5 votes
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Irreducible polynomial in $\Bbb{Z}_2[x]$

The answer is no. The first counterexample is \begin{multline*} x^{23}+x^{21}+x^{19}+x^{17}+x^{15}+x^{13}+x^{11}+x^9+x^7+x^5+x^3+x+1 \\= \left(x^3+x^2+1\right) \left(x^4+x+1\right) \left(x^{16}+x^{...
Greg Martin's user avatar
4 votes

Using Eisenstein's Criterion with a transformation

Yes; this works because polynomials (with non-zero constant term) in $R[x]$ are reducible if and only if they are reducible in $R[x,x^{-1}]$. Related (maybe duplicate): Substituting $y=1/x$ to apply ...
Shean's user avatar
  • 927
3 votes
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Prove that $x^4+3x^3+3x^2-5$ is irreducible over $\mathbb{Q}$

First of all, by Gauss's lemma, irreducibility over $\mathbb{Q}$ is here equivalent to irreducibility over $\mathbb{Z}$. There are no rational roots, so the only possibility is that the polynomial is ...
Mark's user avatar
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3 votes
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Irreducibility of $X^4-\sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$.

Note that $f(x)=x^8-2$ is irreducible over $\mathbb{Q}$ by Eisenstein. Also, $x^8-2=(x^4-\sqrt{2})(x^4+\sqrt{2})$. So if $\alpha\in\mathbb{C}$ is any root $x^4-\sqrt{2}$ then $f$ is the minimal ...
Mark's user avatar
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2 votes
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Investigate whether the polynomial $q(x) = 2x^5 - 78x^3 + 39x + 21$ is irreducible in $\mathbb{F}_{13}[x]$.

It is a classical result that for any field $K$ and any $P\in K[X]$, we have $$ P(a)=0\iff X-a\mid P. $$ In particular, an irreducible polynomial in $K$ of degree $\geqslant 2$ has no roots in $K$. We'...
Tuvasbien's user avatar
  • 8,977
2 votes

Investigate whether the polynomial $q(x) = 2x^5 - 78x^3 + 39x + 21$ is irreducible in $\mathbb{F}_{13}[x]$.

The claim is that it decomposes into two smaller polynomials neither of which is a unit, as a factor $x - a$ is not a unit in $\mathbf{F}_{13}[x]$. To answer your second question, think about what ...
Anton Odina's user avatar
2 votes

If $p(x) = a_3 + a_2x + a_1x^2 + a_0x^3$ is irreducible in $\mathbb{Q}[x]$, then $q(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ is also irreducible?

$p(x) = a_3 + a_2x + a_1x^2 + a_0x^3\Rightarrow p\left(\dfrac 1x\right)=\dfrac{a_3x^3 + a_2x^2 + a_1x + a_0}{x^3}=\dfrac{q(x)}{x^3}$ If $q(x)$ were reducible then for some $\beta\ne0$ (because $a_0\...
Piquito's user avatar
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2 votes
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is $x^4-x+1$ irreducible in $\mathbb{Z}_3$

In order to check if a polynomial of degree $4$ is irreducible you need to check first if it can be divided by polynomials of degree $1$ like $x+1$. if you evaluate your polynomial at $x=-1$ you get: ...
B.A.M's user avatar
  • 431
2 votes

Irreducible polynomial in $\Bbb{Z}_2[x]$

Some simple (non-computer) ways of seeing the existence of non-trivial factors. In the end I prove that every primitive polynomial $\in \Bbb{F}_2[x]$ of degree $>2$ is a factor of infinitely many ...
Jyrki Lahtonen's user avatar
1 vote
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Proving the irreducibility of a polynomial in a field extension.

For each $q$, there is at most one field of size $q$ in any field, since they are all the roots of the polynomial $x^q-x$. Therefore you don't have to do any calculation to conclude $x^3+x^2+2$ is ...
Just a user's user avatar
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1 vote
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Irreducible polynomials with complex root.

