The splitting field of $x^4 - 2x^2 - 6$ is $\mathbb{Q}(\alpha, \sqrt{-6}) = \mathbb{Q}(\alpha, \beta)$ where $\alpha = \sqrt{1+\sqrt{7}}$, $\beta = \sqrt{1-\sqrt{7}}$. From the first representation, $x^4 - 2x^2 - 6$ being irreducible (say, Eisenstein for $2$) and $\sqrt{-6}$ being non-real it follows that the extension (and hence the Galois group) has order $8$.   Now one can sneakily show that the group is $D_8$ as follows -- it is not Abelian, as then by FTGT any intermediate extension must be Galois, whereas $\mathbb{Q}(\alpha) : \mathbb{Q}$ is not. On the other hand, $(\alpha \to -\alpha)$ (and fix the rest) and $(\sqrt{-6} \to -\sqrt{-6})$ (and fix the rest)   are two distinct morphisms of order $2$, hence the group cannot be quaternion. So it must be $D_8$.  But what about a set of generators? I think that $(\sqrt{-6} \to -\sqrt{-6})$ can be the reflection one but I can't think of a suitable rotation one.  Any help appreciated!
6984
From Hungerford, section V, chapter 4 exercise 9:     Let $x^4+ax^2+b$ in  $K[x]$ (with char $K\neq $2) be irreducible with   Galois group $G$.      (a) If $b$ is a square in $K$, then $G =  \mathbb{Z}_2\times\mathbb{Z}_2$.       (b) If $b$ is not a square in $K$   and $b(a^2-4b)$ is a square in $K$, then $G = \mathbb{Z}_4$.       (c)   If neither $b$ nor $b(a^2-4b)$ is a square in $K$, then $G = D_4$   (dihedral group).   What I've tried  With (a):  Let be $u_1,u_2,u_3,u_4$ the roots of the quartic, then $b=u_1u_2u_3u_4$ is in $K$ and there exists $a\in K$ with $a^2=b$. On the other hand, $G = \mathbb{Z}_2\times\mathbb{Z}_2$ if and only if $\alpha=u_1u_2+u_3u_4$, $\beta=u_1u_3+u_2u_4$, $\gamma=u_1u_4+u_2u_3$ are in $K$, but I don't find the way to rely those results.   With (b) I know that $b=u_1u_2+u_1u_3+u_1u_4+u_2u_3+u_2u_4+u_3u_4$ but again I dont know how to continue.  Any help? Thanks.
100654
Given the galois extension $L/K$, where $L=\mathbb{Q}(\zeta, \sqrt[3]{2})$. I've already calculated that $[L:\mathbb{Q}]=6$. I'm trying to prove that the number of intermediate fields is also 6, including $L$ and $\mathbb{Q}$ as intermediate fields. I'm told that the fundamental theorem of galois theory can be useful here, but I can't find a use for it.
58506
The following facts are standard: an irreducible quartic polynomial $p(x)$ can only have Galois groups $S_4, A_4, D_4, V_4, C_4$. Over a field of characteristic not equal to $2$, depending on whether or not the discriminant $\Delta$ is a square and whether or not the resolvent cubic $q(x)$ is irreducible, we can distinguish four cases:   If $\Delta$ is not a square and the resolvent cubic is irreducible, then the Galois group is $S_4$. If $\Delta$ is a square and the resolvent cubic is irreducible, then the Galois group is $A_4$. If $\Delta$ is a square and the resolvent cubic is reducible, then the Galois group is $V_4$. If $\Delta$ is not a square and the resolvent cubic is reducible, then the Galois group is either $D_4$ or $C_4$.   How can we resolve the ambiguity in the last case? For $p$ a monic polynomial in $\mathbb{Z}[x]$, I know the following approaches:   In the simplest cases one can work directly with the splitting field. But this is rare, although it can work if $p = x^4 + ax^2 + b$ for some $a, b$.  If $p$ has two complex roots (equivalently, if the discriminant is negative), there is a transposition in the Galois group, so the Galois group is $D_4$. If $p$ factors as the product of two linear factors and an irreducible quadratic factor modulo a prime, there is a transposition in the Galois group, so the Galois group is $D_4$.    In practice, the last two often work to identify a Galois group of $D_4$ (and in principle it must eventually work by the Frobenius density theorem). But I don't know a corresponding practical way to identify a Galois group of $C_4$. There is a criterion due to Kappe and Warren which I learned about from one of Keith Conrad's expository notes . However, in an upcoming exam I'm taking, I'll only be able to cite results proven in the course, and this criterion isn't one of them.   So what are my other options in general?
