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

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 $$\mathbb{Z}[X]$$.

• Why should $g$ and $h$ have degree $2$? May 5 '19 at 17:26
• Wouldn't it need to be a product of a degree $2$ and a degree $\bf 3$ polynomial?
– user403337
May 5 '19 at 17:27
• May 5 '19 at 17:33
• Indeed, it follows immediately from your previous question, since modulo $2$ it is $X^5+X^3+1$. May 5 '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 . May 5 '19 at 18:18

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 $$X+1$$.

• Equivalently: $\, \gcd(f,\,x^3-1) = 1,\,$ which is done by the Euclidean algorithm in my answer. May 5 '19 at 18:37

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

Remark  Above is 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: $\,x^5-x-1\pmod 3$ May 5 '19 at 18:14

$$\bar f(x)=x^5+x^3+1$$ 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.