Can the norm of a non-algebraic integer be an integer? 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.
 A: $\newcommand{\Q}{\mathbb{Q}}$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, $1$.
A: 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).
A: 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.
