The title pretty much sums it up: do people know of nice parametrized families of polynomials (with integer coefficients) with square discriminant. I should say that one such family consists of trinomials of the form $x^n + a x^k + b,$ where $(n, k)$ is even (there are some conditions on $a, b$ which I don't remember), but that is not satisfying, since the Galois group of such things is very special (since they have the form $f(x^2).$)
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asked Nov 24, 2015 at 21:44
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2$\begingroup$ $x^n + n x^{n-1} + y$ with $n$ odd and $y = \frac{(n-1)^{(n-1)}}{(-1)^{(n-1)/2} n \cdot t^2 - 1}$. The Galois group will generically be $A_n$. (You can clear denominators to make this integral.) $\endgroup$– Electric PenguinCommented Nov 25, 2015 at 1:29
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1$\begingroup$ Must the polynomials be irreducible? $\endgroup$– joroCommented Nov 25, 2015 at 5:43
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$\begingroup$ @user78204 Where does this come from? And why will the Galois group be $A_n$ generically (I believe you, but for specific families, that is usually not trivial...) $\endgroup$– Igor RivinCommented Nov 25, 2015 at 11:12
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$\begingroup$ @joro irreducible is certainly best... $\endgroup$– Igor RivinCommented Nov 25, 2015 at 11:13
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2$\begingroup$ The Galois group of $x^n + a x + b$ over $\mathbf{Q}(a,b)$ is $S_n$ (easy to find specializations where inertia at a prime $p$ gives a transposition). Solving $\Delta = z^2$ over $\mathbf{Q}(a,b)$ thus gives a family with Galois group $A_n$. Clearly one can scale $a$ to any constant, so this gives an $A_n$ family over (what turns out to be) a ramified genus zero curve over $\mathbf{Q}(b)$, which is $\mathbf{Q}(t)$ for some $t$. Do the calculation and you get the family I wrote down. $\endgroup$– Electric PenguinCommented Nov 25, 2015 at 20:59
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3 Answers
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You want families of polynomials with Galois group contained in $A_n$. A generic polynomial for a group $G$ is a polynomial with indeterminates with Galois group $G$ so that every extension of the base field with that Galois group arises from evaluating the indeterminates. Of course, some evaluations will have smaller Galois groups. Two examples:
$A_3: x^3-tx^2+(t-3)x+1,$ discriminant $(t^2-3t+9)^2$
$D_5: x^5+(t-3)x^4+(s-t+3)x^3+(t^2-t-2s-1)x^2+s x + t,$ discriminant $t^2(4t^5-4t^4-24st^3-...+14st-4t)^2$
answered Nov 25, 2015 at 1:59
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$\begingroup$ Nice, but where does this come from? $\endgroup$ Commented Nov 25, 2015 at 11:04
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$\begingroup$ Do you mean the technique of generic polynomials? That goes back to Noether's approach to the Inverse Galois Problem or before. Hilbert's Irreducibility Theorem applies to generic polynomials. If you mean how one determines generic polynomials for particular groups, I don't know the main techniques used. Noether's approach creates some algebraic problems that may take an algebraist or algebraic geometer to solve. A review of Generic Polynomials by Jensen, Ledet, and Yui says they construct a lot of examples for small and medium-sized groups. math.ubc.ca/~reichst/review.pdf $\endgroup$ Commented Nov 25, 2015 at 18:36
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$\begingroup$ Here is the book itself: library.msri.org/books/Book45/files/book45.pdf $\endgroup$ Commented Nov 25, 2015 at 18:38
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$\begingroup$ Ah, ok, so the family comes from JLY?! I thought you might have had some simple reason why this family would work... $\endgroup$ Commented Nov 25, 2015 at 19:17
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1$\begingroup$ These families are on the linked Wikipedia page for generic polynomials. I checked that the discriminants are squares, but not that the second polynomial has a Galois group within $D_5$. $\endgroup$ Commented Nov 25, 2015 at 21:20
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There is a beautiful construction by Mestre [1] (see also Prop. I.5.12 in [2]) which, for fixed odd degree $n$, yields an $n+1$-parametric family of such polynomials: Let $z,t_1,t_2,\dots,t_n$ be indeterminates and $K=\mathbb Q(t_1,\dots,t_n)$ and $g(X)=(X-t_1)\dots(X-t_n)$. Then there is a polynomial $h(X)\in K[X]$ of degree $n-1$ and relatively prime to $g(X)$ such that $g(X)-zh(X)$ has Galois group $A_n$ over $K(z)$. Here $h(X)$ can be computed by solving a system of linear equations.
So specializing the $t_i$'s and $z$ in $\mathbb Q$ give polynomials over $\mathbb Q$ with square discriminants, and usually with Galois group $A_n$.
In order to handle even degree $n-1$, take $g(X)$ and $h(X)$ as above and set $F(X)=\frac{g(X)h(z)-h(z)g(X)}{X-z}$. Then $F(X)$ has Galois group $A_{n-1}$ over $\mathbb K(z)$.
[1] Mestre, Jean-François: Constructions polynomiales et théorie de Galois. Proceedings of the International Congress of Mathematicians, Vol. 1, 2 (Zürich, 1994), 318–323, Birkhäuser, Basel, 1995.
[2] Malle, Gunter; Matzat, B. Heinrich: Inverse Galois theory. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 1999.
answered Nov 25, 2015 at 8:18
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Here is parametrization involving rational numbers for the family $x^{2n+1}+a x^{2n}+b$.
The discriminant factors as $\pm b^{2k}b(Ca^m+Db)$ for integers $k,m,C,D$. To make is square, consider $\pm b(Ca+Db)=\square$.
Solve either $b=Ca^m+Db$ or $-b=Ca^m+Db$ for $b$, which is linear in $b$ depending on the sign.
For $n=3$, we get $b=-{\frac {5832}{102943}}\,{a}^{7}$.
The discriminant is ${\frac {39346408075296537575424}{1190093060316451166187740221249}}\,{a}^{42}$, which is sixth power.
Here are the first few discriminants
2n+1 discriminant
3 (-1) * b * (4*a^3 + 27*b)
5 b^3 * (256*a^5 + 3125*b)
7 (-1) * b^5 * (46656*a^7 + 823543*b)
9 b^7 * (16777216*a^9 + 387420489*b)
11 (-1) * b^9 * (10000000000*a^11 + 285311670611*b)
13 b^11 * (8916100448256*a^13 + 302875106592253*b)
15 (-1) * b^13 * (11112006825558016*a^15 + 437893890380859375*b)
17 b^15 * (18446744073709551616*a^17 + 827240261886336764177*b)
Another approach is to parametrize the genus $0$ curve, quadratic in $b$, but this appears more complicated to me.
