# Tag Info

Accepted

### 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. ...
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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 ...

### "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 ...
• 426k
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• 17.4k

### 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 $I_1>I_2>\cdots>I_m$ ...
• 17.4k
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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 ...
• 47.5k
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 ...
• 132k
1 vote
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• 3,035
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 ...
• 1,983
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 ...
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• 473
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 $X^2$. ...
• 17.4k
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.
• 63.2k
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 $e$...
• 21k

• 306
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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 ...
• 47.5k
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 ...
• 1,788
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 ...
• 434
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### 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 ...
• 38.7k
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 ...
• 1,303
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### 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 ...
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### 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 ...
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 ...
• 426k
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. ...
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