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Let $F= \mathbb{F}_{3}(a)$ a field with $a$ verifying the equation $a^{3}+a^{2}-1 = 0$.

  1. Determine the cardinality of $F$.
  2. Determine the degree of $Irr(a^{2}, \mathbb{F}_{3})$
  3. Compute $Irr(a^{2}, \mathbb{F}_{3})$

I´ve already solved (1). The polynomial $f = x^{3}+x^{2}-1 \in \mathbb{F}_{3}[x]$ is irreducible because it has now roots and is of degree 3. Since $F$ contains a root, it must be the splitting field of $f$ over $\mathbb{F}_{3}$ and this field is $\mathbb{F}_{27}$, so $|F| = 27$.

I think I´ve solved (2) but I´m not 100% sure. This is my attempt: since $a$ is a root of $f$ and $F$ is the splitting field, the remaining roots are obtained by applying the Frobenius map of the extension. Namely, the map $\tau : F \to F$, determined by $\tau(a) = a^{3}$. So the roots of $f$ are $a, a^{3}, (a^{3})^{3} = a^{9}$. From the relation $a^{3}+a^{2}-1 = 0$ we have $a^{3} = 1-a^{2}$,and since $a^{3}$ is another root of $f$ it generates the same extension (this is the point where I´m not 100% sure). It follows that $\mathbb{F}_{3}(a^{2}) =\mathbb{F}_{3}(1-a^{2}) = \mathbb{F}_{3}(a^{3}) = \mathbb{F}_{3}(a)$ and so $deg (Irr(a^{2}, \mathbb{F}_{3})) = 3$

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    $\begingroup$ See this for a standard trick answer to part 3. $\endgroup$ Commented Jun 7 at 5:14

2 Answers 2

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  1. Depending on the context of the exercise, claiming that $F$ is the splitting field of $p:=x^3+x^2-1$ over $\Bbb{F}_3$ because that cubic is irreducible may be a big leap. You can conclude the same with far less theory by noting that the map $$\varepsilon:\ \Bbb{F}_3[x]\ \longrightarrow\ \Bbb{F}_3(a):\ x\ \longmapsto\ a,$$ is a surjective ring homomorphism with $(p)\subset\ker\varepsilon$. Because $a\notin\Bbb{F}_3$ it follows that $\ker\varepsilon\neq(1)$. Because $p$ is irreducible over $\Bbb{F}_3$ it follows that $\ker\varepsilon=(p)$, and so $$\Bbb{F}_3(a)\cong\Bbb{F}_3[x]/(x^3+x^2-1),$$ from which it immediately follows that $|\Bbb{F}_3(a)|=|\Bbb{F}_3|^3=27$.

  2. You are entirely correct: Because $p(a^3)=0$ and $p$ is irreducible over $\Bbb{F}_3$, it follows that $\Bbb{F}_3(a^3)$ is also a splitting field of $p$ over $\Bbb{F}_3$. Clearly $\Bbb{F}_3(a^3)\subseteq\Bbb{F}_3(a)$ so it follows that $\Bbb{F}_3(a^3)=\Bbb{F}_3(a)$. Alternatively and more directly, the minimal polynomial of $a^2$ over $\Bbb{F}_3$ has degree $3$ because $\Bbb{F}_3\subsetneq\Bbb{F}_3(a^2)\subseteq\Bbb{F}_3(a)$.

  3. The standard approach is to simply compute powers of $a^2$ to find its minimal polynomial by elementary linear algebra: \begin{eqnarray} a^2&=&\hphantom{(-)}1\cdot a^2&+&\hphantom{(-)}0\cdot a^1&+&\hphantom{(-)}0\cdot a^0,\\ (a^2)^2&=&\hphantom{(-)}1\cdot a^2&+&\hphantom{(-)}1\cdot a^1&+&(-1)\cdot a^0,\\ (a^2)^3&=&(-1)\cdot a^2&+&\hphantom{(-)}1\cdot a^1&+&\hphantom{(-)}0\cdot a^0, \end{eqnarray} from which it is immediate that the minimal polynomial of $a^2$ is $x^3+2x^2+2x+2$.

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  • $\begingroup$ In my notes this exercise is after this theorem: if $f \in \mathbb{F}_{q}[x]$ is an irreducible polynomial of degree $n$ then it´s splitting field is $\mathbb{F}_{q^{n}}$. The approach I used before this result was seeing $\mathbb{F}_{3}[x]/(f)$ as an extension of $\mathbb{F}_{3}$ via the map $c \mapsto c+(f)$. That extension is of degree 3 and contains a root of $f$. $\endgroup$ Commented Jun 5 at 21:07
  • $\begingroup$ For (2) my confusion was from this example: the extension $\mathbb{Q} \leq \mathbb{Q}(\sqrt[3]{2}, \omega)$ where $w$ is a primitive third root of unity is the splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$. It contains the subextensions $\mathbb{Q}(\omega\sqrt[3]{2})$ and $\mathbb{Q}(\omega^{2}\sqrt[3]{2})$. The generators are roots of the same polynomial but they generate different extensions. In fact, if one computes the Galois group you can see that they correspond to different subgroups. $\endgroup$ Commented Jun 5 at 21:12
  • $\begingroup$ Every algebraic extension of a finite field is normal $\endgroup$ Commented Jun 5 at 21:43
  • $\begingroup$ @MrGran Then by showing that $x^3+x-1$ is irreducible over $\Bbb{F}_3$, the theorem tells you that $\Bbb{F}_3(a)$ is a subfield of $\Bbb{F}_{3^3}$. Of course there are only two subfields, so you can quickly conclude that $\Bbb{F}_3(a)=\Bbb{F}_{3^3}$. $\endgroup$ Commented Jun 6 at 8:43
  • $\begingroup$ @MrGran And yes, your example here is fundamentally different. Every finite field is normal over its prime field, so Galois theory is particularly clean and simple in this setting. $\endgroup$ Commented Jun 6 at 8:45
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Here is a solution without Galois theory.

We need the base field $K$ to be $\mathbb F_3$ to know $x^3+x^2-1$ is irreducible.

But if irreducibility is a given, we don't need any specific knowledge of $K$. In particular, we don't need to know if $K(a)$ is Galois over $K$.

Because $[K(a):K]=3$ is a prime, there is no proper intermediate extension, that is $K(a^2)=K$ or $K(a^2)=K$. In the former case, $a^2\in K$, then $[K(a):K]\le 2$, a contradiction.

To find the minimal polynomial of $b=a^2$, we know it's of degree $3$, hence it must be a nontrivial linear combination of $1, b, b^2=a^4, b^3=a^6$. So we may,

$$\begin{align}a^3=-(a^2-1) & \Longrightarrow a^6=(a^2-1)^2=a^4-2a^2+1 \\ & \Longrightarrow b^3=b^2-2b+1 \\ &\Longrightarrow b^3-b^2+2b-1=0\end{align}$$

More generally, Jyrki Lahtonen gave a trick to find the minimal polynomial of $a^2$ by considering $f(a)f(-a)$. Equivalently, we can rewrite $f(x)=0$ as $\sum_{i \text{ odd }} c_ix^i = \sum_{i\text{ even}} c_ix^i$, then square both sides.

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