Show that $\mathbb{Q}(\sqrt{2 +\sqrt{2}})$ is a cyclic quartic field i.e. is a galois extension of degree 4 with cyclic galois group  with some elementary algebra,  $x - \sqrt{2 +\sqrt{2}} = 0 \implies x^2 = 2+\sqrt{2} \implies (x^2 - 2)^2 = 2 \implies x^4-4x^2+2=0$ is the minimal polynomial and is Eisenstein at 2 so irreducible which gives seperable.  and the roots of $x^4-4x^2+2=0$ are $\pm \sqrt{2 \pm \sqrt{2}}$ again found with elementary algebra. So the splitting field is clearly a degree 4 extension.  Now I've confused myself a bit trying to work out all the automorphisms  since the polynomial is seperable I know $\mid AUT(\mathbb{Q}(\sqrt{2 +\sqrt{2}})/\mathbb{Q}\mid = [\mathbb{Q}(\sqrt{2 +\sqrt{2}}) : \mathbb{Q}]$ so there are 4. So it needs to be ismorphic to $\mathbb{Z}_4$  I think the only automorphisms are id, $\sigma : \sqrt{2 +\sqrt{2}} \mapsto -\sqrt{2 +\sqrt{2}}, \tau : \sqrt{2 -\sqrt{2}} \mapsto -\sqrt{2 +\sqrt{2}} $
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?
89703
Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain.
51163
I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it &lt;.&lt;  Let's consider a polynomial $f(x)= x^4+ax^2+b \in \mathbb{Z} $. Let $ \pm\alpha , \pm\beta$ denote the roots of $f(x)$. Note that $f$ is irreducible iff $\alpha^2, \alpha\pm\beta \notin \mathbb{Q}$. So let's suppose that $f(x)$ is irreducible. Let's denote $G$ the galois group of $f$ (the Galois group of the splitting field $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$)  $i)$ $G\cong V$ the Klein 4-group iff $\alpha\beta \in \mathbb{Q}$  $ii)$ $G\cong \mathbb{Z}_4$ iff $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)$  $iii)$ $G\cong D_8$ the dihedral group of order 8, iff $\alpha\beta \notin \mathbb{Q}(\alpha^2)$  I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that $u\in K/F$ a galois extension, the other roots of the minimal polynomial of $m_u(x)\in F[x]$ are precisely the elements $ \sigma(u)$ , where $\sigma \in Gal(K/F)$. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on $\mathbb{Q}(\alpha,\beta)$, so I have to know the values of both. But I have to respect the algebraic relations involving $\alpha , \beta$ I proved that in $i),ii)$ $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) $ So I have to know the values on $\alpha$.   My first question is: If in both cases $i),ii)$, I have an extension of degree 4 (so $|G|=4$) I have to map $\alpha$ to determine the automorphism, but I have also to map $\alpha$ to all of it's four roots. Why the galois group in this case are different?  My second question is about $iii)$ I have no idea How to attack it.
46139
Suppose $L=\mathbb{Q}(\sqrt{a+b\sqrt{d}})$,($d$ and $a+b\sqrt{d}$ are square free algebraic integers) when is $L/\mathbb{Q}$ a normal extension? When does $Aut(L/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}$ or $Aut(L/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$?
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.
77911
I was asked to prove or disprove that if a field extension is not normal then it is separable. I can't see why this would be true so I want to disprove it with an example of a non-normal inseparable field extension. I am wondering if there is a simple example of such an extension, or even any examples at all. I know that any inseparable extension $L/K$ has to be such that $K$ is infinite and of positive characteristic, so a good place to start would be $K=\mathbb F_p(T)$, but I'm unsure how to prove that an extension of this is inseparable, and even unsure how to come up with a non-normal example, since my usual approach is to exploit that $K$ is a subfield of $\mathbb R$ (so e.g. $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal as $x^3-2$ has $\sqrt[3]2$ as a root but the other two roots are in $\mathbb C \setminus \mathbb R$ and $\mathbb Q(\sqrt[3]2)\subset \mathbb R$). However with $K$ as above, $K \nsubseteq \mathbb R. $
102648
I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible.
72551
Following my  the book then asks      Use this to determine when the field extension  $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ over $\mathbb{Q}$  is biquadratic (where $a,b\in\mathbb{Q}$)   "this" means what proved in the linked post.  My thoughts: First, I don't know what to say when $a^2-b$ is not a square. When $a^2-b$ is a square I think that the extension is of degree $2$ if $a=b$, this leaves the case $a\neq b$ which I am also having problems with and I don't know what to do (I have some cases like if $m,n$ are different primes then the degree of the extension is $4$, but I don't have something that is general)  Can someone please help my do this exercice  ?
93537
Let $L/K$ be a field extension. Let be $\alpha, \beta \in \mathbb{C}$, such that $[\mathbb{Q}(\alpha):\mathbb{Q}] = p$, and $[\mathbb{Q}(\beta):\mathbb{Q}] = q$, for some prime numbers $p$ and $q$. Assume $p \neq q$. Prove that:  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\mathbb{Q}(\alpha,\beta):Q] = [\mathbb{Q}(\alpha):\mathbb{Q}] \cdot [\mathbb{Q}(\beta):\mathbb{Q}]$  I have no idea how to proceed on this, besides proving that the left hand side is $pq$.