13733
I have to compute Galois group of $\mathbb F_q/\mathbb F_p$ where $q=p^n$. I know already that $\mathbb F_q/\mathbb F_p$ is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&amp;\longrightarrow \mathbb F_q\\ x&amp;\longmapsto x^p\end{align*} is an automorphism that fix elements of $\mathbb F_p$ and thus $\sigma \in Gal(\mathbb F_q/\mathbb F_p)$.  Now, I know that if $k&lt;p^n$ then $\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1$ and thus $\sigma $ is of order $p^n$. Now if I can prove that $|Gal(\mathbb F_q/\mathbb F_p)|=p^n$, then $Gal(\mathbb F_q/\mathbb F_p)=\left&lt;\sigma \right&gt;$ and it would be finish, but I can't prove it since even $X^q-X$ split over $\mathbb F_q$, theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree $q$ that split over $\mathbb F_q$ ?
74153
Suppose $K = \mathbb{Q}(\alpha)$ with $\alpha = a + b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ with $D_1,D_2 \in \mathbb{Z}$. Prove that the minimal polynomial $m_\alpha(x)$ for $\alpha$ over $\mathbb{Q}$ is irreducible of degree 4 over $\mathbb{Q}$ but is reducible modulo every prime $p$. In particular show that the polynomial $x^4 - 10x^2 +1$ is irreducible in $\mathbb{Z}[x]$ but is reducible modulo every prime. [Use the fact that there are no biquadratic extensions over finite fields.]  So far I have established the following:  $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(m_\alpha(x))=4$  $Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic, hence no biquadratic extension (which is iso to $V_4$) exists over finite fields.  I'm having a problem proving the reducibility mod every prime though. Any hints?
114611
I need to show that the extension $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$, and compute its Galois group.   I am learning Galois theory by myself and got stuck in this exercise. I know the fundamental theorem of Galois theory. Any help would be useful  Thanks
33881
Let $f(x) \in \mathbb Q[x]$ be irreducible of degree 4. Let $\alpha$ be a root of $f(x)$. Let $E = \mathbb Q(\alpha)$ and let $K$ be the splitting field of $f(x)$ over $\mathbb Q$. Prove that there is no field properly between $\mathbb Q$ and $E$ if and only if $G(K/\mathbb Q) \cong A_4$ or $S_4$.  At this point in the course we've just gone through the Fundamental Theorem of Galois Theory. I'm fairly lost, so any hints, sketches of a proof would help. Thanks!
21628
I've been learning about Galois theory recently on my own, and I've been trying to solve tests from my university. Even though I understand all the theorems, I  seem to be having some trouble with the technical stuff. A specific example would be how to find the Galois group of a given polynomial. I know some tricks, and I manage to solve some of those questions, but some not.  For example, one of the tests asks to find the Galois group of $x^{4}-4x+2$.  I can see that it is irreducible over $\mathbb Q$ (Eisenstein), but I have no clue as to how to find its Galois group over $\mathbb Q$. Can someone tell me how to do this? General techniques concerning this sort of problems are also welcome :).  Thanks!
61639
I read that if $F$ is the field of algebraic numbers over $\mathbb{Q}$, then every polynomial in $F[x]$ splits over $F$. That's awesome! Nevertheless, I don't fully understand why it is true. Can you throw some ideas about why this is true?
117076
How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$   My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when restricted to $\mathbb Q.$  Ya! I can see that: $\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle$ and $\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.$ Now if $\phi$ be such an isomorphism then $\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.$ Hence the restriction of $\phi$ over $\{q+\langle x^2-2\rangle:q\in\mathbb Q\}$ remains the identity.  Now what?
28718
Let $A$ be a UFD, $K$ its field of fractions, and $f$ an element of $A[T]$ a monic polynomial.  I'm trying to prove that if $f$ has a root $\alpha \in K$, then in fact $\alpha \in A$.  I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.
15247
Determine the Galois group of the polynomial $(x^3-2)(x^3-3)(x^2-2)$ over $\mathbb{Q}(\sqrt {-3})$.   I tried to decompose the splitting field into a tower, but then it is very complicated in each step and since the degree is large, I cannot determine the group by guessing. It seems to me that Kummer theory might work, but there are $2$-degree and $3$-degree factors.
22573
So I want to show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$ and determine its Galois group.   My thoughts are as follows:  Define $\alpha := \sqrt{2+\sqrt{2}}$. Then it is easily shown that $\alpha$ satisfies $\alpha^4-4\alpha^2+2=0$.  Define $f(x) := x^4-4x^2+2$. Then $f$ is irreducible over $\mathbb{Q}$ by Eisenstein with $p=2$.  So we have that $f$ is the irreducible polynomial for $\alpha$ over $\mathbb{Q}$.   Further $|\mathbb{Q}(\alpha):\mathbb{Q}|=4$.  For $\mathbb{Q}({\alpha})$ to be Galois, it must contain all roots of $f$.   Define $K=\mathbb{Q}(\alpha)$ for convenience.  Define $\alpha := \alpha_1$.  Since $f$ has only even powers, we know that $-\alpha := \alpha_2$ is a root, and therefore contained in $K$ since $K$ is a field.  We note that the other two roots are $\alpha_3=\sqrt{2-\sqrt{2}}$ and $\alpha_4=-\sqrt{2-\sqrt{2}}$.  So in order to show $K$ is Galois, it must be shown that $\alpha_3$ and $\alpha_4$ lie in $K$.  Now $\alpha_1^2=2+\sqrt2$ and so $\sqrt2 \in K$. Thus $-\sqrt2 \in K$ since $K$ is a field.  Can somebody explain why $\alpha_3$ and $\alpha_4$ lie in $K$?  Next we are to determine the Galois group of $K$.  Assuming $K$ is Galois, since it has degree $4$ over $\mathbb{Q}$ (shown earlier), we know that its Galois group has size $4$. There are only two groups of size $4$, namely $V_4$ and $C_4$, the Klein four group and the cyclic group of order $4$.  How do we determine which of these choice is in fact the Galois Group?
