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4 votes
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Galois group of $\mathbb{C}(t)$ over $\mathbb{C}(t+t^{-1})$

This is one of the rare cases in which it's eaiser to compute Galois group than the degree first. Without considering the Galois group, it's not easy to show $x^{2n}-(t^n+t^{-n})x+1$ is irreducible. ...
Just a user's user avatar
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Determine the degree of a root of $X^p-X-\alpha$ (Serge Lang Algebra exercise VI.29)

I'll give you a new proof for this question. Proof. Observe that $F$($θ$) is the splitting field of a separable polynomial, hence a Galois extension of $F$. We also know that [$F$($θ$) : $F$] = $p$ so ...
Hyperplane lover's user avatar
0 votes

"Abelianicity" of the group of automophisms of the fundamental functor

I think there ought to be something along the following lines, but I don't know how the details work out and I don't know if anyone has found it worthwhile to write anything like this down. A Galois ...
Qiaochu Yuan's user avatar
3 votes
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A question about Hilbert Theorem 90 and Artin-Schreier Theorem

Here is a more hands-on argument avoiding the normal basis theorem. I assume you know this, but just to state the difficulty clearly for a general audience: we of course always have $\text{tr}_{K/F}(1)...
Qiaochu Yuan's user avatar
4 votes

A question about Hilbert Theorem 90 and Artin-Schreier Theorem

That is because, since $K/F$ is Galois, there exists $\vartheta\in K$ such that the family $(\sigma(\vartheta))_{\sigma\in{\rm Gal}(K/F)}$ is an $F$-basis of $K$. Now, $$ {\rm Tr}_{F/K}(\vartheta)=\...
Tuvasbien's user avatar
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0 votes

Does the uniformizer $\pi$ generate the extension $K/T$ where $T$ is the inertia field?

I'll show how to do this when $K/T$ is a totally ramified extension of local fields, and I'll leave you to figure out how to translate this to the exact statement you've given. The key fact is that $\...
CJ Dowd's user avatar
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2 votes
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Can restriction map trivialize a element in $H^1(G_K,M)$ by finite extension?

$\DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\res}{res}$ Here is a simple explicit construction of such an $L$. Since $M$ is finite, the map $G_K\to \Aut M$ ...
Alex B.'s user avatar
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1 vote
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In general, nested and unnested radical extensions have distinct Galois groups?

Yes, the Galois groups of the two cases differ in general. The nicest possible case is in the setup of Kummer theory where $F$ has all $(nm)^{th}$ roots of unity. Say that $\alpha$ is a root of an ...
Qiaochu Yuan's user avatar
0 votes
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Two different embeddings $ \mathbb R \to \mathbb C $ over $ \mathbb R $?

For a field extension $K\subset L$, the separable degree is usually defined as $$[L:K]_s:=|\operatorname{Hom}_K(L,\overline K)|,$$ i.e. the $K$-linear field morphisms from $L$ to an algebraic closure ...
anankElpis's user avatar
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3 votes

Galois group of $X^3-X+1$ over $\mathbb{Q}$ and $\mathbb{R}$ without discriminant.

To be clear, $f$ is not strictly increasing, but its derivative is $f'(x)=3x^2-1$, hence it's strictly increasing over $(-\infty, -\sqrt{1/3}]$ and $[\sqrt{1/3}, \infty)$, but decrease over $[-\sqrt{1/...
Just a user's user avatar
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0 votes

Alternative method : $\mathbb{Q}(\zeta_3+\sqrt[3]{7})=\mathbb{Q}(\zeta_3,\sqrt[3]{7})$

Also true: let $$ a = \zeta_3+\sqrt[3]{7} $$ THEN $$ \frac{-8}{9} - \frac{5a}{72} + \frac{23a^2}{72} + \frac{a^3}{36} + \frac{a^4}{72} - \frac{a^5}{72} = \zeta_3 $$ $$ \frac{8}{9} + \frac{77a}...
Will Jagy's user avatar
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2 votes
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Alternative method : $\mathbb{Q}(\zeta_3+\sqrt[3]{7})=\mathbb{Q}(\zeta_3,\sqrt[3]{7})$

