(adsbygoogle = window.adsbygoogle || []).push({}); "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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# A≡b mod n true in ring of algebraic integers => true in ring of integers

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