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I am attempting to solve part (b) of problem 4.2 in Chapter 12 of Artin's Algebra textbook. I have already solved part (a).

Prove that the following polynomials are irreducible:

(a) $x^2+1$ in $\mathbb{F}_7[x]$

(b) $x^3-9$ in $\mathbb{F}_{31}[x]$

My attempt

Assume $x^3-9$ is reducible. It is easy to see that it is not a unit, so it must have a proper divisor. So we can write $x^3-9=a(x)b(x)$ for some $a,b\in \mathbb{F}_{31}[x]$ where $a,b$ are not units, hence are not constant. One of them has degree 2 and the other degree 1. WLOG assume $a$ has degree 2 and $b$ has degree 1. Then we can write $a(x)=a_0+a_1x+a_2x^2$ and $b(x)=b_0+b_1x$. Where $a_2,b_1\neq0.$

We obtain the equations

  • $a_0b_0=-9$
  • $a_0b_1+a_1b_0=0$
  • $a_1b_1+a_2b_0=0$
  • $a_2b_1=1$

Now, I would like to obtain a contradiction by playing with these four equations, but I am not sure how to do this. Please help me. I would prefer a solution which deduces a contradiction from the four equations rather than a solution using some other method. I am aware that this math problem has already been asked on stackexchange, but I do not believe that my question is a duplicate since I desire to approach the problem from a different angle.

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Pascal's Wager
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1 Answers1

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As I noted in the comments, your approach is not the best for this problem.

Instead, take advantage of the fact that the polynomial is cubic . . .

If $p(x) = x^3-9$ was reducible in $F_{31}[x]$, then since $p$ is cubic, it would have to have a linear factor.

But from a linear factor, we would get a root, $r$ say, in $F_{31}$.

Working in $F_{31}$, not only do we have $r^3 = 9$, but also, since $r$ is nonzero, we have $r^{30}=1$. \begin{align*} \text{Then}\;\;&r^3=9\\[4pt] \implies\;&r^3=3^2\\[4pt] \implies\;&(r^3)^{10}=(3^2)^{10}\\[4pt] \implies\;&r^{30}=3^{20}\\[4pt] \implies\;&3^{20}=1\\[4pt] \end{align*} but $3^{20}$ is not congruent to $1$, mod $31$, contradiction.

quasi
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  • How did you obtain that $r^{30}=1$? – Pascal's Wager Feb 28 '18 at 18:39
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    The set of nonzero elements of a field is a group under multiplication, hence for $F_{31}$, the multiplicative group of nonzero elements has order $30$. Recall that the order of an element of a finite group divides the order of the group. It follows that $r^{30}=1$. – quasi Feb 28 '18 at 18:50
  • @Pascal'sWager Alternatively (because we work in the prime field) the equation $r^{30}=1$ follows from Little Fermat: $n^{30}\equiv1\pmod{31}$ for all integers $n$ not divisible by $31$. – Jyrki Lahtonen Feb 28 '18 at 20:43