Let $K/F$ be a Galois extension. Also, let $E_1$ and $E_2$ be two Galois extensions of $F$ in $K$ with $E_1E_2 = K$ and $E_1 \cap E_2 = F$. Prove that $G = \text{Gal}(K/F) \cong \text{Gal}(K/E_1) \times \text{Gal}(K/E_2)$.  I think this is just an application of results of the Fundamental theorem of Galois theory (and maybe the Chinese Remainder Theorem?), but I'm uncertain on how to prove it.   Since we know $E_1/F$ and $E_2/F$ are Galois extensions, we have $\text{Gal}(K/E_1)$ and $\text{Gal}(K/E_1)$ are normal subgroups of $G$. We get $\text{Gal}(E_1/F) \cong G/\text{Gal}(K/E_1)$ and similarly $\text{Gal}(E_2/F) \cong G/\text{Gal}(K/E_2)$. I don't know what to do from here, or if these statements if even help. Thoughts?
122655
I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory.    Theorem Let $E_1$ and $E_2$ be Galois field extensions of $F$ with trivial intersection ($E_1 \cap E_2 = F$), then $E_1 E_2$ has Galois group $\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.  Proof   $E_1$ and $E_2$ are splitting fields (over $F$) of some polynomials $f_1$ and $f_2$ so $E_1 E_2$ is the splitting field of $f_1 f_2$ over $F$ and therefore it is Galois and we are justified using the "Gal" notation. Let $\sigma_1 \in \text{Gal}(E_1/F)$ and $\sigma_2 \in \text{Gal}(E_2/F)$ then by the fundamental theorem of Galois theory we can lift these both into elements of $\text{Gal}(E_1 E_2/F)$. The claim is that $\sigma_1 \sigma_2 = \sigma_2 \sigma_1$, to see this consider an element $\alpha \in E_1$, we have both $\sigma_2 \alpha = \alpha$ and $\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha$ because it is invariant with respect to $\sigma_2$, but we can combine these equalities to get $\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha$. Both groups $\text{Gal}(E_i/F)$ ($i = 1,2$) are included in $\text{Gal}(E_1 E_2/F)$ and the degrees say there are no more elements, furthermore since they commute we can conclude $\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.     So hopefully that theorem is, if not correct, fixable.. and can be used to prove that $\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ has Galois group $C_2^m$ ($m \le n$) over $\mathbb Q$?
90626
Algebraic class field theory tells us that $\text{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})$ is isomorphic to the group of connected components of the quotient $\mathbb{Q}^{\times}\backslash \mathbb{A}_{\mathbb{Q}}^{\times}\cong \prod_p \mathbb{Z_p}^{\times}\times \mathbb{R}_{&gt;0}$, where $\mathbb{A}_{\mathbb{Q}}$ is the ring of adèles of $\mathbb{Q}$.   It's then said that the group of connected components is given by $\prod_p \mathbb{Z_p}^{\times}$, how can I see this?  Thank you very much in advance!
86188
Let $E$ be an extension field of $F$ and $\alpha \in E$. Then $\alpha$ is transcendental over $F$ if and only if $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions of $F[x]$.   This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand.  Thanks!
119971
Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.  This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.  Thanks!
96746
$K/F$ is a field extension of finite degree $n$. How can I prove that $K$ is isomorphic to a subfield of the ring of $n\times n$ matrices over $F$, so the ring of $n\times n$ matrix over $F$ contains an isomorphic copy of every extension of $F$ of degree $\leq n$?
108046
Letting $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and letting $n \geq 3$ and $m$ be natural numbers, it is known that $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m=n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$ (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967").  So, given a field $\mathbb{F}$, if $n \neq m$, then $\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})$, and thus $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$.  Letting $R$ be a commutative ring (with unity), and letting $n, m \in \mathbb{N}$ be such that $n \neq m$, is there a simple "algebraic" way of proving that $\text{GL}_{m}(R)$ and $\text{GL}_{n}(R)$ are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$ for a field $\mathbb{F}$?  Certain special cases of this problem trivially hold, for example in the case whereby $\mathbb{F}$ is finite, in which case $|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|$.
55663
Given a field $F$, can you necessarily construct a field extension $E \supset F$ such that $\operatorname{Gal}(E/F) = S_n\,$?
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 $pq$.
