<s>[INST] 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 ${C}_{4}$.
 [/INST] 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.
</s>