How can I compute a fixed field over the p-adics

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SageMath does not yet support general extensions of p-adics yet unfortunately, it's something that we're actively working on, e.g., at https://trac.sagemath.org/ticket/28466. If you want to get involved a bit, please join us at https://sagemath.zulipchat.com/#narro....

There's a package that builds on SageMath to provide some Henselizations here that might work for such problems since it can compute in towers of fields and provides valuations. I am not sure how well that package currently works, but please let us know in the above zulip chat if you find it useful or have suggestions. Anyway, the following works on this binder with this package installed…

from henselization import *

K = QQ.henselization(3)
R.<x> = K[]
F.<ξ> = K.extension(x^4 + x^3 + x^2 + x + 1)
# Extension defined by ξ^4 + ξ^3 + ξ^2 + ξ + 1 of Henselization of Rational Field with respect to 3-adic valuation

R.<x> = F[]
f = x^4 - 3*x^2 + 18
LF.<α> = F.extension(f.factor()[0][0])
# Extension defined by α^2 + O(x^(3/4))*α + 3*ξ^3 + 3*ξ^2 + O(x^(3/4)) of Extension defined by ξ^4 + ξ^3 + ξ^2 + ξ + 1 of Henselization of Rational Field with respect to 3-adic valuation

R.<x> = LF[]
# We need to pass to approximations so we can perform arithmetic.
# Unfortunately, we cannot say ξ = ξ.approximation(20) directly. :(
ξ = -(x^4 + x^3 + x^2 + x + 1).factor()[0][0][0].approximation(20)
α = -(x^4 - 3*x^2 + 18).factor()[0][0][0].approximation(20)
basis = [ξ^i*α^j for i in range(4) for j in range(2)]

R.<x> = LF[]
# Note that if you change the first 0 to a 1, you get a different result.
sqrt27 = -((x^2 + 2/7).factor()[0][0][0].approximation(20))

# Write φ in terms of the basis given by ξ and α.
φα = (2*α^2 - 3)* sqrt27 / α
φξ = ξ^3
image = [φξ^i*φα^j for i in range(4) for j in range(2)]

# We write sqrt(3) in terms of that basis
sqrt3 = -(x^2 - 3).factor()[0][0][0].approximation(20)
A = matrix([b._vector_(base=K) for b in basis])
coefficients = A.transpose().solve_right(sqrt3._vector_(base=K))

# Map sqrt(3) through φ
φsqrt3 = sum([b * c for (b, c) in zip(image, coefficients)])
(φsqrt3 - sqrt3).valuation() # big number

# We write sqrt(-3) in terms of that basis
sqrt_3 = -(x^2 + 3).factor()[0][0][0].approximation(20)
A = matrix([b._vector_(base=K) for b in basis])
coefficients = A.transpose().solve_right(sqrt3._vector_(base=K))

φsqrt_3 = sum([b * c for (b, c) in zip(image, coefficients)])
(φsqrt_3 - sqrt_3).valuation() # small number

So, the answer is $\sqrt{3}$ since $\varphi(\sqrt{3}) - \sqrt{3}$ has "infinite" valuation but $\varphi(\sqrt{-3}) - \sqrt{-3}$ has valuation 1/2 or so. However, this depends on the choice of $\sqrt{\frac{-2}{7}}$. Choosing the other root, gives $\sqrt{-3}$.

It's also possible to do the calculation in plain SageMath but we need to help it because creating arbitrary extensions of p-adic fields is currently not implemented.

First, we can create the field $F$:

sage: F.<t> = Qq(3^8)
sage: F
3-adic Unramified Extension Field in a
defined by x^8 + 2*x^5 + x^4 + 2*x^2 + 2*x + 2

Then we create $LF$. For this, we need to find an uniformizer. For this, we factor the polynomial $f(x) = x^4 - 3x^2 + 18$ over $F$.

Unfortunately, factorization is also not implemented. So we do it by hand as follows:

sage: S.<x> = F[]
sage: f = x^4 - 3*x^2 + 18
sage: g = x^2 - x + 2   # f(x) = 9 * g(x^2/3)
sage: b, c = g.roots(multiplicities=False)
sage: f0 = x^2 - 3*b
sage: f % f0
0

Now we can build the extension:

sage: LF.<alpha> = F.extension(f0)
sage: LF
3-adic Eisenstein Extension Field in alpha
defined by x^2 + 4236577851*a^7 + 26772/7805*a^6
- 117429/7789*a^5 - 175194/1681*a^4 - 111303/22931*a^3
+ 5787134532*a^2 + 8260845189*a + 109059/13295 over its base field
sage: f(alpha)
O(alpha^44)

In order to define $\varphi$, we first define the automorphism of $F$; it is actually just the Frobenius:

sage: Frob = F.frobenius_endomorphism()
sage: zeta5 = F.primitive_root_of_unity(5)
sage: Frob(zeta5) == zeta5^3
True

As noticed by saraedum, there are two possibilities for $\varphi$ depending on the choice of the square root of $-\frac 2 7$.

sage: beta = (2*alpha^2 - 3) * sqrt(F(-2/7)) / alpha
sage: phi1 = LF.hom([beta], base_map=Frob)
sage: phi2 = LF.hom([-beta], base_map=Frob)

Finally, we can check whether $\varphi_1$ (resp. $\varphi_2$) fixes $\sqrt 3$ or $\sqrt{-3}$:

sage: sqrt3 = sqrt(LF(3))
sage: phi1(sqrt3) == sqrt3
False
sage: phi2(sqrt3) == sqrt3
True

sage: sqrtm3 = sqrt(LF(-3))
sage: phi1(sqrtm3) == sqrtm3
True
sage: phi2(sqrtm3) == sqrtm3
False