Let $G$ be the Galois group of an irreducible polynomials $f(x)$ in $\mathbb{Q}[x]$. Let $K$ be the splitting field of $f(x)$.

From the fundamental theorem of Galois theory we have that the intermediate fields between $K$ and $\mathbb{Q}$ are in bijective correspondence with the subgroups of $G$.

Is there any way we can get the polynomials whose Galois groups are the subgroups of $G$, given $f(x)$ ?

To put it in an other way is there any relation between $f(x)$ and the polynomials corresponding to the normal extensions of $\mathbb{Q}$ contained in $K$ ?

By polynomial corresponding to normal extensions I mean, given a normal extension $E$ of $\mathbb{Q}$ contained in $K$, The polynomial $p(x)$ whose splitting field is $E$.

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    $\begingroup$ The way you have stated the question is, I think, not what you want. For each subgroup $H$ of $G$, there are many polynomials whose Galois group is $H$. Also, in the last sentence, surely you don't want $f$, and surely you don't want to settle for isomorphism (as there are polynomials whose groups are isomorphic to the group of $E$ without the polynomial having anything to do with $E$). I think the question you want to ask is, given $f$ and a normal subfield $E$ of the splitting field of $f$, find $p$ such that $E$ is the splitting field of $p$. $\endgroup$ – Gerry Myerson Oct 21 '11 at 23:10
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    $\begingroup$ I do not understand what you mean by "..surely you don't want f, and surely you don't want to settle for isomorphism" but I am aware that there are many polynomials for E. By the way does the polynomial has got nothing to do with $E$? Of course $E$ is the splitting field of it right? But is the condition strong enough to talk something about the polynomials of $E$ in terms of $f(x)$? I am not sure!(I am actually presuming the person answering the question will understand any fallacies in formulation of this question, if it means anything) ...cont $\endgroup$ – Dinesh Oct 22 '11 at 5:00
  • $\begingroup$ (cont) And yes, what you thought what I meant to ask is correct, though I can not clearly see a major difference between what you thought and what I asked. @GerryMyerson $\endgroup$ – Dinesh Oct 22 '11 at 5:00
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    $\begingroup$ OK, I'll spell out some of the differences. You write "$G(f,{\bf Q})$ is isomorphic to $G(E/{\bf Q})$" but you've already used $f$ for a polynomial whose splitting field is $K$ - that's why I said you don't want $f$. Also, say $G(E/{\bf Q})$ is ${\bf Z}_2$, well, there are tons of polynomials $p$ such that $G(p,{\bf Q})$ is isomorphic to ${\bf Z}_2$ and most of them have nothing to do with the field $E$ but with other fields entirely. That's not what you want, but it is what you have asked, in the last sentence of your question. $\endgroup$ – Gerry Myerson Oct 22 '11 at 5:38
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    $\begingroup$ @Dinesh: I think your final paragraph is still wrong. You want to find $p(x)$ whose splitting field is $E$, no? Otherwise, say $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then for any proper intermediate field, I could say "take $p(x) = x^2-5$", because the splitting field of $p(x)$ has degree 2 over $\mathbb{Q}$, hence will have Galois group isomorphic to the Galois group of any normal $E$ with $\mathbb{Q}\subset E\subset K$. What you really want, I think, is that if I give you $E$, you want to construct a $p(x)$ with splitting field $E$, starting from the polynomial defining $K$. $\endgroup$ – Arturo Magidin Oct 22 '11 at 6:17

I'm going to (try to) answer the question the way I phrased it in my first comment. It depends a bit on just what one is given. I'll take the easy way out and assume one is given a basis $\lbrace\alpha_1,\dots,\alpha_r\rbrace$ for $E$ over $\bf Q$ (I might also assume given a basis for $K$ over $\bf Q$, and the actions of the elements of $G$ on that basis, and which elements of $G$ are in the subgroup corresponding to the field $E$). Then with a little experimentation one can find $\beta=c_1\alpha_1+\cdots+c_r\alpha_r$ with $c_i$ integers such that $E={\bf Q}(\beta)$. Then you can find all the conjugates of $\beta$ over $\bf Q$ and use that to find the minimal polynomial $p$ for $\beta$ over $\bf Q$. And $p$ is what you want.

If what you really want is some way of just staring at $f$ long enough and hard enough for $p$ to appear, I suspect you're asking for too much.

  • $\begingroup$ Yes..I kind of knew that I was asking for too much but I just wanted to know what can be said about it, if not getting a complete solution. One of the motivations for this question is that, given a prime $p$ there exists infinitely many primes of the form $q=1+pk$, by considering $K=\mathbb{Q}(\zeta_{q})$, the Group $G:=Gal(K/\mathbb{Q})$ will contain a subgroup(cyclic group of order $k$) which is in turn Galois group of a $p-extension$, so I want to know how this $p$ extension will look like hence the minimal polynomial of the $\beta$ you were talking about or the other way. $\endgroup$ – Dinesh Oct 23 '11 at 8:20
  • $\begingroup$ more over finding $c_i$ or getting some explicit information about them was the major problem I was facing when I was thinking about this. $\endgroup$ – Dinesh Oct 23 '11 at 8:23
  • $\begingroup$ Maybe you should ask the question you really want answered, then. Ask a question about the $c_i$. Or ask a question about the $p$-extension. $\endgroup$ – Gerry Myerson Oct 23 '11 at 11:53
  • $\begingroup$ I just wanted to know what can be said in a more general situation (a mathematical habit inherently in me from years, don't know if it is good/bad) than the $p$-extensions, nevertheless I shall ask a question about it separately sooner or later. Not much seems to be working well this question. Thanks for your answer and patient comments and criticism. $\endgroup$ – Dinesh Oct 23 '11 at 17:08

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