Show that the polynomial $7X^5 + 71X^3 - 9$ is irreducible in $\mathbb{Z}[X]$

Question

Show that the polynomial $f(X) = 7X^5 + 71X^3 - 9$ is irreducible in $\mathbb{Z}[X]$

My solution:

Using the irreducibility test: "Reduction Mod p Test"

$f(X)$ is clearly primitive and that the prime number 2 does not divide the leading coefficient.

It is therefore enough to prove the polynomial $\bar{\pi}_2(f(X)) = X^5+X^3-\bar{1} \in (\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible.

$\bar{\pi}_2(f(X))$ has no roots in $(\mathbb{Z}/2\mathbb{Z})[X]$ (to verify this evaluate the polynomial at the two elements $\bar{0}$ and $\bar{1}$ of $(\mathbb{Z}/2\mathbb{Z})[X]$

Thus, if it were reducible then it would have to be of the form $g(X) \cdot h(X)$, where $g(X), h(X) \in (\mathbb{Z}/2\mathbb{Z})[X]$ are both irreducible of degree $2$. Which is not possible.

Concluding that $\bar{\pi}_2(f(X))$ is irreducible in $(\mathbb{Z}/2\mathbb{Z})[X]$, whereby $f(X)$ is irreducible in $\mathbb{Z}[X]$.

Wouldn't it need to be a product of a degree $2$ and a degree $\bf 3$ polynomial? — , May 05 '19 at 17:27
Indeed, it follows immediately from your previous question, since modulo $2$ it is $X^5+X^3+1$. — Dietrich Burde, May 05 '19 at 18:13
@DietrichBurde That's a different question, which yields one way to prove the titled question here. It is not a dupe of the prior question. So I have reopened it . — Bill Dubuque, May 05 '19 at 18:18

score 1 · Answer 1 · answered May 05 '19 at 17:57

1

Modulo $2$, this polynomial is nothing else than $X^5+X^3+1$. As it has no root in $\mathbf F_2$, if it could be factored in $\mathbf F_2[X]$, it would the product of an irreducible quadratic factor and an irreducible cubic factor. Now there's only one irreducible quadratic polynomial in $\mathbf F_2[X]$: $\;X^2+X+1$, and you can check the result of the Euclidean division of $X^5+X^3+1$ by $X^2+X+1$ has a remainder of $X+1$.

answered May 05 '19 at 17:57

Bernard

175,478

Equivalently: $, \gcd(f,,x^3-1) = 1,,$ which is done by the Euclidean algorithm in my answer. – Bill Dubuque May 05 '19 at 18:37

score 1 · Answer 2 · 2019-05-05T18:40:37.923

$\bar f(x)=x^5+x^3+1$ would have to be divisible by $x^2+x+1$, since the other three degree two polynomials are reducible.

So suppose $\bar f(x)=g(x)(x^2+x+1)$. Then let $g(x)=x^3+ax^2+bx+c$. Then $\bar f(x)=x^5+(a+1)x^4+(1+a+b)x^3+(a+b+c)x^2+(b+c)x+c$. So $a=1, c=1,b=1$ which implies $a=0$, a contradiction.

That is, the system of equations is inconsistent, and $f(x)=7x^5+71x^3-9$ is irreducible.

Bill Dubuque · Answer 3 · 2023-02-06T19:53:34.320

Hint $f\,$ has no roots in $\Bbb F_2$ so no linear factors so if $f$ splits it has an irred. quadratic factor $g$, so in $\,\Bbb F_2[x]/g \cong \Bbb F_4\!:\,$ $\,\color{#c00}{x^3\! =\! 1}\,$ so $\ \color{#0a0}0 \!=\! f\! =\! \color{#0a0}{x^2}(\color{#c00}{x^3})+\color{#c00}{x^3}\!-1 = \color{#0a0}{x^2}\,$ so $\,\color{#c00}{1\! =\! x^3}\! =\! \color{#0a0}{x^2}\color{c00}{x} \!=\!\color{#0a0}0,\,$ contradiction.

Remark $ $ We used $\,0\neq f\in \Bbb F_4\Rightarrow f^{\large \color{#c00}3}\! = 1\,$ (for $\,f\!=\!x),\,$ an analogue of Fermat's little Theorem, which is true because $\,\Bbb F_4$ has multiplicative group $\,\Bbb F_4^*$ of size $\,\color{#c00}3 =4\color{darkorange}{-1} \,$ ($\rm non\color{darkorange}{zero}$ elements are invertible in a field), so Lagrange's Theorem $\Rightarrow f^{\large\color{#c00}3}\! = 1\,$ for all $\,f\neq 0.$

Above is essentially a special case of a general polynomial irreducibility test over finite fields - which is an an efficient analog of the impractical Pocklington-Lehmer integer primality test.

See here for another example done this way: $,x^5-x-1\pmod 3$ — Bill Dubuque, May 05 '19 at 18:14

Show that the polynomial $7X^5 + 71X^3 - 9$ is irreducible in $\mathbb{Z}[X]$

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