So I want to find a degree $7$ polynomial whose Galois group is cyclic of order $7$. I know that to do this, I need to find a polynomial which is irreducible mod every prime (not dividing the discriminant). But this is where I'm stuck, since I don't know how to prove that any particular polynomial is irreducible mod every prime. Could I have some hints?
[Math] Polynomial of degree $7$ with cyclic Galois group
abstract-algebragalois-theory
Related Solutions
To construct an Galois extension of $\mathbb{Q}$ of order $p$, take an integer $N$ such that $(\mathbb{Z}/N\mathbb{Z})^*$ maps surjectively onto $\mathbb{Z}/p\mathbb{Z}$, then the field $\mathbb{Q}(\zeta_N)$ contains a subfield $K$ such that $[K:\mathbb{Q}]=p$ and $K$ is Galois over $\mathbb{Q}$ (Here $\zeta_N$ is a primitive $N$-th root of unity). For example, to find a galois extension of $\mathbb{Q}$ of degree 5, you may look at the subfield of $\mathbb{Q}(\zeta_{11})$ that's fixed by the complex conjugation. Kronecker-weber Theorem states that every finite abelian extensions of $\mathbb{Q}$ is a subfield of cyclotomic fields, so every Galois extension of $\mathbb{Q}$ of prime order arises this way.
Continue the example with prime $p=5$, to find all of such extensions, we will look at the sequence $5n+1$. By Dirichlet's theorem on primes in arithmetic progressions, there are infinitely many primes $q$ in this arithmetic progression, each $\mathbb{Q}(\zeta_q)$ contains a subfield which is a degree 5 extension of $\mathbb{Q}$. All these subfields are distinct because they ramifies at different primes. Last, note that that there is also a subfield of $\mathbb{Q}(\zeta_{25})$ which is of degree 5 over $\mathbb{Q}$.
Edit: I was wrong about my complete list statement. For example, $\mathbb{Q}(\zeta_{341})$ (composite of $\mathbb{Q}(\zeta_{11})$ and $\mathbb{Q}(\zeta_{31})$) has 6 subfields which are degree 5 abelian extensions of $\mathbb{Q}$. Apologies.
I did the calculation… This was presumably not what was intended...
Recall from earlier in the assignment that $x^3-2$ has Galois group $S_3$, so it would suffice to find an irreducible polynomial of degree 6 in $\mathbb Q[x]$ with the same splitting field as that of $x^3-2$.
We know more, in fact, since we created the specific action of each group element on the splitting field $L$. Therefore, choose some $\alpha$ which is in no proper subfields of $L$, which is possible because $L$ is a vector space over $\mathbb Q$ and so is not a union of finitely many of its linear subspaces; in particular it is not the union of its proper subfields. Now, let $f(x)$ be the polynomial: $$f(x) = \prod_{\pi\in S_3} (x-\pi(\alpha)).$$ Evidently this has degree six. It is also irreducible, since $\mathbb Q(\alpha)$ cannot be any proper subfield of $L$ and so $\deg_\mathbb Q(\alpha)=[L:\mathbb Q]=6$. Therefore since $L$ contains the splitting field of $f$, it is the splitting field of $f$. But since $L$ is a subfield of the splitting field of the minimal polynomial of $\alpha$ over $\mathbb Q$.
Finally, we claim that its coefficients are rational. At least they are real, since if we write $S_3=\langle\sigma,\tau\rangle$ with $\tau$ representing conjugation and $\sigma$ representing the three-cycle of the roots of $x^3-2$, then $(x-\sigma^n(\alpha))$ and $(x-\tau\sigma^n(\alpha))$ are both factors of $f$ representing conjugate pairs of roots, which implies that $f\in \mathbb R[x]$. We now wish to show that they are in $\mathbb Q(\omega)$, where $\omega$ is a cube root of unity. Since any real in $\mathbb Q(\omega)$ is rational, we must have that the coefficients of $f$ are rational.
We being by observing that $\omega^m$ is a spanning set for $\mathbb Q(\omega)$ and $\sqrt[3]2^n$ is a spanning set for $\mathbb Q(\sqrt[3]2)$, so $L$ is spanned by the products $\sqrt[3]2^n\omega^m$, where $0\leq n,m<3$. Let $\sigma(\sqrt[3]2)=\sqrt[3]2\omega$, then we know by considering ratios that $\sigma(\omega)=\omega$.
