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A conjecture about polynomials $P(x)$ and $Q(x)$ defining the same field is believed to be true, but unproven:

Let $P(x)$ be an $n$ degree polynomial.

There exists a polynomial $Q(x)$ of degree $n$, with fixed of terms from degrees $n$ to $(n/2)$ if $n$ is even, and $(n+1)/2$ if $n$ is odd, and varied terms degrees from $(n/2)-1$ to $0$ if $n$ is even, and degrees from $((n+1)/2)-1$ to $0$ if $n$ is odd such that $Q(x)$ defines the same field as $P(x)$ and also infinitely many polynomials of this form, but does not have the same discriminant as $P(x)$. Note that the fixed coefficients in $Q(x)$ do not have to necessarily match those in $P(x)$.

Example of conjecture (assuming true):

Let $P(x)$ $=$ $x^3-16x^2+8x+1$, since $P(x)$ is a $3$ degree, or cubic polynomial, the first $2$ coefficients can be fixed in $Q(x)$, and there exists infinitely solutions $(a, b)$ such that $P(x)$ and $Q(x)$ define the same field where:

$P(x)$ $=$ $x^3-16x^2+8x+1$

$Q(x)$ $=$ $4x^3+7x^2+ax+b$

Note the coefficients in any such $Q(x)$ may be chosen at random, and infinitely many polynomials of this form $Q(x)$ would define the same field $P(x)$.

Second, can someone please explain an algorithm for solving the example problem assuming conjecture is true?

Thanks for looking into this.

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    Define the same field over what base field? Over the rationals? – 2017-02-25
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    The cubic case is somewhat trivial (in characteristic not 3), since you can easily adjust the leading two coefficients of a polynomial to anything you want by a substitution $x = at + b$. – 2017-02-25

1 Answers 1

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A counterexample is defining the field extension $\mathbf{F}_{16} / \mathbf{F}_2$. The only irreducible quartics over $\mathbf{F}_2$ are:

  • $x^4 + x + 1$
  • $x^4 + x^3 + 1$
  • $x^4 + x^3 + x^2 + x + 1$

In particular, this field extension cannot be defined by a polynomial of the form

$$ x^4 + x^2 + \ldots $$

(I conjecture that this counterexample can be adapted to the $2$-adics, to provide characteristic zero counterexamples)

Furthermore, any finite field example would give counterexamples to the assertion that there are infinitely many polynomials of a given form.

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    I don't know about the first two "counter examples" but the third, my conjecture hols true with Q(x) = x^4+x^3+x^2+61541x+1579481 defining the same field as P(x) = x^4+x^3+x^2+x+1. They have the same properties in Z5, fixing the first degrees x^4, x^3, x^2 to coefficients of 1. – 2017-02-25
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    @J.Linne: There is only one counterexample here: there does not exist any polynomial $Q(x) = x^4 + x^2 + ax + b$ defining this field extension – 2017-02-25