# Questions tagged [irreducible-polynomials]

Often called prime polynomials. Polynomials that have no polynomial divisors.

297 questions
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### How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?

Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$? Right now I'm able to prove that it has no roots and that it is separable, but I have not ...
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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 ...
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### Number of monic irreducible polynomials of prime degree $p$ over finite fields

Suppose $F$ is a field s.t $\left|F\right|=q$. Take $p$ to be some prime. How many monic irreducible polynomials of degree $p$ do exist over $F$? Thanks!
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### Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than 5

Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than $5$. This is what I tried: It's evident that $x,x+1$ are irreducible. Then, use these to find all ...
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### When is $X^n-a$ is irreducible over F?

Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a$ in $F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show ...
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### Irreducibility of $X^{p-1} + \cdots + X+1$

Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ? Our professor gave us already one, namely to substitute $X$ with $X+1$, ...
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### Eisenstein Criterion with a twist

As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a ...
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### Simple extension of $\mathbb{Q} (\sqrt[4]{2},i)$
I know the simple extension of $\mathbb{Q} (\sqrt[4]{2},i)$ is $\mathbb{Q} (\sqrt[4]{2}+i)$, but how do I show this? One direction is easy, but I have trouble with the other direction.