## § Primitive element theorem

• Let $E/k$ be a finite extension. We will characterize when a primitive element exists, and show that this will always happen for separable extensions.

#### § Try 0: Naive attempt

• We try to find some element $\theta \in E$ such that the multiplicative subgroup generated by $\theta$, $\{1, \theta, \theta^2, dots\}$ generate $E$.
• However, to find such an element is our mandate!
• What we can do, however, is to take arbitrary elements of the form $\alpha, \beta, \gamma, \delta$ and so on, and try to write one of them in terms of the other. If we can show that $\gamma = f(\beta, \alpha)$ and $\beta = g(\alpha)$where $f, g$ are polynomials, then we have reduced everything to powers of $\alpha$.
• Now, clearly, if we can do this for two elements, ie, given $E = k(\alpha, \beta)$ , find a polynomial $f$ such that $\beta = f(\alpha)$, we have won. So the question is in finding this $f$.

#### § Forward: Finitely many intermediate subfields implies primitive

• If $k$ is a finite field, then $E$ is a finite extension and $E^\times$ is a cyclic group. The generator of $E^\times$ is the primitive element.
• So suppose $k$ is an infinite field. Let $E/k$ have many intermediate fields.
• Pick non-zero $\alpha, \beta \in E$ such that $E = k(\alpha, \beta)$. We will generalize to arbitrarily many generators via recursion.
• As $c$ varies in $k$, the extension $k(\alpha + c\beta)$ varies amongst the extensions of $E$.
• Since $E$ only has finitely many extensions while $k$ is infinite, pigeonhole tells us that there are two $c_1 \neq c_2$ in $E$ such that $k(\alpha + c_1 \beta) = k(\alpha + c_2 \beta)$.
• Define $L \equiv k(\alpha + c_1 \beta)$. We claim that $L = E$, which shows that $\alpha + c_1 \beta$ is a primitive element for $E$.
• Since $k(\alpha + c_2 \beta) = k(\alpha + c_1 \beta) = L$, this implies that $\alpha + c_2 \beta \in L$.
• Thus, we find that $\alpha + c_1 \beta \in L$ and $\alpha + c_2 \beta \in L$. Thus, $(c_1 - c_2) \beta in L$. Since $c_1, c_2 \in k$, we have $(c_1 - c_2)^{-1} \in K$, and thus $\beta \in L$, which implies $\alpha \in L$.
• Thus $L = k(\alpha, \beta) = k(\alpha + c_1 \beta)$.
• Done. Prove for more generators by recursion.

#### § Backward: primitive implies finitely many intermediate subfields

• Let $E = k(\alpha)$ be a simple field (field generated by a primitive element). We need to show that $E/k$ only has finitely many subfields.
• Let $a_k(x) \in k[x]$ be the minimal polynomial for $\alpha$ in $k$. By definition, $a$ is irreducible.
• For any intermediate field $k \subseteq F \subseteq E$, define $a_F(x) \in F[x]$ to be the minimal polynomial of $\alpha$ in $F$.
• Since $a_k$ is also a member of $F[x]$ and $a_k, a_F$ share a common root $\alpha$ and $a_F$ is irreducible in $F$, this means that $a_F$ divides $a_k$.
• Proof sketch that irreducible polynomial divides any polynomial it shares a root with (Also written in another blog post): The GCD $gcd(a_F, a_k) \in F[x]$ must be non constant since $a_F, a_k$ share a root). But the irreducible polynomial $a_F$cannot have a smaller polynomial ( $gcd(a_F, a_k)$) as divisor. Thus the GCD itself is the irreducible polynomial $a_F$. This implies that $a_F$ divides $a_k$since GCD must divide $a_k$.
• Since $a_k$ is a polynomial, it only has finitely many divisors (upto rescaling, which does not give us new intermediate fields).
• Thus, there are only finitely many intermediate fields if a field is primitive.

#### § Interlude: finite extension with infinitely many subfields

• Let $K = F_p(t, u)$ where $t, u$ are independent variables. This is an infinite field extension of $F_p$, since each of the $t^i$ are independent.
• Recall that the equation $t^p \equiv t (mod p)$ does not help here, as that only tells us that upon evaluation, the polynomials agree at all points. However, they are still two different polynomials / rational functions.
• Consider the subfield $k \equiv F_p(t^p, u^p)$. This too is an infinite field extension of $F_p$.
• Now consider the tower $K/k$, ie, $F_p(t, u)/F_p(t^p, u^p)$. This is of finite degree, as we can only have $t^i$ of power upto $p$. $t^p$ lies in the base field $k$. Exactly the same with $u$.
• So the extension $F/k$ has basis $\{ t^i u^j : 0 \leq i, j < p \}$, and so has degree $p^2$.
• We will first show that $K$ cannot be generated by a single element $\theta \in k$.
• Suppose we have $\alpha \in K = F_p(t, u)$. Then we must have that $\alpha^p \in F_p(t^p, u^p)$frobenius fixes the base field. Thus, frobenius sends $\alpha \in K$ to $\alpha^p \in k$.
• Thus, if we now had that $K = k(\alpha)$, this could not be, since by the previous argument, the extension has degree $p$. But we previously saw that $K$ has degree at least $p^2$. Thus, this field extension $K/k$ does not have a primitive element .
• We will now show that $K/k$ has infinitely many intermediate subfields.
• Pick elements of the form $\{ t^p + \beta u^p \in K : \beta \in K \}$. We claim that $K_\beta \equiv k(\beta) = F_p(t^p, u^p, \beta)$ are all different fields for different $\beta \in K$.
• Suppose for contradiction that $C = K_\beta = K_\gamma$ for $\beta \neq \gamma$ ( $\beta, \gamma \in K$). [ $C$ for contradictory extension ]
• This means that $t^p + \beta u^p - (t^p + \gamma u^p) \in C$, or that $(\beta - \gamma) u^p \in C$, which implies that $u^p \in C$, since we know that $(\beta - \gamma)^{-1} \in C$, as $C$ must contain both $\beta$ and $\gamma$.
• Since $u^p \in C$, $\beta \in C$, and $t^p + \beta u^p \in C$, we must have $t^p \in C$.
• This is a contradiction, since thus now means that $C = K$, where $C = k(t^p + \beta u^p)$, which makes $t^p + \beta u^p$ a primitive element, somthing that we saw is impossible.
• The key idea is that the extension is generated by a minimal polynomial $X^p - 1$, which factorizes as $(X - 1)^p$. We lose the connection between minimality and irreducibility, which makes the extension inseparable, since the minimal polynomial now has repeated roots.

