§ Separable Polynomials and extensions

§ Separable polynomial

• I didn't really study galois theory over char. $p$ all that well the first time I studied it, so let's review.
• Let $J \subseteq K$ be an inclusion of fields, so $K$ is a field extension of $J$.
• An irreducible polynomial $f \in J[x]$ is separable iff it has distinct roots in the algebraic closure of $J$, $\overline{J}$.
• Said differently, the polynomial $f$ has no repeated roots in any extension of $J$.
• Said differently, the polynomial $f$ has distinct roots in its splitting field over $J$. The roots as separable since we can separate all the roots from each other --- they are all distinct.
• Said differently, the polynomial derivative $f'$ of $f$ is not the zero polynomial.

§ Forward: $p$ is not separable implies $p, p'$ do not share a root.

• Let $p$ have a repeated root $\alpha \in \overline K$. Thus $p(x) \equiv (x -\alpha)^2 g(x)$ in $\overline K$.
• Computing $p'$ by product rule, we see that it is $p'(x) = 2(x- \alpha)g(x) + (x - \alpha)^2 g'(x)$ which can be written as $p'(x) = (x - \alpha)(2g(x)+ g'(x)$.
• This shows that $p'(x)$ has $(x - \alpha)$ as a root.

§ Backward: $p, p'$ share a root implies $p$ is not separable

• Let $\alpha \in \overline K$ be such that $p(\alpha) = p'(\alpha) = 0$.
• Write $p(x) \equiv \prod_i (x - r_i)$ for roots $r_i \in \overline K$.
• Let $\alpha = r_1$ [WLOG ].
• We know by product rule of calculus that $p'(x) \equiv \sum_i \prod_{j \neq i} (x - r_i)$.
• Computing $p'(\alpha) = p'(r_1)$, only the first term survives, which is $\prod_{j \neq 1}(r_i - r_j)$ [all other terms have an $(x - r_1)$ term which vanishes.
• For this to vanish, we must have some $j$ such that $r_i = r_j$.
• This implies that $p$ has a repeated root $r_i, r_j$ and is thus not separable.

§ Forward: $p$ is separable implies $gcd(p, p') = 1$

• Let $d(x) = gcd(p, p')$.
• Suppose that it is not a unit, and not a constant (ie, a real polynomial).
• Then $d(x)$ has a root $\alpha \in \overline K$ (by previous)
• Since $d$ divides $p, p'$, this means that in $\overline K$, $p(\alpha) = p'(\alpha) = 0$ since $d(\alpha) = 0$ and $d | p, p'$.
• This implies that $\alpha$ is a repeated root of $p(x)$ since it vanishes at both $p$ and its derivative!
• This contradicts $p$ is separable.
• Thus $d(x)$ must be a unit, and $p$, $p'$ are relatively prime.

§ Backward: $gcd(p, p') = 1$ implies $p$ is separable

• Since $gcd(p, p') = 1$, there are polynomials $k, l$ such that $pk + p'l = 1$.
• Suppose $p$ is not separable. Thus it has a repeated root $\alpha$. This means that $p$ and $p'$ vanish at $\alpha$. Thus $(x - \alpha)$ divides $p, p$'.
• Thus means that $(x - \alpha)$ divides $1$, by the eqn $pk + p'l = 1$. This is absurd, and thus $p$ is separable.

§ Separable extension

• For all elements $\alpha \in L$ where $L/K$, there is a polynomial $f_\alpha \in K[x]$ such that $f_\alpha(\alpha) = 0$ is separable .
• Thus, all elements have separable polynomials that they are roots of.

§ Separable extension is transitive

• Claim: If $R/Q$ and $Q/P$ are separable field extensions, then $R/P$ is separable.
• TODO!

