# Units in ring of integers are exactly those with norm {-1,1}

Let $K$ be an algebraic number field and $R$ be the ring of the integers of $K$. Show that an element $u\in R$ is a unit of $R$ if and only if $N_{K/\mathbb{Q}}(u)\in \{-1,1\}$.

It is easy to show units are of norm {-1,1}. But on the other hand, I have no idea.

Without using ideals: - if $u\in R$ is a unit, then $N(u)N(u^{-1}) = 1$, an equation in $\mathbf Z$, hence $N(u) =\pm 1$ - conversely, if $u\in R$ has norm $\pm1$, as an algebraic integer $u$ is a root of a polynomial of the form $X^n + ... + a_1 X\pm1\in \mathbf Z[X]$, hence $\pm(u^{n-1} + ... + a_1)\in R$ is the inverse of $u$ .

If the norm of an element $u\in R$ is $\pm1$, then the ideal $(u)$ has norm $\pm1$. Norm is multiplicative, so in particular $(u)$ cannot have any prime factors. In ideals, to contain is to divide; hence $(u)$ is not contained in a maximal ideal, i.e. it is the unit ideal.

Write the minimal polynomial of $\alpha$, your candidate unit as

$$p(x) = a_0+a_1x+\ldots + x^n$$

Then note that

$$\alpha^{-n}p(\alpha) = a_0\alpha^{-n} + a_1 \alpha^{-(n-1)}+\ldots + a_n=0.$$

This is a polynomial with coefficients over $\Bbb Z$ of smallest degree possible for $\alpha$ to satisfy, and all the coefficients are already known to be coprime, so after dividing everything by $a_0$ this is the minimal polynomial over $\Bbb Q$ for $\alpha^{-1}$, and by definition this is an algebraic integer, i.e. $\alpha$ is a unit, iff the minimal polynomial has coefficients in $\Bbb Z$, i.e. $a_0|a_i$ for all $1\le i\le n$, but if they are coprime and still $a_0$ divides all the others we know that $a_0=\pm 1$ by the fundamental theorem of arithmetic.

• Do you have a reference for the claim that the coefficients are coprime? Commented Sep 30, 2023 at 14:02