Now, suppose that $\alpha=\sum a_{m,n}\sqrt[3]2^n\omega^m$, then $\sigma(\alpha) = \sum a_{m,n}\sigma(\sqrt[3]2^n)\sigma(\omega^m) = \sum a_{m,n}\sqrt[3]2^n\omega^{n+m}$. Similarly, $\sigma^2(\alpha) = \sum a_{m,n}\sqrt[3]2^n\omega^{2n+m}$. Again, these may not be well-defined, because we have not guaranteed that this spanning set is a basis (and indeed it is not). However, define $\sigma(0)=\sigma^2(0)=0$, suppose that $\alpha$ also equals $\sum b_{m,n}\sqrt[3]2^n\omega^m$, and consider: $$0= \sigma(0)= \sigma\left(\sum_{m,n=0}^2 (b_{m,n}-a_{m,n})\sqrt[3]2^n\omega^m\right)=\sum_{m,n=0}^2 b_{m,n}\sigma(\sqrt[3]2^n\omega^m)-\sum_{m,n=0}^2 a_{m,n}\sigma(\sqrt[3]2^n\omega^m).$$
So this automorphism is well-defined after all; the same trick shows $\sigma^2$ is well-defined as well.
Now we return to our product, and observe that we may split it as follows: $$f(x) = \prod_{k=0}^2 (x-\sigma^k(\alpha))\prod_{k=0}^2 (x-\overline{\sigma^k(\alpha)}).$$ Therefore, it suffices to show that the first product has terms in $\mathbb Q(\omega)$, since the terms in the second product are simply their conjugates. The reader now has two options: we may wade through the algebra together, or you may simply avert your eyes for the next four pages and trust that the calculations are correct. The author sympathizes, and will not think less of you if you choose the second option. \begin{align*} \prod_{k=0}^2 (x-\sigma^k(\alpha)) &= (x-\alpha)(x-\sigma(\alpha))(x-\sigma^2(\alpha)) \\ &= [x^2-(\alpha+\sigma(\alpha))x +\alpha\sigma(\alpha)](x-\sigma^2(\alpha)) \\ &=x^3-[\alpha+\sigma(\alpha)+\sigma^2(\alpha)]x^2 +[\alpha\sigma(\alpha)+\alpha\sigma^2(\alpha)+\sigma(\alpha)\sigma^2(\alpha)]x + [\alpha\sigma(\alpha)\sigma^2(\alpha)]. \end{align*}
The first term falls relatively easily: \begin{align*} f_1=\alpha+\sigma(\alpha)+\sigma^2(\alpha) &= \sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^m + \sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{n+m} + \sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{2n+m} \\ &= \sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^m(1+\omega^n+\omega^{2n}). \end{align*} When $n\neq 0$, the parenthetical sum is of the three third roots of unity, so these terms cancel, leaving only $\alpha+\sigma(\alpha)+\sigma^2(\alpha) = \sum_{m=0}^2 3a_{m,0}\omega^m\in\mathbb Q(\omega)$.
The second term is much worse than it looks, which is a rather extraordinary feat: \begin{align*} f_2=\alpha\sigma(\alpha)&+\alpha\sigma^2(\alpha)+\sigma(\alpha)\sigma^2(\alpha)= \\ & \left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^m\right)\left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{n+m}\right) \\ &\qquad + \left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^m\right)\left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{2n+m} \right) \\ &\qquad\qquad + \left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{n+m}\right)\left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{2n+m} \right). \end{align*} At risk of tearing out our eyes, we will only consider the $\sqrt[3]2$ and $\sqrt[3]4$ terms, as we wish to show they vanish to leave us with only terms in $\omega$. Even this will be too unwieldy, so we split the algebra in pieces for it. To keep the whole thing somewhat readable, we leave the powers of $\sigma$ that we are using in subscripts below each term. For each of the two calculations and for each of the three pairs of terms there are three distinct summations: for $\sqrt[3]2$ we choose the $n$ to be $1$ and $0$, or to be $2$ and $2$; the first one is not symmetric so we may choose them in either order. Similarly, for $\sqrt[3]4$ we choose the $n$ to be $2$ and $0$, or to be $1$ and $1$.