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
25044
I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$.   I managed to find that it is a root of the polynomial $x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17$. This is where I run in to some problems. I don't know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?
99074
Let $\mathbb{F}_p$ be the finite field with $p$ elements, where $p$ is a prime number. Let $x$ and $y$ be transcendental and algebraically independent over $\mathbb{F}_p$. The extension $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite of degree $p^2$. To see this consider the polynomials $T^p-x^p$  and $T^p-y^p\in\mathbb{F}_p(x^p,y^p)[T]$.  So, if $K$ is any (proper) intermediate field of this extension, we must have $[K:\mathbb{F}_p(x^p,y^p)]=p$. For each prime $q\neq p$, consider the polynomial $T^p-x^{qp}\in\mathbb{F}_p(x^p,y^p)[T]$. It is an irreducible polynomial of degree $p$ with $x^q$ as root. So the extension $\mathbb{F}_p(x^p,y^p)(x^q)/\mathbb{F}_p(x^p,y^p)$ has degree $p$ and hence, $\mathbb{F}_p(x^p,y^p)(x^q)$ is a proper intermediate field of the original extension. Note that we can do the same with $y$ instead of $x$. Since there are infinite primes $\neq p$, there will be infinite (proper) intermediate fields of the original extension. These fields are distinct, because they are obtained by adjoining a root of an irreducible polynomial, and their irreducible polynomials are distinct.  This shows that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite but not simple, hence, it is not separable.      My question is: are there other intermediate fields other than that pointed out previously?   I first tried to show that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by supposing the existence of a primitive element then trying to get a contradiction, specifically, that $x$ and $y$ are algebraically dependent. The problem is, they are algebraically dependent over $\mathbb{F}_p(x^p,y^p)$, because they vanish this (non-zero) polynomial $T_1^p-x^p+T_2^p-y^p\in\mathbb{F}_p(x^p,y^p)[T_1,T_2]$. If you know someway of proving that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by contradiction, please answer it too.
13965
I am trying to prove the following proposition:     Let $L$ be an algebraically closed field, $g \in Aut(L)$ and $K=\{x \in L \; | \; g(x)=x\}$. Show that every finite extensions $E/K$ is a cyclic Galois extension.    So far my thoughts have led me to the following:  If $E=K$, then clearly $E/K$ is Galois and $Gal(E/K)=Aut(K/K)=i$, where $i$ is the identity automorphism, so it is trivially cyclic. Assume $[E:K]=n$ for some $n&gt;1$. Since $E/K$ is a finite extension, it is algebraic. Note that if $\beta \in E\setminus F$ then we have the minimal polynomial $m_{\beta}(X)$ over $K[x]$. Since $K$ is fixed by $g$, we know that $g(\beta)$ is also a root of $m_{\beta}(X)$. We can also conclude that $g(\beta)\not \in K$ since $\beta \not \in K$.   Where to take any of these observations, I have yet to determine. I feel I must be missing something elementary, for I cannot see how to conclude this is a Galois extension from here.   Any tips would be greatly appreciated, especially over full solutions.
13715
Let $n≥1$ be an integer. I would like to prove (or disprove) the existence of a subfield $K \subset \Bbb R$ such that $K/\Bbb Q$ is Galois and has degree $n$.   It is easy to construct such a subfield for $n=2^k$: one  take $K=\Bbb Q(\sqrt 2, \sqrt 3, \sqrt 5, ..., \sqrt{p_k})\;$ where $p_k$ is the $k$-th prime number.  I found that the irreducible polynomial $x^3-6x+2 \in \Bbb Q[X]$ (Eisenstein's criterion with $p=2$) has three real roots, so its splitting field $K$ satisfies my conditions for $n=3$. Then I'm done with $n=3\cdot 2^k$ thanks to compositum of fields.  , a Galois extension of odd degree must be . In all cases, I don't know how to construct, in general or in particular cases ($n$ prime for instance), such a totally real Galois extension $K/\Bbb Q$ of degree $n$.  Thank you for your help!
98573
I'm trying to show that $Q[\sqrt3]$ is a field. I am particularly struggling with showing that every element is a unit.   So for some element, we want to show that for some element $a+b\sqrt3$, there exists $c+d\sqrt3$ in $Q[\sqrt3]$ such that $(a+b\sqrt3)(c+d\sqrt3) = 1$.  Is that the right direction? I am not really sure of how to come up with that inverse though. Any help would be appreciated.
54698
What are the conditions for integers $D_1$ and $D_2$ so that $\mathbb{Q}[\sqrt {D_1}] \simeq \mathbb{Q}[\sqrt {D_2}]$ as fields.  Here $\mathbb{Q}[\sqrt {D}] := \{a + b \sqrt D \mid a,b \in \mathbb{Q} \}$  Really not sure where to begin with this sort of problem. I was thinking that I should split into cases where the integer is a square or not.
80032
Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?   I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.