I have been doing a lot of work with quadratic fields and I am attempting to generalize the results to abstract fields and rings and I am having trouble showing that a certain set is a subring (I don't even know for sure if it is, but my intuition tells me it is and I cannot find any counter examples.

Let $F$ be a field and let $N: F \to \mathbb{Q}$ be a totally multiplicative function (i.e. $N(ab)=N(a)N(b)$ for all $a,b \in F$). Now construct the set $I=\{x \in F \mid N(x) \in \mathbb{Z} \}$. My intuition tells me that the set $I$ is a subring of $F$, but I am having trouble proving it. The fact that $I$ is closed under multiplication is fairly easy to show:

If $a$ and $b$ are in $I$ then $N(a)$ and $N(b)$ are in $\mathbb{Z}$ and thus $N(a)N(b)$ is in $\mathbb{Z}$ and since $N$ is totally multiplicative, we have $N(a)N(b)=N(ab)$ and thus $N(ab) \in \mathbb{Z}$ and therefore $ab \in I$.

In order to fully show that $I$ is a subring, I need to show that $a-b \in I$ whenever $a$ and $b$ are in $I$, and I cannot figure out how to show this since $N$ has no specific properties regarding addition or subtraction. My intuition tells me it's a subring, however it is also entirely possible that it is not necessarily closed under subtraction and there are counterexamples that I just can't find.

Any help would be appreciated.


Let our field be $\mathbb Q$ and define $N$ by $N(x)=1/x$ if $x\neq 0$ and $N(0)=0$. Then the set you mention consists of 0 and all reciprocals of integers and hence is not closed under addition.

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  • $\begingroup$ Wow, remarkably simple. Thanks! $\endgroup$ – ASKASK May 10 '15 at 19:23
  • $\begingroup$ @ASKASK you're welcome. $\endgroup$ – Matt Samuel May 10 '15 at 19:24

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