53117
Every finite group is isomorphic to some Galois group for some finite normal extension of some field.   I'm trying to write a proof, but I know this is incorrect. Can anyone point me in the write direction?  Proposition: Every finite group is isomorphic to some Galois Group $\text{Aut}_F(K)$ for some finite normal extension K of some field F.   My hack at a proof: For every $n \in \mathbb{Z}^+$ where $n = |\text{Aut}_E K|$ there exists a normal extension $K$ of $E$ such that $\text{Aut}_E K \cong S_n$. Then consider a subgroup $H \leq \text{Aut}_E K$ where $H \cong \text{Aut}_E K$. It follows that $H$ is the Galois group of $K$ over $K_H$.
21715
I've been wondering about this for a while, is it the case that if $K$ is a finite Galois extension of $\mathbb{Q}_p$, then there exists some field $L$ with $$\mathbb{Q} \subset L \subset K,$$ such that $[L:\mathbb{Q}]&lt;\infty$ and $L\cdot \mathbb{Q}_p = K$. I can see that this is true for $K$ unramified, because we can just attach the appropriate roots of unity to $\mathbb{Q}$ so it remains the case where $K$ totally ramifies. I guess it would be enough to show that there exists some $\alpha \in K$ which is algebraic over $\mathbb{Q}$ but not an element of $\mathbb{Q}$. I have tried to use Hensel's lemma but that fails spectacularly because the uniformizer is the root of some Eisenstein polynomial which reduces to be $x^n$.
37895
Let $E/F $ be a field extension and $f,g$ $\in$ $F[x]$ (the polynomial ring with coefficients in $F$ ). Let's denote with $\gcd_F(f,g)$ the greatest common divisor of $f$ and $g$ in $F[x]$. Is it true that $\gcd_F(f,g)$ $=$ $\gcd_E(f,g)$ ?
79364
Suppose $K$ is a field and $F$ is an algebraic extension of some degree $n=(F:K)$. It is stated that the field of rational functions $F(x)$ is in fact an algebraic extension of the field $K(x)$ and moreover $(F(x):K(x))=n$.    How do I approach this exercise? I'm new in this area so any help would be greatly appreciated!
72934
I'm trying to prove that every finite abelian group is the Galois group of of some finite extension of the rationals. I think I'm almost there.   Given a finite abelian group $G$, I have constructed field extensions whose Galois groups are the cyclic groups occurring in the direct product of $G$. How do I show that the compositum of these fields has Galois group $G$.  Cheers
43458
I think I remember seeing a theorem about this in my Abstract Algebra class, but I cannot seem to find the reference for it anymore.   Let $G$ be a group and let $H$ and $K$ be non trivial subgroups of $G$. Are there any sufficient conditions on $H, K$ and $G$ such that $G = HK$? (In other words $\forall g \in G$, there exist $h \in H$ and $k \in K$ such that $g = hk$)
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!
118198
I want to prove that $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})=\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ for $p_{1},p_{2}$ primes. I know this was proved before for the generalized case of $p_{1},....,p_{n}$ primes here:    And every proof of this kind of equality involves Galois theory but I still don't know Galois theory and I'm supposed to prove this with only basic  Field theory(field extensions, irreducible polynomials, algebraic extensions, etc).  By definition $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt p_{1}+\sqrt p_{2}$, also $\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ is the smallest field containing $\sqrt p_{1}$ and $\sqrt p_{2}$. Or  $$\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}})=\{a+b\sqrt{p_{1}}+c\sqrt{p_{2}}+d\sqrt{p_{1}p_{2}} \mid a,b,c,d\in\mathbb{Q}\}$$   $$\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}) = \lbrace a+b(\sqrt{p_{1}}+\sqrt{p_{2}}) \mid a,b \in \mathbb{Q} \rbrace  $$  Also I know that $[\mathbb{Q}(\sqrt p_{1},\sqrt p_{2}):\mathbb{Q}]=4$ from this:    Still don't know how to proceed proving this two extensions are the same.
115087
Let $K/F$ be a field extension, I know that if $K/F$ is a finite extension then a simple argument from linear algebra shows that since every homomorphism of fields from $K$ to $K$ that fixes $F$ is 1-1 it is also onto, i.e. an automorphism of $K$.  Can someone please give an example that this is not the case if $K/F$ is not a finite extension ?