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

In summary, the conversation discusses the statement "a≡b mod n" being true in the ring of algebraic integers (Ω) and how this also implies its truth in the ring of integers (\mathbb Z) as long as a, b, and n are integers. The proof given involves a theorem, "blabla", which states that if x is an algebraic integer and a rational, it is also an integer. The relevance of Ω being a ring is questioned, but it is explained that it is simply a shortcut for saying that "n divides a-b in Ω". However, the argument only mentions "because Ω is a ring" and not "because Ω is a ring without zero div
  • #1
nonequilibrium
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"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 [itex]a\equiv b \mod n[/itex] is true in [itex]\Omega[/itex], being the ring of the algebraic integers, then the modular equivalence (is that the right terminology?) it also true in [itex]\mathbb Z[/itex], at least if [itex]a,b,m \in \mathbb Z[/itex].

Fair enough, but as a prove it states: "This is a direct consequence of theorem blabla, because [itex]\Omega[/itex] 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 [itex]\Omega[/itex] is a ring"? I don't see the relevance of [itex]\Omega[/itex] being a ring. After all, if [itex]a\equiv b \mod n[/itex] is true in [itex]\Omega[/itex], it means that [itex]\frac{a-b}{n} \in \Omega[/itex], and since a,b,n are integers, this fraction is a rational number and hence by theorem blabla it's an integer. Done(?)
 
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  • #2


mr. vodka said:
Hello,

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

Fair enough, but as a prove it states: "This is a direct consequence of theorem blabla, because [itex]\Omega[/itex] 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 [itex]\Omega[/itex] is a ring"? I don't see the relevance of [itex]\Omega[/itex] being a ring. After all, if [itex]a\equiv b \mod n[/itex] is true in [itex]\Omega[/itex], it means that [itex]\frac{a-b}{n} \in \Omega[/itex], 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 [itex]a\equiv b\mod n[/itex] actually means [itex]a-b=rn\,,\,r\in\Omega[/itex] , as the equivalence is in [itex]\Omega[/itex] , and from

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

the RHS is an alg. integer, we deduce that in fact [itex]\,\,r\in\mathbb Z[/itex] .

DonAntonio
 
  • #3


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 [itex]\frac{a-b}{n}=r[/itex] using Omega is a ring?

I would think (and I realize I can be wrong) that given a-b=rn, [itex]\frac{a-b}{n}[/itex] 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 [itex]\Omega[/itex] is a ring" and not "because [itex]\Omega[/itex] 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.
 
  • #4


mr. vodka said:
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 [itex]\frac{a-b}{n}=r[/itex] using Omega is a ring?


*** It's just a short way to say "n divides a-b in [itex]\Omega[/itex]" 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, [itex]\frac{a-b}{n}[/itex] 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 [itex]\Omega[/itex]" means that there exists [itex]r\in\Omega\,\,s.t.\,\,a-b=nr\,,\,\,r\in\Omega[/itex] . 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 [itex]R:=\mathbb Z/12\mathbb Z[/itex] because there is an element [itex]r\in R\,\,[/itex] s.t [itex]\,\,6=2r[/itex] , namely [itex]r=3[/itex] , and I couldn't care less that ALSO [itex]r=9[/itex] works in this case...

DonAntonio ***




This is of course true for Omega, but the argument simply mentions "because [itex]\Omega[/itex] is a ring" and not "because [itex]\Omega[/itex] 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.

...
 
  • #5


Okay, but then it seems like your first post doesn't make sense: you said (shortening) "we have [itex]a-b = rn, r \in \Omega[/itex] and from here, SINCE WE'RE IN A RING, we conclude that [itex]\frac{a-b}{n} = r \in \Omega[/itex]", 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?
 

What is the definition of "A≡b mod n true in ring of algebraic integers"?

The notation "A≡b mod n" means that A and b are congruent (have the same remainder) when divided by the integer n. In the context of the ring of algebraic integers, this means that A and b belong to the same residue class modulo n.

What does it mean for "A≡b mod n true in ring of integers"?

If "A≡b mod n" is true in the ring of algebraic integers, it means that A and b are also congruent when considered as elements of the ring of integers. In other words, the residue class of A modulo n is the same as the residue class of b modulo n.

Why is it important for "A≡b mod n true in ring of algebraic integers" to also be true in the ring of integers?

The ring of algebraic integers is a subring of the ring of integers, meaning that it contains a smaller set of elements. If "A≡b mod n" is true in the ring of algebraic integers, it means that the congruence relation is preserved even when considering a smaller set of elements. This is important in certain mathematical proofs and applications.

Can "A≡b mod n true in ring of algebraic integers" be true if it is not true in the ring of integers?

No, if a congruence relation is true in the ring of algebraic integers, it must also be true in the larger ring of integers. This is because the ring of algebraic integers inherits its algebraic properties from the ring of integers and must follow the same rules for congruence.

What are some examples of "A≡b mod n true in ring of algebraic integers" being true in the ring of integers?

One example is 2≡5 mod 3. This is true in the ring of algebraic integers because both 2 and 5 belong to the residue class [2] modulo 3. In the ring of integers, this congruence can be written as 2≡5 mod 3 because 2 and 5 have the same remainder when divided by 3 (both have a remainder of 2). Another example is 4≡7 mod 5, which is true in both the ring of algebraic integers and the ring of integers because both 4 and 7 belong to the residue class [4] modulo 5.

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