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

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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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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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$\newcommand{\Q}{\mathbb{Q}}$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).