This is totally fine, but you have to check $\alpha$ indeed have six distinct Galois conjugates. If you have done so, then you don't have to form the minimal polynomial. Since the only element in $\...
Just a user's user avatar
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-1 votes

The $n$-th root in the ring of polynomial over a field containing $\mathbb{C}$

It suffices to assume the characteristic of $K$ is $0$ (not necessarily an extension of $\mathbb C$). Let $g=a_{I_1}X^{I_1}+a_{I_2}X^{I_2}+\cdots+a_{I_n}X^{I_m}$, where $I_1>I_2>\cdots>I_m$ ...
Just a user's user avatar
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3 votes
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Why cyclic subgroup of $\text{Gal}(L/K)$ is achieved by some decomposition group?

Let $G = {\rm Gal}(L/K)$. Chebotarev says each conjugacy class in $G$ is a conjugacy class of Frobenius elements at infinitely many primes in $L$. Once one element in a conjugacy class in $G$ is a ...
KCd's user avatar
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1 vote

Simplest unsolvable quintic with one real root

Another simple example is $f = x^5 + 2x + 2$ over $\Bbb Q$. By Eisenstein, $f$ is irreducible over $\Bbb Q$. It has exactly one real root, namely $x \approx -0.817471019001$. Furthermore it has the ...
Dietrich Burde's user avatar
1 vote
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Yes/No .An extension having Galois group of order $1$ is normal

As noted in the comments OP is following Ian Stewart's book, in which the definition of the Galois group of a field extension $E/F$, denoted $\Gamma(E/F)$, is given as the set of $F$-automorphisms of $...
Warren Moore's user avatar
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0 votes

An exercise about abelian Kummer extensions

I don't know how to finish part (f). To show that $K=F(u_1, ⋯, u_r)$, it suffices to show $\operatorname{Gal}(K/F(u_1, ⋯, u_r))$ is trivial. Suppose $g∈G$ fixes $u_1, ⋯, u_r$, then $g(u_i)=u_i$, so $...
hbghlyj's user avatar
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1 vote

Rationalizing General Denominators

Very old question but for those interested, you can also use matrices for this, although it's arguably not very pretty. Multiplication by some rooty number N can be represented as a linear ...
law-of-fives's user avatar
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1 vote
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Searching Explanation/References for a certain Corollary to Chebotarev Density Theorem

And the Chebotarev Density Theorem tells us that infinitely many primes split totally This is only a corollary of Chebotarev. The full theorem tells us much more than this and we need more of the ...
Qiaochu Yuan's user avatar
4 votes

What are good sources for examples of field extensions?

These kinds of examples (perhaps in the form of exercises) should be in any good Galois theory book. Here are some common examples of extensions $F/K$: $K=\mathbb{Q}$, $F=\mathbb{Q}(\sqrt[3]{2})\cong ...
Alex B.'s user avatar
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2 votes
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Galois group of splitting field of $x^3-5$ over $\mathbb F _7$

You can generalize the idea for characteristic $0$ fields. However, for $\mathbb{F}_7$, you can verify that $x^3-1$ splits over $\mathbb{F}_7$ already, so the field extension $\mathbb{F}_7(t)$ with $t^...
Ubik's user avatar
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1 vote
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why the map $X \mapsto X^2$ induces a $K$-monomorphism?