\begin{align*} f_2[\sqrt[3]2] &= \left(\sum_{x,y=0}^2 a_{x,0}a_{y,1}\sqrt[3]2\omega^{x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,1}\sqrt[3]2\omega^{1+x+y} + \sum_{x,y=0}^2 a_{x,2}a_{y,2}2\sqrt[3]2\omega^{2+x+y} \right)_{1,2} \\ &\qquad + \left(\sum_{x,y=0}^2 a_{x,0}a_{y,1}\sqrt[3]2\omega^{x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,1}\sqrt[3]2\omega^{2+x+y} + \sum_{x,y=0}^2 a_{x,2}a_{y,2}2\sqrt[3]2\omega^{4+x+y} \right)_{1,3} \\ &\qquad + \left(\sum_{x,y=0}^2 a_{x,0}a_{y,1}\sqrt[3]2\omega^{1+x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,1}\sqrt[3]2\omega^{2+x+y} + \sum_{x,y=0}^2 a_{x,2}a_{y,2}2\sqrt[3]2\omega^{6+x+y} \right)_{2,3}; \end{align*}
\begin{align*} f_2[\sqrt[3]4] &= \left(\sum_{x,y=0}^2 a_{x,0}a_{y,2}\sqrt[3]4\omega^{x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,2}\sqrt[3]4\omega^{2+x+y} + \sum_{x,y=0}^2 a_{x,1}a_{y,1}\sqrt[3]4\omega^{1+x+y} \right)_{1,2} \\ &\qquad + \left(\sum_{x,y=0}^2 a_{x,0}a_{y,2}\sqrt[3]4\omega^{x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,2}\sqrt[3]4\omega^{4+x+y} + \sum_{x,y=0}^2 a_{x,1}a_{y,1}\sqrt[3]4\omega^{2+x+y} \right)_{1,3} \\ &\qquad + \left(\sum_{x,y=0}^2 a_{x,0}a_{y,2}\sqrt[3]4\omega^{2+x+y} + \sum_{x,y=0}^2 a_{y,0}a_{x,2}\sqrt[3]4\omega^{4+x+y} + \sum_{x,y=0}^2 a_{x,1}a_{y,1}\sqrt[3]4\omega^{3+x+y} \right)_{2,3}. \end{align*}
We could rearrange this formally, but this seems like a lot of work for not much gain. Each of the nine summations contains nine terms, and these terms cancel in a peculiar way: The $(x,y)$ term of the first summation and the $(y,x)$ terms of the second and fifth summation are identical except they have three different roots of unity; therefore they cancel. Similarly, the $(x,y)$ term of the fourth and seventh summation cancel with the $(y,x)$ term of the eighth summation; and finally the last column of summations cancel term by term.
Hence, all of our hard work in making sure indices were carefully and correctly placed has been rewarded with tremendous cancellation. This is good, and proves that the second term in our polynomial is in $\mathbb Q(\omega)$, but it does leave an inexplicible bitter emptiness.
In case someone is still reading for some reason, the author would like to emphasize how very apologetic he is, for what has been and what is to come. Perhaps you would enjoy one of the two other solutions better. They take less than a page each, and are generally quite enjoyable (although the author does not guarantee their accuracy, of course.)
We now turn our attention to the third term. Not willingly, but we do so. $$ f_3=\alpha\sigma(\alpha)\sigma^2(\alpha) = \left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^m\right)\left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{n+m}\right)\left(\sum_{m,n=0}^2 a_{m,n}\sqrt[3]2^n\omega^{2n+m} \right). $$
As a general principle, we can see from the above work that what we are really seeking is an assurance that any $a_{m,n}\omega^k$ lies in a triple that has the same indices on the coefficients and distinct indices on the omegas. The third term has at least one bit of good news for us. Namely, that we can use the cancellation that we discovered in the second term to make our problems look slightly more reasonable. But to do this, we will need the constant-term coefficients for the product of the first two terms; we use $n=2,1$ for the non-symmetric cases and $n=0$ for the third term. $$(f_2[1])_{1,2} = \left(\sum_{x,y=0}^2 a_{x,2}a_{y,1}2\omega^{1+x+y} + \sum_{x,y=0}^2 a_{y,2}a_{x,1}2\omega^{2+x+y} + \sum_{x,y=0}^2 a_{x,0}a_{y,0}\omega^{x+y} \right)_{1,2}$$
We do this, of course, because we are going to need to multiply $$\alpha\sigma(\alpha)\sigma^2(\alpha) = \Big((f_2[1])_{1,2}+(f_2[\sqrt[3]2])_{1,2}+(f_3[\sqrt[3]4])_{1,2}\Big)\sigma^2(\alpha)$$
The sinking feeling you have right now is your vomit reflex. It can be ignored, or if you prefer, you may use the page to prevent any accidents while running to the wastebin.
