# Linked Questions

3answers
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### 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 ...
1answer
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### 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 $f(X)\in A[X]$....
3answers
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### 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 ...
4answers
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### 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 ...
3answers
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3answers
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### Are there any quadratic subfields of $\mathbb{Q}(\sqrt{1+\sqrt{2}})$?

I suspect that the field $\mathbb{Q}(\sqrt{1+\sqrt{2}})$ 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 ...
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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