According to the universal property of polynomial rings, there exists a unique homomorphism $\varphi$ from $K[X]$ to $K(X)$ that extends the embedding $K\hookrightarrow K(X)$ and sends $X$ to $X^2$. ...
Just a user's user avatar
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1 vote

In finite fields, generators of $F^*$ under automorphisms in Galois group are also generators

An automorphisms in the Galois group induces an automorphism of the (cyclic) multiplicative group. An automorphism of a cyclic group carries generators to generators.
Lubin's user avatar
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1 vote
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In finite fields, generators of $F^*$ under automorphisms in Galois group are also generators

We say an element $a$ generates $(F)^×$ if every element $y \in (F)^×$, or equivalently, every nonzero element $y$ in $F$, can be written $y=a^r$ for some intrger $r$. And here, by a proper divisor $e$...
Mike's user avatar
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0 votes

Prove that $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{3}, \sqrt[5]{3}, \xi_3,\xi_5) \colon \mathbb{Q}] = 240$.

Consider the two subfields $K_1=\mathbb{Q}(\sqrt3, \sqrt[3]3, \sqrt[5]3)$ and $K_2=\mathbb{Q}(\zeta_3,\zeta_5)$ of $L=\mathbb{Q}(\sqrt3, \sqrt[3]3, \sqrt[5]3,\zeta_3,\zeta_5)$. It is easily seen that $...
nor's user avatar
  • 121
3 votes

The Frobenius Endomorphism is Surjective iff the field is perfect

See what happens when the $p$th power map on a field $K$ with characteristic $p$ is not surjective: if $K$ contains an $a$ that is not a $p$th power in $K$ then $x^p - a$ is irreducible in $K[x]$ and ...
KCd's user avatar
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1 vote
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Constructing a primitive element for each fixed field of a Galois extension

I think I have an answer. Let $G=Gal(K/F)$ Since $N=\{\sigma(\alpha):\sigma\in G\}$ is a normal basis, then $|N|=[K:F]=|G|$ since $K/F$ is Galois. Consider now the map $G\rightarrow N,\sigma\mapsto\...
A Name's user avatar
  • 306
9 votes
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Isomorphism of topological groups

You never showed $\psi$ is surjective, so you can't say you showed yet that the Galois group (merely as a group, ignoring the topology) is isomorphic to the infinite product of $\{\pm 1\}$ that you ...
KCd's user avatar
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1 vote

Show Galois Group of a polynomial is isomorphic to $S_n$

I'm assuming you already proved that $f$ is irreducible and has exactly two non-real roots over $\mathbb C$. Since it is irreducible, the size of the Galois group of $f$ is divisible by $p$ and by ...
Derivative's user avatar
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1 vote

Roots of $3x^5 - 15x +5$

The analytical answer to your problem is simple if you use the Lambert-Tsallis function. The solution is: $$x= \frac{1/3}{1 + W_r\bigg(-\frac{r}{5}(\frac{1}{3})^{r} \bigg)}$$ with $r=4$. The video ...
K Z Nobrega's user avatar
3 votes
Accepted

I don't understand the proof that extensions that aren't separable have 0 trace in Keith Conrad's norm and trace notes.

"The integer $[L:k(a)]$ is $0$ in $k$" is just a shorthand for "the canonical image in $k$ of this integer is $0$". It implies that $\operatorname{Tr}_{L/k}(a)$ is $0$ because (as ...
Anne Bauval's user avatar
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1 vote

Show that $\mathbb{Q}(\sqrt[3]{5} \cdot \sqrt{2}) \subset \mathbb{Q}(\sqrt[3]{5}+ \sqrt{2})$

If you want to prove$ \mathbb{Q}(\sqrt[3]{5} \sqrt{2})\subseteq \mathbb{Q}(\sqrt[3]{5}+ \sqrt{2} )$, then: Let $u= \sqrt[3]{5},v=\sqrt{2}$. $M_{1}=u+v \in \mathbb{Q}(\sqrt[3]{5}+ \sqrt{2}$ In the ...
Mahmoud albahar's user avatar
1 vote
Accepted

Show that $\mathbb{Q}(\sqrt[3]{5} \cdot \sqrt{2}) \subset \mathbb{Q}(\sqrt[3]{5}+ \sqrt{2})$

