Linked Questions

2
votes
3answers
1k views

Integral polynomial irreducible over $\mathbb Q$ but reducible mod $2$, $3$, and $5$ [duplicate]

Possible Duplicate: Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$ Anyone know an $f\in \mathbb{Z}[x]$ that is irreducible over $\mathbb{Q}$ but ...
1
vote
1answer
157 views

converse of reduction criterion? [duplicate]

By the reduction criterion, I mean the following test for the irreducibility of polynomial with Dedekind domain coefficients. Let $\mathfrak{m}$ be maximal in Dedekind domain A and $f(X)\in A[X] $....
33
votes
3answers
9k views

Irreducible polynomial which is reducible modulo every prime

How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$? For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not ...
40
votes
4answers
4k views

Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

I have read the following theorem: If $p_1,p_2,\dots,p_n$ are distinct prime numbers, then$$\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.$$ I have tried to prove a ...
32
votes
3answers
4k views

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[...
32
votes
2answers
4k views

“Standard” ways of telling if an irreducible quartic polynomial has Galois group C_4?

The following facts are standard: an irreducible quartic polynomial $p(x)$ can only have Galois groups $S_4, A_4, D_4, V_4, C_4$. Over a field of characteristic not equal to $2$, depending on whether ...
11
votes
3answers
700 views

Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?

We know that $f(x)=x^4+1$ is a polynomial irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$. My question is: Is it true that if $f(x)$ has a linear factor over $\...
11
votes
2answers
720 views

Proving that $x^4 - 10x^2 + 1$ is not irreducible over $\mathbb{Z}_p$ for any prime $p$.

So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but ...
6
votes
4answers
244 views

Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$

Is there any way to determine the Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$ not using the discriminant? Thanks!
4
votes
2answers
808 views

Minimal polynomial reducible modulo every prime $p$

Suppose $K = \mathbb{Q}(\alpha)$ with $\alpha = a + b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ with $D_1,D_2 \in \mathbb{Z}$. Prove that the minimal polynomial $m_\alpha(x)$ for $\alpha$ over $\mathbb{Q}$...
10
votes
2answers
261 views

Galois group of $P(X)=X^6−4X^3+2$

I would like to find Galois group of $P(X)=X^6−4X^3+2$. I have that $(2+\sqrt{2})^{1/3}$ is a root of $P$, denote it $\alpha$. By Eisenstein $P(X)$ is irreducible over $\Bbb{Q}$ then $[\Bbb{Q}(\alpha)...
13
votes
1answer
514 views

Families of Polynomials Irreducible in $\mathbb{Z}$ but reducible in $\mathbb{Z}/p\mathbb{Z}$ for all primes $p$.

I am wondering if there exist classification of polynomials that are irreducible in $ \mathbb{Z}$ but reducible $\pmod p$ for all primes $p$. I am aware that $\Phi_n$ has this property if $(\mathbb{Z}...
0
votes
2answers
283 views

Solution for $x^5-x+1=0$ [closed]

I'm looking for solution for this equation: $$x^5-x+1=0$$ I know There are not solution with radicals. But, I can not find possible solution. (In MSE and internet resource) And Can you show me why ...
3
votes
3answers
181 views

Are there any quadratic subfields of $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$?

I suspect that the field $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$ does not contain any subfield $K$ with $[K:\mathbb{Q}]=2$, but I'm not sure how to prove it. More generally, if $L$ denotes the splitting ...
4
votes
3answers
299 views

Reducibility of polynomials modulo p [duplicate]

How do we show that $X^4-10X^2+1$ is reducible modulo every prime $p$? I've managed to show it for all primes less than 10, for primes greater than 10 we have $X^4+(p-10)X^2+1$. Where do I go from ...

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