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I know that on $\mathbb Z[\sqrt n]$, $|N(a)|=1$ iff $a$ is a unit in $\mathbb Z[\sqrt n]$.

Note: $N$ is the norm function i.e., $N(0)=0$ and $N(a)N(b)= N(ab)$

However, does this property still hold for any norm on any Integral Domain?

Edit: A "norm" $N$ on any integral domain $D$ is a function $N:D\rightarrow \mathbb Z$ that satisfies the following properties: $N(0)=0$ and $N(a)N(b)= N(ab)$

Jhon Doe
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    What does $N$ mean on an arbitrary integral domain? – Eric Wofsey Sep 12 '19 at 15:06
  • @EricWofsey I have edited the post – Jhon Doe Sep 13 '19 at 01:17
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    It's still wrong to ask whether this property holds for arbitrary integral domains. The question doesn't make sense unless you choose a specific norm, and there is not a unique choice for most integral domains. What would make sense would be to ask if it holds for any norm on any integral domain. – Eric Wofsey Sep 13 '19 at 01:24

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No. For instance, on any integral domain you can define a norm by $N(0)=0$ and $N(x)=1$ for all $x\neq 0$. This will not satisfy the property you mention unless the integral domain is actually a field.

Eric Wofsey
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Arbitrary integral domains don't come equipped with a norm.

One can do the following: consider the class of (commutative) rings $R$ which are infinite, not fields, and such that if $I$ is a nonzero ideal then $R/I$ is finite (Pete Clark calls these abstract number rings). Then we can define the norm $N(I)$ to be the cardinality $|R/I|$, and we can define the norm $N(r), r \in R$ to be $N((r)) = |R/(r)|$. Then it's straightforward to verify that $N(I) = 1$ iff $I$ is the unit ideal, hence that $N(r) = 1$ iff $(r)$ is the unit ideal iff $r$ is a unit.

However, it's no longer true that $N(IJ) = N(I) N(J)$ in general.

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Qiaochu Yuan
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