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For a parameter $t=(t_{0},t_{1},t_{2},t_{3},t_{4},t_{5},)\in\mathbb F_{5}^{6}$ with $t_{0}\ne 0$ and {$t_{i},i>0$} are ordering of elements in $\mathbb F_{5}$ (t1~t5 is a permutation of [0]~[4] here at least as I think), define a polynomial $$P_{t}(x)=(x-t_{1})(x-t_{2})(x-t_{3})+t_{0}(x-t_{4})(x-t_{5}).$$

  1. Show that $P_{t}(x)$ is irreducible in $\mathbb F_{5}[x]$.

  2. Prove that two parameters $t,t'$ give the same polynomial over $\mathbb F_5$ if and only if $t_{0}=t_{0}'$ and $\{t_{4},t_{5}\}=\{t_{4}',t_{5}'\}$.

  3. Show that every irreducible cubic monic polynomial over $\mathbb F_{5}$ is obtained in this way.

After trying $x,x-1,x-2,x-3,x-4$ the first question can be solved. But I have no idea about where to start with the remaining two. Expanding the factor seems failed for proving two polynomials are equal to each other.

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I don't understand what stops you from taking $t_0=\cdots=t_5=1$, say, in which case $P_t$ is assuredly not irreducible. Or are you saying $t_1,\dots,t_5$ are distinct? – Gerry Myerson Jul 12 '13 at 9:43
@Gerry: You're right. But I think the sentence "$\{t_i,i>0\}$ are ordering of elements in $\mathbb{F}_5$" is trying to convey exactly the requirement that they are distinct. In that case 1. is easy, and 3. follows from 2. by a counting argument, but I don't have a useful way of getting 2. – Jyrki Lahtonen Jul 12 '13 at 9:46
@Jyrki, yes, it's trying, and almost succeeding. But I (mis)read it as "$t_0\ne0$ and $\{{t_i,i\gt0\}}$ are ordering..." and wondered how 6 terms could be an ordering of 5. – Gerry Myerson Jul 12 '13 at 9:50
@GerryMyerson $t_{1}$~$t_{5}$ should be a permutation of [$0$]~[$4$] here. – Jebei Jul 12 '13 at 10:21
Yes, I get that --- now. Could you edit the question to make that clearer? – Gerry Myerson Jul 12 '13 at 10:31
up vote 2 down vote accepted


  1. A cubic is reducible, only if it has a linear factor. But then it should have a zero in $\mathbb{F}_5$, so it suffices to check that none of $t_1,t_2,t_3,t_4,t_5$ is a zero of $P_t(x)$.
  2. This part is tricky. I would go about it as follows. Let $t$ and $t'$ be two vectors of parameters. Consider the difference $$ Q_{t,t'}(x)=P_t(x)-(x-t'_1)(x-t'_2)(x-t'_3). $$ It is a quadratic. Show that if $\{t_1,t_2,t_3\}=\{t'_1,t'_2,t'_3\}$ then $Q_{t,t'}$ has two zeros in $\mathbb{F}_5$, but otherwise it has one or none. This allows you to make progress.
  3. Count them! The irreducible cubics are exactly the minimal polynomials of those elements of the finite field $L=\mathbb{F}_{125}$ that don't belong to the prime field. The number of such elements is $125-5=120$. Each cubic has three zeros in $L$ (it's Galois over the prime field), so there are a total of 40 irreducible cubic polynomials over $\mathbb{F}_5$. How many distinct polynomials $P_t(x)$ are there?
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So the third one is actually a combination problem based on some basic Galois theory? – Jebei Jul 12 '13 at 10:55
@frame99: Not necessarily. There are more elementary methods of calculating the number of monic irreducible cubic polynomials. You can count the total number and then subtract the number of product of three linear factors and also the number of products of a linear and an irreducible quadratic. To that end you first need to count the number of irreducible quadratics (by the same process: count all the monics, subtract the reducible ones). – Jyrki Lahtonen Jul 12 '13 at 10:58
Thanks. I will try it later. – Jebei Jul 12 '13 at 11:14

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