§ All Polynomials over character 0 is separable

• Let $L/K$ and $char(K) = 0$. Then we claim that $L$ is separable.
• Let $f$ be an irreducible polynomial in $K[x]$. (the minimal polynomial of some element in $L$)
• Recall that a polynomial $f \in K[x]$ is irreducible over a field $K$ iff it cannot be written as the product of two non-constant polynomials.
• We wish to show that $f$ is separable (has no repeated roots).
• For contradiction, suppose that $f$ has a repeated root $r$ in the algebraic closure. so $f(x) \equiv (x - r)^2 g(x)$ for $r \in \overline K[x]$, $g(x) \in \overline K[x]$.
• Thus, $f(x)$ and $f'(x)$ share a common factor in $\overline K[x]$.
• But the GCD algorithm works in $K[x]$, thus $f(x)$ and $f'(x)$ share a common factor in $K[x]$ [SID: I find this dubious! ]
• Hence, this means that $gcd(f, f') \in K[x]$ is not a constant polynomial.
• If $gcd(f, f') \neq f$, then $f$ can be factored, which contradicts its irreducibility.
• Thus, $gcd(f, f') = f$ [to prevent contradiction ].
• However, $f'$ has smaller degree than $f$. Thus the only way it can be divided by its GCD ( $f$) which has larger degree than it is if $f'(x) = 0$.
• This means (in characteristic zero) that $f(x)$ is linear, and thus cannot have repeated roots!
• That this means that $f(x)$ is zero can be seen by computing te derivative. Suppose $f(x) \equiv \sum_{i=0}^n a_i x^i$ with $a_n \neq 0$. Then $f'(x) = \sum_{i=1}^n i a_i x^{i-1}$. Since $a_n \neq 0$, $n a_n \neq 0$, and thus the derivative of an $n$th degree polynomial is $(n-1)$ degree.

§ All Polynomials over character 0 is separable, alternative proof.

• Let $f$ be an irreducible polynomial in $K[x]$ (the minimal polynomial of some element in $L$). We claim that $f$ is separable.
• The key lemma is to show that if $g$ is ANY polynomial which shares a root $r$ with $f$, then $f | g$.
• Idea: since $f(r) = g(r) = 0$, this means that $(x - r)$ divides $gcd(f, g)$. Thus, $gcd(f, g)$ is non-constant.
• Further, $gcd(f, g) | f$ since the gcd divides both its factors.
• But since $gcd(f, g)$ divides $f$ while $f$ is irreducible, we must have $gcd(f, g)$ equals $f$.
• Since $gcd(f, g) = f$ divides $g$, we have $f | g$.
• Now, going back to our claim, let $f$ be some irreducible in $K[x]$. Suppose for contradiction that $f$ is not separable. Then $f, f'$ share a common root. By the above lemma, this implies that $f$ divides $f'$. But this is absurd, since $deg(f) > deg(f')$.
• Hence, no irreducible polynomial in $f$ can share a root with its derivative, which implies $f$ is always separable.
• This breaks down for character $p$ since $f'$ can simply "die out".

§ All finite field extensions over character 0 is separable

• Can write any finite field extension $L/K$ as $L = K(\alpha_1, \dots, \alpha_n)$. This is the same as $K(\alpha_1)(\alpha_2)\dots(\alpha_n)$.
• Since separability of field extensions is transitive, and at each step, we add an element with separable minimal polynomial (all polynomials over char. 0 are separable), the full extension is separable.

§ All field extensions over character $p$ is separable

• Consider $F_{p^m} \subseteq F_{p^n}$.
• build the "Fermat's little theorem" polynomial $x^{p^n} - x = f(x)$.
• All elements of $F_{p^n}$ satisfy this, thus $f(x)$ has $p^n$ roots, which means all of its roots are distinct.
• Alternatively, see that $f'(x) = p^n x^{p^n - 1} - 1 = 0 - 1 = -1$ so $f(x)$ and $f'(x)$ don't share a commmon root.

§ Purely inseparable extensions

• $K \subseteq L$ of char. p. Then the field $L$ is purely inseparable iff any $\alpha \in L$ is a root of $x^{p^n} - k$ for some $k \in K$, with $n \geq 1$.
• Over the algebraic closure, this can be written as $x^{p^n} - (k^{1/p^n})^{p^n}$ which over finite fields factorizes as $(x - \sqrt{a}{p^n})^{p^n}$ by freshman's dream .
• Thus, over the algebraic closure, this has many copies of one root, which is "as far as possible" from being separable.

