I am new in field theory and i began studying it recently.  I have a question about a finite field extension  The basis of the extension $\mathbb{Q}$$(\sqrt{1+\sqrt{3}})$/$\mathbb{Q}$ is the set $\{1,\sqrt{1+\sqrt{3}},\sqrt{1+\sqrt{3}}^2,\sqrt{1+\sqrt{3}}^3\}$ becase of the monic polynomial $(x^2-1)^2-3=0$ ?  Or am i missing something?  Thank you in advance.
72551
Following my  the book then asks      Use this to determine when the field extension  $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ over $\mathbb{Q}$  is biquadratic (where $a,b\in\mathbb{Q}$)   "this" means what proved in the linked post.  My thoughts: First, I don't know what to say when $a^2-b$ is not a square. When $a^2-b$ is a square I think that the extension is of degree $2$ if $a=b$, this leaves the case $a\neq b$ which I am also having problems with and I don't know what to do (I have some cases like if $m,n$ are different primes then the degree of the extension is $4$, but I don't have something that is general)  Can someone please help my do this exercice  ?
109899
$\mathbb R$ is an extension of $\mathbb Q$.  Why the elements of $\mathbb Q(2^{1/3})$ have the form $a+b\sqrt[3]2$?  More specifically why $\mathbb Q(2^{1/3})=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c \in\mathbb Q\}$?
11021
I just need a check for my reasoning:  If I have a ring $R=\mathbb{F}_p[X]$ and elements $x,y\in R$ it is true $$(x+y)^p=x^p+y^p?$$ It is true since every other element of expansion has a multiple of $p$ as coefficient.  What about field $R=\mathbb{F}_p[X]/(f)$ where $f$ is polynomial of degree $n$. I think rule still holds for the same reason.   Also what can be said about $x^p$? Is there any general rule about that element? It seems to me that $x^p=x$ since $(\mathbb{F}_p,\cdot)$ has order $p-1$. Am I missing something?
88414
$F$ is an extension field of a field $K$. Let $F$  be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that $(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$ $(ii)$  The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$. $(iii)$  The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$    So, here is my attempt for (i). Let  $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$.  Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well.  Thanks.  ADDED:  Attempt at (iii).   but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$ hints...  The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$. Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...
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!
102239
Proof: Since $\mathbb{Q}$ is a field, then $\mathbb{Q}$ is a domain.  (Theorem: if $R$ is a domain, then $R[x]$ is a field.)  By the theorem, $\mathbb{Q}[x]$ is a field.  So, letting $x = \sqrt{2}$, we get $\mathbb{Q}[\sqrt{2}]$ is a field. QED.  I am wondering if this is an acceptable proof? Also, I am wondering if it can be proven using facts of irreducibility? Isn't this what Galois Theory is about? I am curious, thanks.
32199
Question is to solve for Galois Group of $x^4-2x^2-2$ over $\mathbb{Q}$.  I know the roots of this polynomial are $\sqrt{1+\sqrt{3}},-\sqrt{1+\sqrt{3}},\sqrt{1-\sqrt{3}},-\sqrt{1-\sqrt{3}}$.  But, $\sqrt{2}i=\sqrt{1+\sqrt{3}}.\sqrt{1-\sqrt{3}}$. So, I concluded that splitting field would be $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$ (for some reason i am not very sure about, i have not written splitting field to be $\mathbb{Q}(\sqrt{1-\sqrt{3}},\sqrt{2}i)$).  So, i have splitting field as $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$.   What i do usually after i know that extension is of the form $F(a,b)$ with $F(a)\cap F(b)=F$, I calculate Galois group of $F(a,b)/F(b)$ and Galois group of $F(a,b)/F(a)$ and write their product as in $G=Gal(F(a,b)/F)$  I do the same here for $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))$, this is of order 2, sending $\sqrt{2}i\rightarrow -\sqrt{2}i$ (fixing $\sqrt{1+\sqrt{3}}$) So, I have $Gal(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))\cong \mathbb{Z}_2$.  The problem is with $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$, I Have to prove that Galois group of $x^4-2x^2-2$ is dihedral group of order 8.I already have an element of order 2, I have to get an element of order 4 from $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$ which is becoming more cumbersome.  Any help would be appreciated. Thank You.
121085
Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$  So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.  This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.  Could someone maybe give me a hint about how to proceed? Thanks a lot!!!
48253
I had an example in the book given as follows:        Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .       Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$      $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$       Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$       Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.       Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$   So degree of $p(x)$ is atleast $4$.      But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .      So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$   But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$   Kindly help with this.
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
16633
I have a question about simple field extensions.  For a field $F$, if $[F(a):F]$ is odd, then why is $F(a)=F(a^2)$?
124213
Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?  ($\cong$ denotes an isomorphism that is the identity on restriction to $K$).  Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
70708
I would like to verify the validity of the following line of thought:  Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).  Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.  Is there any other simpler argument than that?  Added:  Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge &lt;l$. Then $x^p-a=(m_b(x))^q (x-b)^r$. Can i conclude that $(x-b)^r \in K[x]$ and so $r=0$ by the minimality of $l$?  Thanks.
82410
Let $h$ the irreducible polynomial over $\Bbb F_2 = \{0, 1\}$  $$h  =  x^4 +x +1$$  Let $E=\Bbb F_2(\alpha)$, i.e. the field $\Bbb F_2$ extended with $\alpha$, where $\alpha$ is a root of $h$.   How do we determine the numbers of elements in the extension $E$? Can every non-zero element in the extension be expressed in the form $(\alpha)^n$ for some $n\in\Bbb N$ ? How do we find all the roots of $h$ over the extension $E$?
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.
104050
My problem is "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension"  I met a similar problem "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is a simple extension" and I showed that it is equal to Prove $\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}$. I let $u=\sqrt{2}+\sqrt{3}$ and deduce that $u-\sqrt{2}=\sqrt{3})$, then square it and prove $\sqrt{2} \in Q(\sqrt{2}+\sqrt{3})$. However, the same technique cannot be applied in the first problem since we have 3 square root.  I actually found a proof for the first problem, by letting $u=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and calculating $u,u^2,..., u^5$ and prove $\sqrt{2},\sqrt{3}\in Q(u)$. Though it's not so nice. Looking for a simpler answer
118984
If I understand correctly, Arturo Magidin says in  that the following is true.     If $E\subset F$ is a finite field extension, then $|\operatorname{Aut}(F/E)|\leq (F:E).$   I understand why this is true when the extension is simple, but the general case eludes me. (I also understand that my particular extension in that question is simple, so this is irrelevant to the problem, but I would like to understand this.)  Let $E\subset F$ be a finite extension. Any finite extension is finitely generated, so let $$F=E(a_1,\ldots,a_n).$$ Let $$E=F_0\subset F_1\subset\ldots\subset F_n=F,$$ with $F_i=F_{i-1}(a_i)$ for $i=1,\ldots,n.$ Let $f_i$ be the minimal polynomial of $a_i$ over $F_{i-1}$, and let $g_i$ be the minimal polynomial of $a_i$ over $E$. Then $$(F:E)=(F_n:F_{n-1})\cdots (F_1:F_0)=\deg(f_n)\cdots\deg(f_1).$$  I think I can't use $f_i$ to bound the order of $\operatorname{Aut}(F/E).$ Surely, $$|\operatorname{Aut}(F/E)|\leq \deg (g_n)\cdots\deg (g_1),$$ because the generators must be mapped to roots of their minimal polynomials over $E$ (taking non-minimal polnomials here would only make things worse), but this doesn't work because the inequality we have is $$\deg (f_i)\leq\deg (g_i),$$ not the other way around.