However, all is not lost; I mentioned this earlier. Recall that the in the first three summations there were two that helped to cancel each other; we can therefore replace two of the summations with a single summation having sign reversed. In fact, we may do this for all of them: \begin{align*} (f_2[1])_{1,2} &= -\sum_{x,y=0}^2 a_{x,2}a_{y,1}2\omega^{x+y} + \sum_{x,y=0}^2 a_{x,0}a_{y,0}\omega^{x+y} =\sum_{x,y=0}^2 (a_{x,0}a_{y,0}-2a_{x,2}a_{y,1})\omega^{x+y} \\ (f_2[\sqrt[3]2])_{1,2} &= -\sum_{x,y=0}^2 a_{y,0}a_{x,1}\sqrt[3]2\omega^{2+x+y} + \sum_{x,y=0}^2 a_{x,2}a_{y,2}2\sqrt[3]2\omega^{2+x+y} =\sum_{x,y=0}^2 (2a_{x,2}a_{y,2}-a_{x,1}a_{y,0})\sqrt[3]2\omega^{2+x+y} \\ (f_2[\sqrt[3]4])_{1,2} &= -\sum_{x,y=0}^2 a_{y,0}a_{x,2}\sqrt[3]4\omega^{1+x+y} + \sum_{x,y=0}^2 a_{x,1}a_{y,1}\sqrt[3]4\omega^{1+x+y} =\sum_{x,y=0}^2 (a_{x,1}a_{y,1}-a_{x,2}a_{y,0})\sqrt[3]4\omega^{1+x+y} \end{align*}
This looks almost tolerable. In particular, we will use the same trick of not acutally writing out the entire summation but only the parts which interest us, the $\sqrt[3]2$ and $\sqrt[3]4$ coefficients. But some things look promising here. We will remind the reader that $\sigma^2(\alpha)$ is not an interminably long expression, but only a single double summation. Furthermore, we no longer have variable exponents on our roots in the expression for $\alpha\sigma(\alpha)$ which we have spent this page deriving. So we have reason to be thankful and to think that things will be simpler.
The algorithm, at least, is simpler: we fix some $n$ in the summation for $\sigma^2(\alpha)$ and then choose the complementary root value in the three expressions above. Still, to make things as clear as possible for our now-certainly imaginary reader, we put subscripts denoting which choice of $n$ we took to get a particular summation. Just as the $x$ and $y$ played the roles of replacing $m$ once multiplication would make it redundant, we now must enlist some other unsuspecting index variable. There is a natural choice, $z$, and this is by design: a small reminder that the author is not doing this calculation because mental illness has rendered it enjoyable, but because all other remotely reasonable options were considered and exhausted.