47020
Let $F$ be the field $\Bbb Q(\alpha_1,...,\alpha_k)$ where  $\alpha_i^2\in\Bbb Q\ \forall i$  I want to show that $\sqrt[3]{2}\notin F$  I thought if $\sqrt[3]{2}\in F$ then $\sqrt[3]{2}\in\Bbb Q\cup(\alpha_1,...,\alpha_k)$ but this is not a disjoint union so I'm not sure what to do... taking the square of $\beta:=\sqrt[3]{2}$ gives $2^{2/3}=\sqrt[3]{4}\notin\Bbb Q$ but that doesn't really advance things  So I'm looking for a hint
3614
Show that the splitting field of $X^n-a$ over a field $K$ is $K(\alpha, \zeta_n)$, where $\alpha$ is a $n$-th root of $a$ and $\zeta_n$ is a primitive $n$-root of unity.
14478
First of all I am new to this topic, algebraic number theory, so I only know a decent (not great) amount of abstract algebra.     The question I have is that, given the imaginary quadratic field $\mathbb{Z}[\sqrt{-2}]$, I want to find;   (1) all irreducible elements of it,   (2) show that it is a Euclidean domain, and   (3) show that for an odd prime number $p,\; \exists \;x,y\; \in \mathbb{Z}$ s.t. $p = x^2+2y^2$ iff $p=1,3(\textrm{mod}\; 8)$.    I have been reading and have books but there are some things I am not getting.  (a) My attempt at finding the units (I read that there are only ${}^{\pm}1$ for this integral domain (ID));  A unit is an element with an inverse, so for an element $p_1 \in \mathbb{Z}[\sqrt{-2}]$, there is another element $p_1&#39;$ s.t. $p_1\,p_1&#39; = p_1&#39;\,p_1 = 1$ (it is integral domain, not just domain).  Let $p_1 := a+b\sqrt{-2}$ and $p_1&#39;:=x+y\sqrt{-2}$ and so $p_1\,p_1&#39; = 1$ becomes $(a+b\sqrt{-2})(x+y\sqrt{-2}) = 1  = 1+0\sqrt{-2}$ and into the two equations, $1=ax-2by$ and $0=ay+bx$. Solving these leads to $x=\frac{a}{a^2+2b^2}$, $y=\frac{-b}{a^2+2b^2}$ and $p_1&#39;=\frac{a}{a^2+2b^2} + \left( \frac{-b}{a^2+2b^2} \right)\sqrt{-2}$, which can only belong to $\mathbb{Z}$ if $b=0,\;a={}^{\pm}1$.   Is there a better way of determing the units of an ID?  I read that the units, $\epsilon$, of a quadratic ID of the general form $R[\sqrt{d}]$, where $d$ was square-free, were determined by $Norm(\epsilon) = {}^{\pm}1$ Is this general ? Is this for any $d$ that is square-free (though I see little difference between $d=d$ and $d=z*d$, as $z\in \mathbb{Z}$) ?  (1) I know, procedurally, how to do this for a given element, but I do not know of a better way to do it in general. Here is my attempt:  I read that: An element $p$ of an ID is irreducible in R if it satisfies: (i) $p \neq 0$ and $p$ is not a unit, (ii) if $p=ab$ in R, then $a$ or $b$ is a unit in R.  (maybe because I know the units in this ID I can say that all other non-zero elements are irreducible ? )  So if $p = ab$, with $a = m+n\sqrt{-2}$, then using the as-of-yet-unproved-homomorphism-norm-map, $N(ab) = N(a)\,N(b)$, $N(p=x+y\sqrt{-2}) = x^2+2y^2 = N(a)\,N(b) = (m^2+2n^2)\,N(b)$. Now if I had a specific element, to determine if it was irreducible, I could then determine what values of $N(a)$ and $N(b)$ were valid so that their product equaled $N(p) = N(x+y\sqrt{-2})$, which this latter term would be an integer ($N: \mathbb{Z}[\sqrt{-2}] \mapsto \mathbb{Z}$).  The latter two questions I haven't got far with either but am wanting to get this initial question(s) understood first.  Thanks all for your time reading my rather lengthy question!
1406
Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in   $K[x]$ or has a root in $K$.   My strategy was to split this up into a case for each characteristic.   The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.  The characteristic $0$ case  and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.   Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.