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I've (mostly) not studied these topics in Uni, so please point out any nonsense you see.

A couple of days ago I found the intriguing fact that for irreducible $f \in \mathbb{Z}[x]$ $$\text{Gal}(f) \subseteq G \text{ up to conjugacy} \iff R_G(f) \text{ has a root in $\mathbb{Q}$}$$ if $G \leq S_n$ is a resolvent invariant of some symmetric polynomial $g \in \mathbb{Z}[X_1,\ldots,X_n]$, i.e. $\text{Stab}_{S_n}(g) = G$, and where $R_G$ is the resolvent polynomial defined by $$R_G(X) = \prod_{i = 1}^{m}(X-P_i)$$ where $P = S_n(g) = \{P_1,\ldots,P_m\} \subseteq \mathbb{Z}[X_1,\ldots,X_n]$ is the orbit of $S_n$ on $g$ ($\sigma \in S_n$ acts on $\mathbb{Z}[X_1,\ldots,X_n]$ by permuting the corresponding indices of the indeterminates). By then employing the Fundamental Theorem of symmetric polynomials and Vieta's formulae we can write this polynomials in terms of the coefficients of our starting polynomial (making it a really neat tool without knowing the roots).

Given this fact, you can try to start searching for such $g$ given a Galois group $G$, e.g. after some bit of tinkering you might notice that $$\text{Stab}_{S_4}(X_1X_2 + X_3 X_4) = D_4$$ or more visually $$g := X_1X_2 + X_3X_4 \text{ is invariant under } \{(),(24),(12)(34),(1234),(13),(13)(24),(1432),(14)(23)\} \cong D_4$$ and not for any larger subgroup of $S_4$ that includes $D_4$, hence $g$ is a resolvent invariant of $D_4$ (see also the Wiki article linked above). In another thread it was mentioned that we can canonically construct such a $g$:

  • Get the permutation subgroup representation of $G \leq S_n$.
  • Take the monomial $X_1 X_2^2 \ldots X_n^{n}$, sum it under the action of all elements in $G$ and define it as $g$, i.e. $$g := \sum\limits_{\sigma \in G} \sigma(X_1 X_2^2 \ldots X_n^{n})$$
  • Then we have $\text{Stab}_{S_n}(g) = G$.

Note that this procedure would yield a more "complicated" polynomial for $D_4$ than the one given above. Hence the following question:

The canonically defined polynomial $g$ is not always minimal (in the sense of polynomial degree and summed terms involved). Is there an existence result or an algorithm to find "reduced" representations like for $D_4$?

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  • $\begingroup$ Please ask one question at a time. $\endgroup$
    – Shaun
    Jun 12, 2023 at 15:56
  • $\begingroup$ Okay, then I'll delete my first one as this has been answered by ahulpke in the other thread. $\endgroup$
    – TheOutZ
    Jun 12, 2023 at 15:57
  • $\begingroup$ @Shaun This is also the second time someone (you?) has been voting to close my question not even an hour after it has seen the light of day. Give me some time people, I'll do my best to polish it, but I can't read your mind. $\endgroup$
    – TheOutZ
    Jun 12, 2023 at 15:59
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    $\begingroup$ @Shaun Fair enough. I know that close votes can be retracted, sorry if I sounded rude just then. It just gets really frustrating after a while when the first thing happening to my posts is a close vote. $\endgroup$
    – TheOutZ
    Jun 12, 2023 at 16:07
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    $\begingroup$ @Shaun I've added some observations if you're still interested :). $\endgroup$
    – TheOutZ
    Jun 15, 2023 at 18:09

1 Answer 1

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This is not going to be an answer, but I want to share my observations that I have made in the meantime (heavily focused around degree four and five polynomials for testing):

  • Using GAP and brute-force, I have calculated the stabilizers of all sums of monomials $X_1^{a_1}X_2^{a_2}X_3^{a_3}X_4^{a_4}$ under the action of a transitive group $G \leq S_4$ for $a_i = 0, \ldots,3$. Higher degrees are obviously not needed because we are only focusing on the distinctness of the variables. Eyeballing the results I get the following -- in my opinion -- "nicer" generating polynomials up to permutation of indices: $$\begin{array}{rl} C_4:& X_1^2X_2+X_2^2X_3+X_3^2X_4+X_4^2X_1 \\ C_2 \times C_2:& \text{None}\\ D_4:& X_1 X_2 + X_3 X_4\\ A_4:& \text{None}\\ S_4:& X_1+X_2+X_3+X_4 \end{array}$$ This is, of course, not a full analysis if there should be polynomials with non-trivial stabilizers not generated by monomials (examples?). Let's look at the transitive groups of polynomials of degree $5$: $$\begin{array}{rl} C_5:& X_1^2 X_2+X_2^2X_3+X_3^2X_4+X_4^2X_5 + X_5^2X_1\\ D_5:& X_1X_2+X_2X_3+X_3X_4+X_4X_5 + X_5X_1\\ F_5:& X_1^2X_2X_5+X_1^2X_3X_4+X_1X_2^2X_3+X_1X_2X_4^2+X_1X_3^2X_5+X_1X_4X_5^2+X_2^2X_4X_5+X_2X_3^2X_4+X_2X_3X_5^2+X_3X_4^2X_5\\ A_5:& \text{None}\\ S_4:& X_1+X_2+X_3+X_4+X_5 \end{array}$$