\begin{align*} \alpha\sigma(\alpha)\sigma^2(\alpha)[\sqrt[3]2] &= \left( \sum_{x,y,z=0}^2 (2a_{x,2}a_{y,2}-a_{x,1}a_{y,0})a_{z,0}\sqrt[3]2\omega^{2+x+y+z} \right)_{n=0} \\ &\qquad + \left( \sum_{x,y,z=0}^2 (a_{x,0}a_{y,0}-2a_{x,2}a_{y,1})a_{z,1}\sqrt[3]2\omega^{2+x+y+z} \right)_{n=1} \\ &\qquad + \left( \sum_{x,y,z=0}^2 (2a_{x,1}a_{y,1}-2a_{x,2}a_{y,0})a_{z,2}\sqrt[3]2\omega^{6+x+y+z} \right)_{n=2} \end{align*}
\begin{align*} \alpha\sigma(\alpha)\sigma^2(\alpha)[\sqrt[3]4] &= \left( \sum_{x,y,z=0}^2 (a_{x,0}a_{y,0}-2a_{x,2}a_{y,1})a_{z,2}\sqrt[3]4\omega^{4+x+y+z} \right)_{n=2} \\ &\qquad + \left( \sum_{x,y,z=0}^2 (2a_{x,2}a_{y,2}-a_{x,1}a_{y,0})a_{z,1}\sqrt[3]4\omega^{4+x+y+z} \right)_{n=1} \\ &\qquad + \left( \sum_{x,y,z=0}^2 (a_{x,1}a_{y,1}-a_{x,2}a_{y,0})a_{z,0}\sqrt[3]4\omega^{1+x+y+z} \right)_{n=0} \end{align*}
At first glance this makes us very sad, because our usual method of inspection has failed. Although it is possible that some manipulations and value-checking of the three index variables may force things into our favor, we have a better tool. Namely, before we had only positive terms, and now we have minus signs. So it will be traditional cancellation tools for us this time. Our initial sadness then wears off, because these terms fall almost immediately into our hands by inspection. We have intentionally arranged the choices of $n$ such that the following work in both expressions:
- The $(x,y,z)$ term on the positive side of the first summation cancels with the $(x,z,y)$ term on the negative side of the third summation.
- The $(x,y,z)$ term on the negative side of the first summation cancels with the $(z,y,x)$ term on the positive side of the second summation.
- The $(x,y,z)$ term on the negative side of the second summation cancels with the $(z,y,x)$ term on the positive side of the second summation.
We have therefore shown that the third coefficient is in $\mathbb Q(\omega)$, which completes the proof: $f(x)$ is an irreducible polynomial of degree 6 with rational coefficients whose Galois group is the symmetric group on three elements.
At last --- at long last --- we may rest.
Think no more of what you have seen here. It is but sand through the hourglass.
Best Answer
Let $\zeta = \exp(2\pi i/29)$ be a primitive 29th root of unity. Then the field $\mathbb Q(\zeta)/\mathbb Q$ has Galois group $$ G = (\mathbb Z/29\mathbb Z)^\times \simeq \mathbb Z/28 \mathbb Z \simeq \mathbb Z/4 \mathbb Z \times \mathbb Z/7 \mathbb Z . $$ Since 2 is a primitive root mod 29 (a generator of $(\mathbb Z/29\mathbb Z)^\times$), this Galois group is generated by the automorphism $\zeta \mapsto \zeta^2$. Furthermore, an element of order 4 can by found by the seventh power of this automorphism, namely $\zeta \mapsto \zeta^{2^7} = \zeta^{12}$. Let $H$ be the subgroup of $G$ generated by $\zeta \mapsto \zeta^{12}$, which is cyclic of order 4. Then $G/H \simeq \mathbb Z/7\mathbb Z$ is the Galois group of the fixed field $K := \mathbb Q(\zeta)^H$ over $\mathbb Q$.
Now all we have to do is find the minimal polynomial of a generator for $K/\mathbb Q$. Since $\zeta$ generates $\mathbb Q(\zeta)/\mathbb Q$, the traces $\operatorname{Tr}_H(\zeta^i)$ generate $K/\mathbb Q$. Let $$ \xi = \operatorname{Tr}_H(\zeta) = \sum_{\sigma \in H} \sigma(\zeta) = \zeta + \zeta^{12} + \zeta^{28} + \zeta^{17} . $$ By solving a large system of linear equations involving the powers of $\xi$ (or using a computer to do so), we find that $\xi$ has minimal polynomial $$ f(x) = x^7 + x^6 - 12x^5 - 7x^4 + 28x^3 + 14x^2 - 9x + 1 . $$
Remark: I chose 29 because it is the smallest prime which is 1 mod 7.
Your comment above about needing the polynomial to be reducible mod every prime except those dividing the discriminant is not true. The only primes dividing the discriminant of $f$ are 17 and 29, but $f$ splits completely mod 59 or 233, for example. This is not a coincidence. These primes are both 1 mod 29, and so they split completely in $\mathbb Q(\zeta)$, and hence also in $K$