Is there a polynomial $f(T,X) \in \mathbb{Q}(T)[X]$ in the indeterminate $X$ over the field $\mathbb{Q}(T)$ with $\mathrm{Gal}(f/\mathbb{Q}(T)) \cong \mathbb{Z}/5\mathbb{Z}$ such that for every Galois extension $K/\mathbb{Q}$ with $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/5\mathbb{Z}$ there exists some $t \in \mathbb{Q}$ such that the splitting field of $f(t,X)$ over $\mathbb{Q}$ is isomorphic to $K$.
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3$\begingroup$ Can one do the analogous problem for degree 3? $\endgroup$ – Gerry Myerson Mar 2 '16 at 22:19
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$\begingroup$ @GerryMyerson for degree $3$ the answer to the question is positive. $\endgroup$ – Pablo Mar 3 '16 at 7:09
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2$\begingroup$ So, where does the story for 3 break down when you try to tell it for 5? $\endgroup$ – Gerry Myerson Mar 3 '16 at 9:51
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3$\begingroup$ @GerryMyerson the short version of the story is that finding such a polynomial for $\mathbb{Z}/3\mathbb{Z}$ amounts to finding a rational point on a curve of genus zero, which is not the case for $5$ (the genus becomes higher). This should be described in the book 'Generic polynomials' and an example of a degree $3$ polynomial $f$ is given there. $\endgroup$ – Pablo Mar 3 '16 at 20:43
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Evidently no. In "Generic Polynomials Constructive Aspects of the Inverse Galois Problem" by Jensen, Ledet, Yui (2002): generic dimension (i.e. minimum number of parameters in a generic polynomial) of $\mathbb{Z}/5$ over $\mathbb{Q} = 2$ (p.203) and moreover essential dimension (i.e. minimum transcendence degree of parameters) of $\mathbb{Z}/5$ over $\mathbb{Q} = 2$ (p.190). They conjecture (p.202) that if generic dimension is finite then essential dimension = generic dimension (over any field). One story for $\mathbb{Z}/3$ that breaks down for $\mathbb{Z}/5$ uses an action of $\mathbb{Z}/3$ on $\mathbb{P}^1_{\mathbb{Q}}$.
Edit: As Pablo noted, this answer is about a generic polynomial over $\mathbb{Q}$ but it doesn't answer the original question about existence of a 1-parameter parametric polynomial over $\mathbb{Q}$. I didn't appreciate this nuance in the original question when I wrote the above answer.
answered Mar 3 '16 at 23:01
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1$\begingroup$ This does not answer my question. $C_5$ has essential dimension $2$ over the rationals. Then the generic dimension of $C_5$ is at least $2$ (this is Proposition 8.5.2 of the book mentioned). Hence there is no one parameter generic polynomial $P(T,Y)$ of group $C_5$ over the rationals. By the definition, it means that there exists some field $k$ of characteristic $0$ and some cyclic Galois extension of $k$ of degree 5 which can't be obtained by specializing $P(T,Y)$. To show that there is no parametric extension (this is what I want), you would have to prove that one can take $k=\mathbb{Q}$. $\endgroup$ – Pablo Mar 4 '16 at 16:39
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1$\begingroup$ @Pablo Thanks for your clarification. I added an Edit about it in the answer. $\endgroup$ – David Lampert Mar 4 '16 at 17:03