I think proving $\mathbb{Q}(\sqrt[3]{5}+ \sqrt{2} ) \subseteq \mathbb{Q}(\sqrt[3]{5} \sqrt{2} )$ is easier combined with the equality of degree of both of them. Let $u=\sqrt[3]{5}, v=\sqrt{2}$, we ...
Mahmoud albahar's user avatar
4 votes

Transforming the reduced sextic $x^6+x^2+ax+b$ into a quintic

Since it has been several days and there is no answer, I guess it is ok to do it myself. To transform the reduced sextic $$x^6+x^2+ax+b=0$$ to the quintic $$K_5x^5+K_4x^4+K_3x^3+K_2x^2+K_1x+K_0=0$$ is ...
Thinh Dinh's user avatar
30 votes
Accepted

"Galois theory" on graphs

There is actually an exact analogue of Galois theory in this context, given by the theory of covering spaces in topology. Covering space theory defines a topological version of a (separable) field ...
Qiaochu Yuan's user avatar
1 vote

Is this extension cyclic? proof of Dummit and Foote

$F'/F$ may not be cyclic, but all we need is abelian. The key is that there simply needs to exist a chain of cyclic intermediate extensions, not that all possible intermediate fields are cyclic. ...
Dheeran Wiggins's user avatar
2 votes
Accepted

Subfields of splitting field of $x^4+25$ over $ℚ$.

I think, you are right. Of course, $r_1+r_2=\sqrt{10}i\in L$ and $r_1+r_4=\sqrt{10}\in L$, so that the splitting field is given by $L=\Bbb Q(\sqrt{10},i)$. This field doesn't contain $\sqrt{5}$, or $\...
Dietrich Burde's user avatar
2 votes

When can we divide a shape into n equal parts (like Gauss and Abel)?

For the cardioid $r=1-\cos\theta$, the counterclockwise arc length from the cusp to any point is determined by calculus to be $4[1-\cos(\theta/2)]; 0\le\theta\le2\pi$ versus a full perimeter of $8$ (...
Oscar Lanzi's user avatar
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1 vote
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If $f(x)\in \mathbb{Z}[x]$ is irreducible (over $\mathbb{Q}$), is it always possible to find $a$ and $b$ in $\mathbb{Q}$ with $f(ax+b)$ Eisenstein?

To check that your proof works for rational numbers as well, you may just look at $f((mx+n)/k)$ where $m,n,k$ are integers. After factoring out $1/k^2$, this becomes $$m^2x^2+2mnx+(n^2+4k^2).$$ So you ...
Dave's user avatar
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5 votes
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Does the Galois group of a polynomial change upon translation?

Yes, because the splitting fields of the polynomials will be equal. Suppose $f,g\in K[X]$ are polynomials where $f(X+1)=g(X)$. If $\alpha_1, \dots, \alpha_n$ are the roots of $f$, then $\alpha_1-1,\...
Yaneda's user avatar
  • 777
2 votes
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Artin-Schreier extension is cyclic of degree 1 or $p$

It is not true that $f$ is irreducible if $a$ equals $\alpha^p - \alpha$ for some $\alpha \in K$, as then $f(X) = (X-\alpha)(X-\alpha-1)\cdots(X-\alpha-(p-1))$ and $L = K$ is of degree $1$. So assume $...
Béranger Seguin's user avatar
4 votes
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Why is the subfield of $\mathbb{Q}(\zeta_p)$ of index $2$ expressible in terms of the sum of $\zeta_p$ to the power of all quadratic residues mod $p$?

It is called a subfield of degree $2$ (over $\mathbf Q$), or more commonly a quadratic subfield. To see a number $\alpha$ in $\mathbf Q(\zeta_p)$ is quadratic over $\mathbf Q$, we can do two things: (...
KCd's user avatar
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