# Generic polynomial for alternating group ${A}_{4}$ is not correct

I was validating the percentage of cases where the generic two parameter polynomial for Galois group ${A}_{4}$ is valid. We have

\begin{equation*} {f}^{{A}_{4}} \left({x, \alpha, \beta}\right) = {x}^{4} - \frac{6\, A}{B}\, {x}^{3} - 8\, x + \frac{1}{{B}^{2}} \left({9\, {A}^{2} - 12 \left({{\alpha}^{3} - {\beta}^{3} + 27}\right) B}\right) \in K \left({\alpha, \beta}\right) \left[{x}\right] \end{equation*}

where

\begin{equation*} A = {\alpha}^{3} - {\beta}^{3} - 9\, {\beta}^{2} - 27\, \beta - 54 \end{equation*}

and

\begin{equation*} B = {\alpha}^{3} - 3\, \alpha\, {\beta}^{2} + 2\, {\beta}^{3} - 9\, \alpha\, \beta + 9\, {\beta}^{2} - 27 \left({\alpha - \beta - 1}\right). \end{equation*}

from Arne Ledet "Constructing Generic Polynomials", Proceedings of the Workshop on Number Theory, Institute of Mathematics, Waseda University, Tokyo, 2001. When testing for $- 100 \le \alpha, \beta \le + 100$ we have 99.990% of the irreducible cases belonging to the ${S}_{4}$ group and 0.005% of the remaining cases belonging to the ${A}_{4}$ and ${D}_{4}$ groups, respectively.

What is the correction if known and are there other two parameter cases known for the ${A}_{4}$ group? I do have from Gene Smith ("Some Polynomials over $\mathbb{Q} \left({t}\right)$ and their Galois groups", Mathematics of Computation, 69(230):775-796, August 1999.) the example ${f}^{{A}_{4}} \left({x, t}\right) = {x}^{4} + 18\, t\, {x}^{3} + \left({81\, {t}^{2} + 2}\right)\, {x}^{2} + 2\, t \left({54\, {t}^{2} + 1}\right) x + 1 \in K \left({t}\right) \left[{x}\right]$ which is valid and I encountered a five parameter case which I have not yet tested. I have validated the other common examples for ${S}_{4}$, ${V}_{4}$, ${D}_{4}$, and ${C}_{4}$.

• I find it hard to figure out what the question actually is... – Igor Rivin Oct 12 '15 at 21:12
• My question this generic polynomial is not correct from what I have computed. Does anyone have the corrected version and are there other examples, say of the two parameter cases, that are known. I have only found these three cases and the main one that I found the error in is repeated in several papers. For example, I have tried sign variation of the three terms-all eight possible cases as a simple search for the error and found none. – Lorenz H Menke Oct 12 '15 at 21:20
• It might be worth checking Jensen, Ledet, and Yui, Generic Polynomials, MSRI Publications 45, Cambrdige (2002), MR1969648 (2004d:12007), if you haven't already. – Gerry Myerson Oct 12 '15 at 22:15

The Galois group is a subgroup of $A^4$ if and only if the discriminant is a perfect square. If you change $x^3$ to $x^2$ you get the discriminant to be: $$\frac{1728^2 \left(b^2+3 b+9\right)^2 \left(a^3 (2 b+3)-3 a^2 \left(b^2+3 b+9\right)+\left(b^2+3 b+9\right)^2\right)^2}{\left(a^3-3 a \left(b^2+3 b+9\right)+2 b^3+9 b^2+27 b+27\right)^4},$$ so that definitely works. The discriminant is definitely NOT a square in the original version.

• Yes I checked the references and made the calculations and it is now correct. Thanks – Lorenz H Menke Oct 13 '15 at 17:19

(A version of) Ledet's paper can be found here. The formula given in that version has $x^2$ where the question here has $x^3$. Perhaps this accounts for the difficulties.

• OK I see it, I will check/compute tonight that is likely the error. Thanks – Lorenz H Menke Oct 12 '15 at 23:33
• The formula with $x^2$ is also the formula given in their book "Generic Polynomials -- Constructive Aspects of the Inverse Galois Problem" (C.U. Jensen, A. Ledet, N. Yui, 2002), p. 37. – Tom De Medts Oct 13 '15 at 7:19