set-theoretically, but I don't care. – Marc van Leeuwen Nov 02 '12 at 15:15

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First, review the definition of what a ring is.

In particular, it's a type of algebraic structure with two different binary operations defined on it (sometimes referred to as "addition" and "multiplication").

Next, review the definition of the ring $\mathbb{Z}_n$ and you will see that the two binary operations on it are just addition mod $n$ and multiplication mod $n$.

These are interrelated concepts, but the former is a type of algebraic structure with operations, and the latter ($+$ mod $n$ and $\times$ mod $n$) are examples of operations; so they are certainly not one and the same.

Benjamin Dickman
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They are of one, but they are different kinds of mathematical objects.

Congruence modulo $n$ is an equivalence relation on $\Bbb Z$, whereas $\Bbb Z_n$ is a ring, namely the quotient ring by the congruence modulo $n$ relation (meaning that two elements of $\Bbb Z$ are regarded identical in $\Bbb Z_n$ iff they are congruent mod $n$).

Berci
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To elaborate on my comment, something positive can be said about the relation between the congrence modulo $n$ and the ring $\Bbb Z/n\Bbb Z$: the latter is the quotient of the the ring $\Bbb n$ by the relation of congrence modulo $n$.

This means the elements of $\Bbb Z/n\Bbb Z$ are classes of the relation of congrence modulo $n$ (which is an equivalence relation; any equivalence relation gives a partition of the set it is defined on into equivalence classes). Moreover the operations of addition and multiplication in $\Bbb Z/n\Bbb Z$ are derived from those operations in $\Bbb Z$: to perform one of these operations, one chooses elements (representatives) from the classes to be added/multiplied, one performs the corresponding operation upon them in $\Bbb Z$, and finally one takes the class (modulo $n$) of the resulting number.

It is a non-trivial fact, which can be proved, that every set of choices of representatives will give the same end result (a class modulo $n$); this is why the quotient ring of $\Bbb Z$ by the relation of equivalence modulo $n$ is defined in the first place. (And it explains why there is no such thing as a quotient ring $\Bbb R/n\Bbb R$; in this case the required property for defining multiplication fails.)

Marc van Leeuwen
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  • Generally it is not true that congruences are defined to be merely a set-theoretic object such as an equivalence relation. That fails to capture the algebraic essence of a congruence. Instead, congruences are more naturally defined algebraically: a congruence of a general algebraic structure $,\rm A,$ is a certain sub-algebra of $\rm,A^2$ (containing the diagonal $\rm,A(1,1),:$ etc). For example, see my answer here, which discusses in detail ring congruences on $,\Bbb Z,$ from this viewpoint. – Bill Dubuque Nov 02 '12 at 14:26
  • @BillDubuque: Doubtless you are right (although I think what you say is essentially the same as saying an equivalence relation that is also compatible with the algebraic operations so that these can be transferred to the quotient; this is just what I said, differently), but I doubt whether OP has the mathematical maturity to understand your answer. Or indeed mine. – Marc van Leeuwen Nov 02 '12 at 14:54