# Irreducibility of $x^4-5$ over $\mathbb{Z}_{17}$

Obviously, it'd be hard to try all the $17$ elements to see if there is some root, and even if there is none, it'd be necessary to verify if it can't be factored into two irreducible 2 degree polynomials. The Eisenstein criterion will just be able to say about irreducibility on $\mathbb{Q}[x]$.

What can I do?

I must also verify it for the polynomials $x^3-5$ and $x^4+7$, both over $\mathbb{Z}_{17}$.

• First of all, it's not that hard to check all the $9$ elements (notice that the degree of the polynomial is even). But the hard part just begins: you also have to show that $x^4-5$ is not a product of two polynomials of degree two. – user26857 Jan 11 '16 at 1:47

## 4 Answers

If $x^4\equiv 5\pmod{17}$, then $x^{16}\equiv 5^4\equiv 13\pmod{17}$, but this contradicts Fermat's Little theorem.

More strongly, $y^2\equiv 5\pmod{17}$ has no solution. By Quadratic Reciprocity:

$$\left(\frac{5}{17}\right)=\left(\frac{17}{5}\right)=\left(\frac{2}{5}\right)=-1$$

• But doesn’t this just show that $X^4-5$ has no root in $\Bbb F_{17}$? – Lubin Jan 11 '16 at 1:39
• I should not use these theorems, I'm in a ring theory course. Could you give a specific answer on this, if possible? – Magritte Jan 11 '16 at 1:41
• You're right. This really didn't deserve the upvotes. – user236182 Jan 11 '16 at 2:19

Assuming you mean the finite field $\mathbb F_{17}$ as opposed to the $17$-adic integers $\mathbb Z_{17}$, here is a method you can use to check for quadratic factors:

There is a unique field extension $\mathbb F_{17^2}$ of $\mathbb F_{17}$ or degree $2$. Thus, every irreducible quadratic polynomial defines the same field extension. Therefore, $x^4 - 5$ has no quadratic factors if and only if it does not have a root in $\mathbb F_{17^2}$.

The order of $5$ in $\mathbb F_{17}^\times$ is at most $16$ by Fermat's little theorem. On the other hand, if the order were $8$ or lower, then $5$ would be a square (because $\mathbb F_{17}^\times$ is cyclic). This is impossibly by user236182's answer. Therefore, $5$ is a primitive unit in $\mathbb F_{17}$.

Since the group $\mathbb F_{17^2}^\times$ is cyclic of order $17^2 - 1 = 288 = 32 \cdot 9$, and $5$ has order $16$, we conclude that $5$ cannot be a fourth power in $\mathbb F_{17^2}^\times$ (for otherwise the order would divide $\frac{288}{4} = 8 \cdot 9$, which it clearly doesn't).

Thus, the polynomial $x^4 - 5$ is irreducible over $\mathbb F_{17}$.

• I really appreciate your help but I can't use these, I should use more ring theory tools and less number theory – Magritte Jan 11 '16 at 1:47
• I think your answer here is clearer than mine. Good. – Lubin Jan 11 '16 at 2:00

You’re looking for a fourth root of $5$ in a field of characteristic $17$. You can show by direct computation that $5$ is a “primitive residue modulo $17$”, that is, that $5$ generates the multiplicative group $\Bbb F_{17}^*$ of nonzero elements of this finite field. In particular, it is not a fourth power, so that $X^4-5$ has no root in $\Bbb F_{17}$. This is not good enough.

But notice that $5$ has period $16$ in $\Bbb F_{17}^*$, and its fourth root must have period $4\cdot16=64$ in its field. What is the smallest field of characteristic $17$ that has elements of period $64$? We need $17^n-1$ to be divisible by $64$, and the smallest such number is $n=4$, for $(17^2-1)/64=9/2$, but $17^4-1=64\cdot1305$. Thus the minimal polynomial for $\sqrt[4]5$ over $\Bbb F_{17}$ must be $X^4-5$ : an irreducible polynomial.

Hint: Prove that the equation on $a,b,c,d$ $$x^4+7=(x^2+ax+b)(x^2+cx+d)$$ has no solutions in $\mathbb{Z}_{17}.$