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Specifically I am interested in the the ring of integers for the field $\mathbb{Q}(\sqrt{-1+2\sqrt{2}})$. Does this ring of integers have a power basis?

More generally, for any Salem number $s$, does the ring of integers for $\mathbb{Q}(s)$ have a power basis?

Does anyone know of a sage command or some clever way to determine if any given ring of integers for an algebraic number field has a power basis?

Thanks!

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  • $\begingroup$ This is a local problem, and I have a hunch that there's a problem at 2. The discriminant of the field is $-2^6 \cdot 7$, and I wonder if you can show that $2^8$ divides the discriminant of any element whose powers form a power basis. $\endgroup$ – Thom Tyrrell Jun 2 '16 at 1:20
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I've written a Python/Sage script that computes the discriminant of a general element in $L = \mathbb{Q}(\sqrt{-1+2\sqrt{2}})$.

#Set up the field L

K.<a> = QuadraticField(2)
S.<y> = K[]
L.<b> = K.extension(y^2 - (-1 + 2*a))
L.<c> = L.absolute_field()

#Set up the general integer theta

T.<i,j,k,l> = QQ[]
U.<x> = T[]
U.<x> = U.quotient_ring(U.ideal(L.polynomial()))
int_basis = map(lambda omega:U(omega.lift()), L.integral_basis())
theta = sum(map(mul,zip(int_basis,T.gens()))) #dot product

#Compute the matrix determinant

m = matrix([(theta*x^i).lift().coefficients() for i in range(4)])
traces = [(m^i).trace() for i in range(7)]
trace_matrix = matrix([traces[i:i+4] for i in range(4)])
disc = trace_matrix.det()

The variable disc should be $-448 = -2^6 \cdot 7$ (the discriminant of $L$) times an integer polynomial in $i,j,k,l$. This doesn't prove that $L = \mathbb{Q}(\sqrt{-1+2\sqrt{2}})$ isn't monogenic, but if we could show that the polynomial is divisible by 2 that would suffice.

Here's how it works. After setting up the field $L$, we find an integral basis for its integers and express a general element $\theta$ in terms of it. In the definition of int_basis we switch from the number $c$ to the variable $x$ so that Sage treats $\sqrt{-1+2\sqrt{2}}$ and the integral basis as polynomials in $x$; this makes it easier to collect coefficients and compute the matrix of $\theta$ acting on $L$ using the coefficients() method. In the last block we compute the matrix $(\text{Tr }\theta^{i+j-2})_{i,j}$ and its determinant.

This script could easily be generalized to handle other number fields.

See this question and its answer for a successful application of this method.

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Let $u = (\sqrt{-1+2\sqrt 2}+\sqrt 2+1)/2$.

The minimal polynomial of $u$ over $\Bbb Q(\sqrt 2)$ is $u^2-u(1+\sqrt 2)+1$, and over $\Bbb Q$ it is $u^4-2u^3+u^2-2u+1$, whose discriminant is $-448$.

Since this is the discriminant of the number field, its ring of integers is $\Bbb Z[u]$.

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  • $\begingroup$ Awesome. How did you find it? $\endgroup$ – Thom Tyrrell Jun 6 '16 at 13:51
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    $\begingroup$ I worked over $\Bbb Z[\sqrt 2]$ and kinda followed what i exposed in there math.stackexchange.com/questions/1445333/…. We can check locally if we have a good candidate by checking if $(\sqrt 2)$ factors as needed. I started with $u_0=\sqrt{-1+2\sqrt 2}$ and obviously it failed. But it lead me to a closer candidate $u_1 = (u_0+1)/\sqrt 2$. The $\sqrt 2$-valuation of his discriminant got smaller but it still wasn't enough but by doing this again i got to the $u$ above and found that $(\sqrt 2)$ doesn't ramify and is inert. $\endgroup$ – mercio Jun 6 '16 at 13:57
  • $\begingroup$ Since the adjustments are translations and divisions by $\sqrt 2$, I think it cannot make the discriminant over $\Bbb Z[\sqrt 2]$ any bigger (though this is a lot less clear when you look at this over $\Bbb Z$). Also I have no idea how well this "method" works in the wild. Here we had that $(\sqrt 2)$ was principal so that's a good help, but for higher degree extensions I'm not sure this is always this simple. $\endgroup$ – mercio Jun 6 '16 at 14:01
  • $\begingroup$ If I read the counter-example in the linked question correctly, it shows that the problem can be local too. $\endgroup$ – mercio Jun 6 '16 at 14:03

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