  • Now for the "fun" part: Let's calculate resolvents for some of these groups! Let $f(x) = x^4+ax^3+bx^2+cx+d \in \mathbb{Q}[x]$ denote an irreducible polynomial, then we get the following resolvents: $$\begin{array}{rl} C_4:& x^6+(-2ab + 6c)x^5+(a^2 b^2+2 b^3+2 a^3 c-17 a b c+24 c^2+a^2 d-8 b d) x^4+(-2 a b^4-2 a^4 b c+10 a^2 b^2 c+8 b^3 c+8 a^3 c^2-54 a b c^2+56 c^3-2 a^5 d+10 a^3 b d-12 a^2 c d-32 b c d+32 a d^2 ) x^3+(b^6+3 a^3 b^3 c-13 a b^4 c+a^6 c^2-13 a^4 b c^2+33 a^2 b^2 c^2+22 b^3 c^2+22 a^3 c^3-116 a b c^3+96 c^4+2 a^6 b d-11 a^4 b^2 d+10 a^2 b^3 d-12 b^4 d-5 a^5 c d+30 a^3 b c d+48 a b^2 c d-36 a^2 c^2 d-120 b c^2 d+7 a^4 d^2-72 a^2 b d^2+48 b^2 d^2+144 a c d^2-64 d^3)x^2+ (-a^2 b^5 c+2 b^6 c-a^5 b^2 c^2+10 a^3 b^3 c^2-19 a b^4 c^2+2 a^6 c^3-19 a^4 b c^3+34 a^2 b^2 c^3+28 b^3 c^3+28 a^3 c^4-128 a b c^4+96 c^5-2 a^5 b^3 d+9 a^3 b^4 d-2 a^8 c d+22 a^6 b c d-60 a^4 b^2 c d-4 a^2 b^3 c d-24 b^4 c d-29 a^5 c^2 d+134 a^3 b c^2 d+120 a b^2 c^2 d-104 a^2 c^3 d-176 b c^3 d-a^7 d^2+24 a^3 b^2 d^2+38 a^4 c d^2-288 a^2 b c d^2+96 b^2 c d^2+336 a c^2 d^2+16 a^3 d^3-128 c d^3)x+ b^6 c^2+2 a^3 b^3 c^3-12 a b^4 c^3+a^6 c^4-12 a^4 b c^4+36 a^2 b^2 c^4+16 b^3 c^4+16 a^3 c^5-96 a b c^5+64 c^6+a^4 b^5 d-4 a^2 b^6 d+a^7 b^2 c d-12 a^5 b^3 c d+29 a^3 b^4 c d+16 a b^5 c d-2 a^8 c^2 d+17 a^6 b c^2 d-6 a^4 b^2 c^2 d-134 a^2 b^3 c^2 d-28 b^4 c^2 d-22 a^5 c^3 d+48 a^3 b c^3 d+296 a b^2 c^3 d-48 a^2 c^4 d-224 b c^4 d+a^10 d^2-11 a^8 b d^2+45 a^6 b^2 d^2-76 a^4 b^3 d^2+32 a^2 b^4 d^2+15 a^7 c d^2-133 a^5 b c d^2+344 a^3 b^2 c d^2-128 a b^3 c d^2+128 a^4 c^2 d^2-648 a^2 b c^2 d^2+176 b^2 c^2 d^2+416 a c^3 d^2-5 a^6 d^3+32 a^4 b d^3-64 a^2 b^2 d^3-48 a^3 c d^3+256 a b c d^3-320 c^2 d^3 \\ C_2 \times C_2:& [TODO] \\ D_4:& x^3-bx^2+(ac-4d)x-a^2d+4bd-c^2\\ A_4:& x^2-\Delta\\ \end{array}$$ where $\Delta$ is the discriminant of $f$. Hm... as we can see, even relatively simple invariants can lead to headache-inducing formulas which are far too unwieldy without a CAS at hand, but it is a nice exercise anyway!

  • By the orbit-stabilizer theorem we immediately get that $\text{deg}(R_G(f)) = \frac{|S_n|}{\text{Stab}_{S_n}(g)} = \frac{n!}{|G|}$ for some resolvent invariant $g$ for $G$. This implies that e.g. resolvents for cyclic groups will tend to be absolutely gigantic since their orbit under $S_n$ will contain many terms that may also need a lot of cross-cancelling with other elementary symmetric polynomials to obtain the desired coefficients.

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