§ Breaking down extension into separable + purely inseparable

• Given any $K \subseteq L$ algebraic, can break it down into $K \subseteq K^{sep} \subseteq L$.
• The extension $K^{sep}/K$ contains all elements of $L$ which have separable polynomials. Can show that this is a field.
• We can show that $L/K^{sep}$ will be purely inseparable.

§ Example of inseparable extension

• Let $L \equiv F_p(t)$, which are rational functions in $t$ over $F_p$, and $K \equiv F_p(t^p)$ which are rational functions in $t^p$.
• Clearly, $K \subseteq L$, and the extension $L/K$ is of degree $p$, because $t \in L$ is a root of $X^p - t^p \in F_p(t^p) = K$.
• See that $X^p - t^p$ is irreducible over $K = F_p(t^p)$.
• Over $L = F_p(t)$, $X^p - t^p$ factorizes as $(X - t)^p$ by freshman's dream , with all roots the same.
• Thus, we have an element whose minimal polynomial is not separable. Here, the function $g(Y) = Y^p - X \in K(X)$ has derivative zero, and is thus inseparable.
• In some sense, the failure is because of freshman's dream, where $X^p - t^p \equiv (X - t)^p$.
• Reference: Borcherds, separable extension

§ Primitive element theorem / Theorem of the primitive element

• Let $J \subseteq K$ be a field extesion. We say $\alpha \in K$ is primitive for the extension $K/J$ (the extension) if $K = J(\alpha)$.
• If $K/J$ is a finite separable extension, then for some $\alpha \in J$, we have $K = J(\alpha)$.
• Recall that $J(\alpha) \simeq J[x]/minpoly(x)$.
• TODO!

§ Tensor product of field extensions

• Let $K$ be a finite separable extension of $J$ and $\Omega$ be an arbitrary extension of $J$. (usually, $\Omega$ is the p-adics, $J$ is $\mathbb Q$, $K$ is a number field).
• Then, $K \otimes_J \Omega$ is a product of finite separable extensions of $\Omega$. So $K \otimes_J \Omega \equiv \prod_i \Omega_i$, where each $\Omega_i$ is a finite extension of $\Omega$.
• If $\alpha$ is a primitive element for the extension $K$ (so $K = J(\alpha)$), then the image of $\alpha \otimes 1$ in $\Omega_i$is a primitive element for $\Omega_i$ over $\Omega$.
• If $f$ is the minimal poly. for $\alpha \in K$ over $J$ and $f_i$ is the minimal polynomial for $\alpha_i \in \Omega_i$ over $\Omega$ then $f(x) = \prod_i f_i(x)$.
• Proof : Start with a primitive element $\alpha$ for $K/J$. Then $K \simeq_\phi J[x]/(f(x))$for $f(x)$ the minimal polynomial of $\alpha$ over $J$. So $\phi$ witnesses this isomorphism.
• Consider $K \otimes_J \Omega$. This is isomorphic to $(J[x]/(f(x))) \otimes \Omega$. Call the map that sends the LHS to the RHS as $\phi \otimes id$
• We claim that the ring $(J[x]/(f(x))) \otimes_J \Omega$ is isomorphic to $\Omega[x]/(f(x))$. Intuition: tensoring by $J$ doesn't do anything useful, and we can re-interpret $f(x)$ as living in $\Omega(x)$. The isomorphism is $\psi((g(x) + K(x)f(x)) \otimes \omega) \equiv \omega \cdot g(x) + \Omega[x]f(x)$.
• Now suppose $f$ factors as $\prod_i f_i$ over $\Omega[x]$. Since $\alpha$ is separable over $J$ and $\Omega$ is an extension of $J$, all the $f_i$ are distinct (otherwise it contradicts separability). Thus the family of ideals $\{ (f_i) \}$ is pairwise coprime.