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It's known that all abelian groups are regularly realizable over $\mathbb{Q}(x)$, but it occurred to me that I don't even have an example of a cyclic regular extension of $\mathbb{Q}(x)$ handy.

So: what is an example of a regular realization of $C_5$ over $\mathbb{Q}(x)$?

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Dear James: Geometrically conn'd finite abelian coverings of geom. conn'd smooth proj. curves over a perfect field $k$ (such as $\mathbf{Q}$, $\mathbf{C}$, $\mathbf{F}_q$, etc.) can be understood nicely via generalized Jacobians. One constructs such covers of the curve by pullback of isogenies to an appropriate generalized Jacobian (depending on ramification). This provides insight into covers much as class field theory does for global fields (with its classical "modulus"), not giving "explicit" eqns for the abelian ext'n. Would you be happy with a construction along such lines? –  BCnrd Dec 12 '10 at 21:46
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Dear James: Jensen, Ledet and Yui in their book Generic polynomials: constructive aspects of the inverse Galois problem describe both generic $C_5$ and $D_5$-extensions. The google book reference is books.google.co.uk/books?isbn=0521819989 and their family for $C_5$ is on page 44 –  Tim Dokchitser Dec 12 '10 at 22:56
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At the start of sect. 3 of the paper math.ucalgary.ca/~aksilves/papers/lecacheux.pdf a messy quintic polynomial f_t(X) is described with coeff. in Z[t] (Lecacheux's quintics). Theorem 2 in the paper says that for rational t, if f_t(X) is irred. over Q then its Galois group over Q is Z/5Z. If instead we let t be an indeterminate and you can show f_t(X) to be irred. in Q(t)[X], then calculations in section 3 of the paper show the Galois group of f_t(X) over Q(t) is either Z/5 or D_5 and probably the calculations that eliminate D_5 if t is in Q also work if t is indeterminate. –  KConrad Dec 12 '10 at 23:17
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A simpler example than the one in my previous comment is Emma Lehmer's quintic, which is given on the first page of Darmon's article jstor.org/pss/2008408. Darmon gives a reference at the bottom of the first page for a proof that Lehmer's quintic defines a regular Z/5Z Galois extension of Q(t). –  KConrad Dec 12 '10 at 23:22
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James, it's not bizarre that an example is complicated. Over Q if you want to prove the existence of a cyclic extension of degree n, the easiest way I know is to first see that (Z/p)* is a Galois group over Q using the explicit and non-complicated cyclotomic polynomial (x^p-1)/(x-1), and then for each n you find a prime p which is 1 mod n and pick out the degree-n subextension of the p-th cyclotomic field. This argument does not give you an explicit polynomial defining the degree-n extension of Q and you can't expect the explicit realization of such a polynomial to be a simple polynomial. –  KConrad Dec 12 '10 at 23:38

2 Answers 2

up vote 8 down vote accepted

Emma Lehmer's quintic \[\begin{align*} y^5 +& x^2y^4 - (2x^3 + 6x^2 + 10x + 10)y^3 +\\ &(x^4 + 5x^3 + 11x^2 + 15x + 5)y^2 + (x^3 + 4x^2 +10x + 10)y + 1 \end{align*}\] has Galois group $C_5$ over ${\mathbf Q}(x)$ and the splitting field over ${\mathbf Q}(x)$ is a regular extension. Her paper is "Connection between Gaussian periods and cyclic units" Math. Comp. 50 (1988), 535--541.

Typing "Lehmer quintic" in Google produces several hits which will give more information.

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I wanted to make a certain comment, but it got too long, so here it is as an "answer" that is constructive in principle (and applicable over any field of characteristic 0, suitable to make cyclic regular covers ramified in a controlled locus).

Let $k$ be a field of characteristic 0 and let $X$ be the projective line over $k$, so its usual Jacobian vanishes. Thus, its generalized Jacobians are affine. Being commutative and connected, they're products of a unipotent $k$-group and a $k$-torus, so isogenies to them from smooth connected commutative $k$-groups must arise from the torus part (as we're in char. 0). Hence, in language of geometric class field theory (see sec. 2 in Ch. I and sec. 3 in Ch. V of Serre's book "Algebraic groups and class fields"), we can stick with reduced moduli; i.e., reduced divisors $D$ on $X$ and their generalized Jacobians $J_D$, which are $k$-tori.

