A≡b mod n true in ring of algebraic integers => true in ring of integers

[nonequilibrium]

"a≡b mod n" true in ring of algebraic integers => true in ring of integers

Hello,

So I'm learning about number theory and somewhere it says that if $a\equiv b \mod n$ is true in $\Omega$, being the ring of the algebraic integers, then the modular equivalence (is that the right terminology?) it also true in $\mathbb Z$, at least if $a,b,m \in \mathbb Z$.

Fair enough, but as a prove it states: "This is a direct consequence of theorem blabla, because $\Omega$ is a ring."

Now theorem blabla, as I called it so eloquently, is basically the statement that if x is an algebraic integer and a rational, that it is also an integer.

But isn't theorem blabla all we need then? Why is there "because $\Omega$ is a ring"? I don't see the relevance of $\Omega$ being a ring. After all, if $a\equiv b \mod n$ is true in $\Omega$, it means that $\frac{a-b}{n} \in \Omega$, and since a,b,n are integers, this fraction is a rational number and hence by theorem blabla it's an integer. Done(?)

 

[DonAntonio]

Hello,

So I'm learning about number theory and somewhere it says that if $a\equiv b \mod n$ is true in $\Omega$, being the ring of the algebraic integers, then the modular equivalence (is that the right terminology?) it also true in $\mathbb Z$, at least if $a,b,m \in \mathbb Z$.

Fair enough, but as a prove it states: "This is a direct consequence of theorem blabla, because $\Omega$ is a ring."

Now theorem blabla, as I called it so eloquently, is basically the statement that if x is an algebraic integer and a rational, that it is also an integer.

But isn't theorem blabla all we need then? Why is there "because $\Omega$ is a ring"? I don't see the relevance of $\Omega$ being a ring. After all, if $a\equiv b \mod n$ is true in $\Omega$, it means that $\frac{a-b}{n} \in \Omega$, and since a,b,n are integers, this fraction is a rational number and hence by theorem blabla it's an integer. Done(?)

We have that $a\equiv b\mod n$ actually means $a-b=rn\,,\,r\in\Omega$ , as the equivalence is in $\Omega$ , and from

here, SINCE WE'RE IN A RING. we can conclude that $\,\,\frac{a-b}{n}=r\in\Omega$ , and since the LHS is a rational and

the RHS is an alg. integer, we deduce that in fact $\,\,r\in\mathbb Z$ .

DonAntonio

 

[nonequilibrium]

Thank you for your reply, but I'm not yet getting your point.

Can you explain more about how you go from a-b=rn to $\frac{a-b}{n}=r$ using Omega is a ring?

I would think (and I realize I can be wrong) that given a-b=rn, $\frac{a-b}{n}$ is by definition equal to r. That being said, it needs to be shown that this definition is consistent, and I suppose for that one needs to know that if a-b=sn, that r = s. In other words Omega needs to be a ring without zero divisor. This is of course true for Omega, but the argument simply mentions "because $\Omega$ is a ring" and not "because $\Omega$ is a ring without zero divisors".

So it seems that due your post I've now gone from "I don't see how being a ring is relevant" to "I don't see how being a ring is enough" haha.

 

[DonAntonio]

Thank you for your reply, but I'm not yet getting your point.

Can you explain more about how you go from a-b=rn to $\frac{a-b}{n}=r$ using Omega is a ring?


*** It's just a short way to say "n divides a-b in $\Omega$" and that's all, when division is as defined in a ring.****

I would think (and I realize I can be wrong) that given a-b=rn, $\frac{a-b}{n}$ is by definition equal to r. That being said, it needs to be shown that this definition is consistent, and I suppose for that one needs to know that if a-b=sn, that r = s.


*** No need at all of this as "n divides a-b in $\Omega$" means that there exists $r\in\Omega\,\,s.t.\,\,a-b=nr\,,\,\,r\in\Omega$ . Nothing here needs to be checked for consistency as it is just a definition in a ring. ***


In other words Omega needs to be a ring without zero divisor.


*** Not at all: "division by" is a term that can be used in any ring, for example: "2 divides 6 in the ring $R:=\mathbb Z/12\mathbb Z$ because there is an element $r\in R\,\,$ s.t $\,\,6=2r$ , namely $r=3$ , and I couldn't care less that ALSO $r=9$ works in this case...

DonAntonio ***



This is of course true for Omega, but the argument simply mentions "because $\Omega$ is a ring" and not "because $\Omega$ is a ring without zero divisors".

So it seems that due your post I've now gone from "I don't see how being a ring is relevant" to "I don't see how being a ring is enough" haha.

...

 

[nonequilibrium]

Okay, but then it seems like your first post doesn't make sense: you said (shortening) "we have $a-b = rn, r \in \Omega$ and from here, SINCE WE'RE IN A RING, we conclude that $\frac{a-b}{n} = r \in \Omega$", yet in your second post you state there is no difference between writing these two expressions, so why did you write when going from the former to the latter expression "since we're in a ring", even in capital letters?