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# Talk:Algebraic integers

Eisenstein integers are quadratic integers? Alonso del Arte 03:03, 4 August 2012 (UTC)

Yes, ${\displaystyle \omega }$ is a root of ${\displaystyle x^{2}+x+1=0.}$ Not obvious, the way they're usually defined, but true. Charles R Greathouse IV 07:11, 4 August 2012 (UTC)
To see that the Eisenstein integers are [quadratic] algebraic integers, note that each ${\displaystyle \scriptstyle z\,=\,a+b\,\omega \,}$ is a root of the monic quadratic polynomial
${\displaystyle z^{2}-(z+{\bar {z}})z+z{\bar {z}}=z^{2}-(a+b\,\omega +a+b\,{\bar {\omega }})z+(a+b\,\omega )(a+b\,{\bar {\omega }})=z^{2}-(2a-b)z+(a^{2}-ab+b^{2}),\,}$
where ${\displaystyle \scriptstyle \omega +{\bar {\omega }}\,=\,-1\,}$ and ${\displaystyle \scriptstyle \omega \,{\bar {\omega }}\,=\,|\omega |\,=\,1.\,}$
Daniel Forgues 09:20, 4 August 2012 (UTC)

## Nonarithmetic algebraic integers

It seems that we may have algebraic integers which are not arithmetic! (See arithmetic numbers.) — Daniel Forgues 06:19, 11 January 2013 (UTC)

I added an example of nonarithmetic algebraic integers. I suspect that almost all algebraic integers are nonarithmetic! — Daniel Forgues 06:37, 11 January 2013 (UTC)