#### § Part 2: If $E/k$ is finite and separable then it has a primitive element

• Let $K = F(\alpha, \beta)$ be separable for $\alpha, \beta \in K$. Then we will show that there exists a primitive element $\theta \in K$ such that $K = F(\theta)$.
• By repeated application, this shows that for any number of generators $K = F(\alpha_1, \dots, \alpha_n)$, we can find a primitive element.
• If $K$ is a finite field, then the generator of the cyclic group $K^\times$ is a primitive element.
• So from now on, suppose $K$ is infinite, and $K = F(\alpha, \beta)$ for $\alpha, \beta \in F$.
• Let $g$ be the minimal polynomial for $\alpha$, and $h$ the minimal polynomial for $\beta$. Since the field is separable, $g, h$ have unique roots.
• Let the unique roots of $g$ be $\alpha_i$ such that $\alpha = \alpha_1$, and similarly let the unique roots of $h$ be $\beta_i$ such that $\beta = beta_1$.
• Now consider the equations $\alpha_1 + f_{i, j} \beta_1 = \alpha_i + f_{i, j} \beta_j$ for $i \in [1, deg(g)]$ and $j \in [1, deg(h)]$.
• Rearranging, we get $(\alpha_1 - \alpha_j) = f_{i, j} (\beta_j - \beta_1)$. Since $\beta_j \neq \beta_1$ and $\alpha_1 \neq \alpha_j$, this shows that there is a unique $f_{i, j} \equiv (\alpha_1 - \alpha_j)/(\beta_j - \beta_1)$ that solves the above equation.
• Since the extension $F$ is infinite, we can pick a $f_*$ which avoids the finite number of $f_{i, j}$.
• Thus, once we choose such an $f_*$, let $\theta \equiv a_1 + f b_1$. Such a $\theta$ can never be equal to $\alpha_i + f \beta_j$ for any $f$, since the only choices of $f$that make $\alpha_1 + f \beta_1 = \alpha_i + f \beta_j$ true are the $f_{i, j}$, and $f_*$ was chosen to be different from these!
• Now let $F_\theta \equiv F(\theta)$. Since $\theta \in K$, $E$ is a subfield of $K$.
• See that $K = F(\alpha, \beta) = F(\alpha, \beta, \alpha + f \beta) = F(\beta, \alpha + f \beta) = F(\theta, \beta) = F_\theta(\beta)$.
• We will prove that $K = F_\theta$.
• Let $p(x)$ denote the minimal polynomial for $\beta$ over $F_\theta$. Since $K = F_\theta(\beta)$, if $p(x)$ is trivial, the $K = F_\theta$.
• By definition, $\beta$ is a root of $h(x)$. Since $p(x)$ is an irreducible over $F_\theta$, we have that $p(x)$ divides $h(x)$[proof sketch: irreducible polynomial $p(x)$ shares a root with $h(x)$. Thus, $gcd(p(x), h(x))$ must be linear or higher. Since $gcd$ divides $p(x)$, we must have $gcd = p(x)$ as $p(x)$ is irreducible and cannot have divisors. Thus, $p(x)$, being the GCD, also divides $h(x)$].
• Thus, the roots of $p(x)$ must be a subset of the roots $\{ \beta_j \}$ of $h(x)$.
• Consider the polynomial $k(x) = g(\theta - f_* \cdot x)$. $\beta$ is also a root of the polynomial $k(x)$, since $k(\beta) = g(\theta - f_* \beta)$, which is equal to $g((\alpha + f_* \beta) - f_* \beta) = g(\alpha) = 0$. [since $\alpha$ is a root of $g$].
• Thus, we must have $p(x)$ divides $k(x)$.
• We will show that $\beta_j$ is not a root of $k(x)$ for $j \neq 2$. $k(\beta_j) = 0$ implies $g(\theta - f_* \beta_j) = 0$, which implies $\theta - f_* \beta_j = \alpha_i$since the roots of $g$ are $\alpha_i$. But then we would have $\theta = \alpha_i + f_* \beta_j$, a contradiction as $\theta$ was chosen precisely to avoid this case!
• Thus, every root of $p(x)$ must come from $\{ \beta_j \}$. Also, the roots of $p(x)$ must come from the roots of $k(x)$. But $k(x)$ only shares the root $\beta_1$with the set of roots $\beta_2, \dots, \beta_j$. Also, $p(x)$ does not have multiple roots since it is separable. Thus, $p(x)$ is linear, and the degree of the field extension is 1. Therefore, $K = E = F(\theta)$.