Let $h(x)=\gcd(f,g)$. Since $f$ is irreducible, either $h$ is a unit, or $h$ is an associate of $f$. But it cannot be a unit: if $\gcd(f,g)=1$, then there exist polynomials $p(x)$ and $q(x)$ in $\...
Arturo Magidin's user avatar
1 vote
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Factorization and irreducibilty for $x^n-2x^m+1$ trinomials.

This is a consequence of known results on irreducibility and factorization of trinomials, e.g. see Schinzel's theorem below, excerpted from this 2020 survey by Koley and Reddy. Beware there is typo ...
Bill Dubuque's user avatar
1 vote

Irreducibility of $X^4-\sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$.

The non-trivial factors of $\ X^4-\sqrt{2}\ $ are $\ X-2^\frac{1}{8},$$\,X+2^\frac{1}{8},$$\,X-i2^\frac{1}{8},$$\,X+i2^\frac{1}{8},$$\,X^2-2^\frac{1}{4},$$\,X^2+2^\frac{1}{4},$$\,X^2-(1+i)X+i2^\frac{1}...
lonza leggiera's user avatar
1 vote

Investigate whether the polynomial $q(x) = 2x^5 - 78x^3 + 39x + 21$ is irreducible in $\mathbb{F}_{13}[x]$.

$q(x) = 2x^5 - 78x^3 + 39x + 21$ is equivalent to $f(x)=2x^5+8=2(x^5+4)$ in $\Bbb F_{13}[x]$ and if $x^5=-4=9$ modulo $13$ then $x=3$ because $3^3=1$ and $3^2=9$ modulo $13$. Then $q(x)$ is not ...
Piquito's user avatar
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1 vote
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If $p(x) = a_3 + a_2x + a_1x^2 + a_0x^3$ is irreducible in $\mathbb{Q}[x]$, then $q(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ is also irreducible?

It's easier to prove the contrapositive: If $q(x)=a_0+a_1x+a_2x^2+a_3x^3$ is reducible, then $p(x)=a_3+a_2x+a_1x^2+a_0x^3$ is reducible. I'm actually just going to prove a generalized version of this, ...
Noble Mushtak's user avatar
1 vote

Proof $f=x^2y^5+x^3+xy^2+y^2+x-1$ is irreducible in $\mathbb{C}[x,y]$

This is basically the same as Rotman exercise 6.28, 6.30. Thus, I write this answer for my own review. Proposition: Let $K$ be a field, $f \in K[x_1, \cdots, x_n]$, and $f$ is primitive in $(K[x_1, \...
Long Song's user avatar
1 vote

Is a polynomial monotone when the first derivative has only imaginary roots?

Consider $f(x):=2x^3-9x^2+12x$. Then $f(x)$ has just one real zero, at $x=0$, the other two being complex. Also, its gradient $f'(x)=6(x^2-3x+2)=6(x-1)(x-2)$ is $12$ at $x=0$ and $x=3$, while it has a ...
John Bentin's user avatar
  • 18.5k
1 vote

Is a polynomial monotone when the first derivative has only imaginary roots?

Not necessarily. The first derivative being positive at the lower and upper bounds of the range indicates that the function is increasing at those points. However, the function could still have other ...
Antony Theo's user avatar
1 vote
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Prove that $f(x)$ is irreducible in $\mathbb{Z}$ with $f(b)$ a prime, $f(b-1) \neq 0$ and $\Re(\alpha_i) < b -1/2$

For the sake of contradiction assume $f(x)=g(x)h(x)$ is a non-trivial factorization. Then from $f(b)$ being a prime we can assume (without loss of generality) that $|g(b)|=1$. Now let $g(x)=a\prod (x-\...
Sil's user avatar
  • 16.6k
1 vote
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irreducibility in p-adic field

I claim that $f$ is irreducible, independent of concrete choice of $u$. Note that we know all the roots of $f$: They are of the form $\zeta_p^k u^{1 / p}$ for $\zeta_p$ a primitive $p$th root of unity,...
Ben Steffan's user avatar
  • 2,947

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