The generalized Jacobian is isomorphic to the Picard variety ${\rm{Pic}}^0_{X_D/k}$ where $X_D$ is the geometrically integral curve over $k$ obtained by "crushing $D$ to a $k$-point" (defined via Galois descent of an analogous procedure over a splitting field of $D$). By 9.2/10 in "Neron Models", this is described rather explicitly as a $k$-torus. It is the quotient of a product of Weil restrictions: $$J_D = (\prod_{x \in D} {\rm{R}}_{k(x)/k}(\mathbf{G}_m))/\mathbf{G}_m.$$
We seek connected commutative extensions of this $k$-torus by (cyclic) finite constant $k$-groups.

A cofinal system of tori equipped with isogenies onto $J_D$ over $k$ is given by the multiplication maps $[n]:J_D \rightarrow J_D$, whose kernels are (by the snake lemma with commutative $k$-groups) $$J_D[n] = (\prod_{x \in D} {\rm{R}}_{k(x)/k}(\mu_n))/\mu_n.$$ as $k$-groups. Let $C_{D,n} = J_D[n]/H_{D,n}$ denote the maximal $k$-group quotient of $J_D[n]$ that is constant (i.e., trivial as a Galois module). Using $\infty$ as a $k$-rational base point and assuming $\infty\not\in D$, there is a canonical $k$-morphism $$f_D:X - D \rightarrow J_D,$$ and geometrc CFT says that the $f_D$-pullbacks of the $C_{D,n}$-coverings $J_D/H_{D,n} \rightarrow J_D$ are a cofinal system of geometrically connected finite abelian coverings of $X$ (i.e., regular finite abelian extensions of $k(X)$) unramified outside of $D$. So if $f_D$ could be made explicit, then we'd have a pretty "explicit" way of making finite abelian regular extensions of $X$ unramified outside of $D$, provided we can understand the structure of the Galois module of geometric points of $J_D[n]$. For $k = \mathbf{Q}$ this Galois module seems like something one can understand pretty systematically, provided that one understands how $\mathbf{Q}(\zeta_n)$ interacts with each of the fields $k(x)$ for $x \in D$.

Example: Suppose $n = p$ is an odd prime and $k = \mathbf{Q}$ and $D$ is the single closed point $x \in \mathbf{A}^1_k$ with degree $p-1 > 1$ corresponding to the primitive $p$th roots of unity. Then ${\rm{R}}_{k(x)/k}(\mu_n)$ has the $p$-torsion Galois module of geometric points given by $\oplus_{0 \le i < p} \omega^i$ where $\omega$ is the mod-$p$ cyclotomic character, with $\mu_p = \omega$ embedded in the evident manner (the factor for $i = 1$). Hence, there's exactly one cyclic quotient of order $p$ with trivial Galois action (projection to the factor for $i = 0$) and so we conclude that $\mathbf{P}^1_{\mathbf{Q}}$ admits a unique degree-$p$ geometrically connected cyclic covering whose ramification is supported at precisely the primitive $p$th roots of unity. This corresponds to a particularly natural degree-$p$ regular abelian extension of $\mathbf{Q}(t)$. What is it? (I have no idea, even for $p = 3$ or $p = 5$.)

In general, here is how I think $f_D$ can be made explicit. Change coordinates so $0 \not\in D$ (no problem, since $X(k)$ is infinite), and increase $D$ (harmless) so that $\infty \in D$. Then $D' := D - \{\infty\}$ is the zero locus of a unique separable $h \in k[t]$ such that $h(0) = 1$, and $J_D = \prod_{x \in D'} {\rm{R}}_{k(x)/k}(\mathbf{G}_m)$. For $x \in D'$, let $h_x|h$ be the unique irreducible factor over $k$ with $h_x(0) = 1$ and $h_x(x) = 0$, and let $y_x = y \bmod h_x(y) \in k[y]/(h_x) = k(x)$ be the "canonical" zero of $h_x$ in $k(x)$. The $x$th component of $f_D$ corresponds to a unit in $k(x)[t][1/h]$. This unit is probably $1 - (t/y_x)$ or its inverse (but I have not checked rigorously).

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