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Let L/K be a finite field extension and define the norm of an element as the product of each K-embedding evaluated at that element.

Can the norm of a non-algebraic integer be an integer?

I know that the norm of an algebraic integer is always an integer as it corresponds to the final term in the minimum polynomial, I was wondering if there was a converse to this- even under special conditions.

EDIT: Sorry yes an element which belongs to L.

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When you say non-algebraic integer, do you mean an algebraic element of the extension which is not an algebraic integer? –  Alex Becker May 16 at 17:15
I think that's what I mean. If there's an example where the rationals are the ground field that would be even better. –  user70147 May 16 at 17:20

3 Answers 3

up vote 9 down vote accepted

The norm of $$ \frac35+\frac45i\in\Bbb Q[i] $$ is $$ \left(\frac35\right)^2+\left(\frac45\right)^2=1 $$ More generally, any primitive pythagorean triple $(a,b,c)$ can be used to construct elements in $\Bbb Q(i)$ of norm 1 which are not integers, just take $z=\frac ac+\frac bci$

Even more generally if $K$ is a quadratic field and $0\neq z\in K$ any, the quotient $z/\bar z$ has always norm 1.

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I was dubious at first that the quotient $z / \overline{z}$ was in fact more general, but indeed every primitive pythagorean triple can be written this way. This follows from the general formula for pythagorean triples and also from reasoning about Gaussian integers, as is done here. (Just in case anyone was interested.) (+1.) –  Goos May 17 at 6:39
@Goos: well, actually one can generalize even further: if $Gal(K/F)=<\sigma>$ is cyclic then every element $x\in F$ of norm 1 is of the form $x=z^\sigma/z$ for some $z\in K$ (Hilbert Theorem 90: en.wikipedia.org/wiki/Hilbert%27s_Theorem_90). –  Andrea Mori May 17 at 8:34
Thanks for the reference! –  Goos May 17 at 14:58

$\newcommand{\Q}{\mathbb{Q}}$I believe you could take a root $\alpha$ of $f = 2x^2 + x + 2$.

Clearly $f$ is irreducible in $\Q[x]$, so $\alpha$ is not an algebraic integer as its minimal monic polynomial $f/2$ over $\Q$ does not have integer coefficients.

But the norm of $\alpha$ in $\Q(\alpha)/\Q$ is the constant coefficient of $f/2$, that is, $1$.

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Hint $ $ Over $\,\Bbb Q,\,$ a quadratic number is an algebraic integer iff its $\,\color{#0a0}{\rm norm}\,$ and $\,\rm\color{#c00}{trace}$ are integers, i.e. iff its monic minimal polynomial has integer coefficients. So your question reduces to finding an irreducible polynomial $\,x^2\! + \color{#c00}b\,x + \color{#0a0}c\in\Bbb Q[x] $ with $\,\color{#c00}{b\not\in \Bbb Z},\ \color{#0a0}{c\in \Bbb Z},\,$ which is quite easy.

Remark $\ $ The same argument works over any integrally-closed domain $\,D,\,$ since then monic minimal polynomials of $D$-integral eements must have coefficients in $\,D\,$ (a property that is equivalent to $\,D\,$ being integrally-closed).

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