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Separable

Let $L/k$ be a splitting field extension for an irreducible polynomial $f\in k[X]$, then $f$ is called separable over $k$ if it has distinct roots in the splitting field $L$. $$f=c(X-\alpha_1)\cdots(X-\alpha_n) \in L[X],\qquad c\in k, \alpha_i\in L$$ Any polynomial $f\in k[X]$ is called separable over $k$ if all its irreducible factors are separable over $k$, else it is called inseparable . For a separable polynomial $\deg f=|\Root_f|$.
Let $\alpha\in E$ where $E/k$ is an algebraic field extension, then $\alpha$ is called separable over $k$ if its minimal polynomial $m_\alpha$ is separable, else it is called inseparable. The extension $E/k$ is separable extension if all $\alpha\in E$ are separable.
$X^2,(X-1)^2$ are not separable since, it has repeated roots. Linear polynomials $X-a$ are always separable. If $f\in k[X]$ is separable then so is any $g|f$.

Formal Derivative

Let $f\in k[X]$ and consider the linear map for all $n\geq 0$ $$D:k[X]\ra k[X]\qquad X^n\mapsto nX^{n-1} $$ Note that $Dc=0$ for $c\in k$ and $Df$ is written as $f'$.

Lemma


Proof
  1. $(f + g)' = f' + g'$ is true by linearity.
    Using linearity, checking $(fg)'=f'g+fg'$ for monomials is enough. If $f=X^m$ and $g=X^n$, then $(fg)'=(m+n)X^{m+n-1}=f'g+fg'$.
$\Rightarrow$ Assume $f$ has a repeated root, that is $f = (X - \alpha)^2 h \in L[X]$ where $\alpha \in L$. Then $$f' = 2(X - \alpha)h + (X - \alpha)^2 h' = (X- \alpha)(2h + (X - \alpha)h').$$ Hence, $f$ and $f'$ have common root $\alpha$ (coming from the common factor $(X-\alpha)\in L[X]$). In order to get a common irreducible factor in $k[X]$ consider $m_\alpha$,the minimal polynomial of $\alpha$ over $k$, then $m_\alpha \mid f$ and $m_\alpha \mid f'$. The common factor is $m_\alpha$.

$\Leftarrow$ Suppose $g$ is a common irreducible factor of $f$ and $f'$ in $k[X]$. Let $\alpha \in \Root_g(L)$, hence, $\alpha \in \Root_f(L) \cap \Root_{f'}(L)$. Since $\alpha$ is a root of $f$, we can write $f = (X - \alpha)h \in L[X]$ for some polynomial $h$. Then $$f' = (X - \alpha) h' + h.$$ Since $\alpha$ is a also a root of $f'$ over splitting field $L$, so $(X - \alpha) \mid f'$, we must have $(X - \alpha) \mid h$ or $(X - \alpha)^2 \mid f$.

Remark for irreducible $f$
If $f$ is irreducible then it is also maximal, and $\lr{f,f'}=k[X]$ iff $f'\nmid f$. Since, $\deg f'<\deg f$ this is equivalent to $f'\neq 0$. Thus for $f$ irreducible, $f$ is separable iff $f'\neq 0$. Another way to see this would be use the above lemma and conclude $f$ has repeated roots iff $f'=0$.

Corollary


By the lemma above $f$ is separable iff $f'\neq 0$, which is true for a non constant polynomial in $k[X],\Char k=0$. In case $f\in k[X^p],\Char k=p$, then $f'=0$ making $f$ not separable.


  1. $\Q(\sqrt{2})/\Q$ is a separable extension. The minimal polynomial is $X^2-2$ and $\Char \Q=0$.
  2. $K/k$ where $K=\F_p(t)$ and $k=\F_p(t^p)$ is not separable. The minimal polynomial of $t$ over $k$ is $X^p-t^p$ which splits as $(X-t)^p$ in $K[X]$, and thus not separable.

Proposition

Proof Since $1\Rightarrow 2\Rightarrow 3$ it suffices to prove 1. For all $\sigma\in\Hom_k(k(\alpha),M)$ we have $f(\sigma(\alpha))=\sigma(f(\alpha))=0$ (since coefficients of $f$ are in $k$ hence fixed by $\sigma$), so $\beta=\sigma(\alpha)\in M$ is a root of $f$ and for all $g\in k[X]$, $\sigma(g(\alpha))=g(\sigma(\alpha))=g(\beta)$.
 Conversely, for all roots $\beta$ of $f$ in $M$ the map $$\sigma_\beta:k(\alpha)\ra k[X]/\lr{f}\ra k(\beta)\subset M $$ is a $k$ homomorphism sending $\alpha\to\beta$.


  1. $L=\Q(\sqrt{2},\sqrt{3})=\Q(\sqrt{2}+\sqrt{3}), k=\Q, \alpha=\sqrt{2}+\sqrt{3}$ and $ M=\C$ $$ \begin{align*} \Hom_\Q(\Q(\alpha),\C)&=\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}\\ \sigma_j:g(\alpha)&\mapsto g(\beta_j),\\ \text{where } \{\beta_1,\beta_2,\beta_3,\beta_4\}&=\{\pm\sqrt{2},\pm\sqrt{3}\} \end{align*} $$
  2. $L=\Q(\sqrt[3]{2})=M$ and $k=\Q$. Since, $\sqrt[3]{2}$ is the only root of the equation $f(X)=X^3-2$ in $L$, there is only one $k$ homomorphism of fields $L\ra L$, the identity map.

Proposition

Proof If $L=k(\alpha)$ is a simple extension of $k$, then the first two items on the list result from the last two items of the proposition above. In general, $L=k(\alpha_1,\ldots,\alpha_n)$, denote $k_i=k(\alpha_1,\ldots,\alpha_i)=k_{i-1}(\alpha_i)$: $$ k=k_0\subset k_1\subset \cdots\subset k_n=L\\ $$
  1. Fix $i\in\{1, \ldots, n-1\}$ and using the second item of the proposition above, all elements $\sigma\in\Hom_k(k_{i-1},M)$ can be lifted to $\tau\in \Hom_k(k_i,M)$ upto $[k_{i-1}(\alpha_i):k_{i-1}]=[k_i:k_{i-1}]$ times.
    k i-1 M k i σ τ
    Hence, $$|\Hom_k(k_i,M)|\leq [k_i:k_{i-1}]\cdot|\Hom_k(k_{i-1},M)| $$ Multiplying the above inequalities one obtains $$|\Hom_k(L,M)|\leq \left(\prod_{i=1}^n[k_i:k_{i-1}]\right)|\Hom_k(k,M)|=[L:k] $$
  2. If the extension $L/k$ is not separable one can choose the elements $\alpha_1,\ldots \alpha_n$ such that $\alpha_1$ is not separable on $k$. Thus, by item 2 of proposition above $$ \begin{align*} |\Hom_k(k_1,M)|=|\Hom_k(k(\alpha_1),M)|&<[k(\alpha_1):k]=[k_1:k]\\ \text{for all extensions }&L/K\quad \\ |\Hom_k(L,M)|&<\left(\prod_{i=1}^n[k_i:k_{i-1}]\right)=[L:k] \end{align*} $$ If the extension $L/k$ is separable, let $f_i\in k[X]$ (resp. $g_i\in k_{i-1}[X]$) be the minimal polynomial (separable polynomial) of $\alpha$ on $k$ (resp. on $k_{i-1}$). Define by recurrence $M_0=k$, $M_i=$ a splitting field of $f_i$ on $M_{i-1(1\leq i\leq n)}. M=M_n$. The polynomial $g_i$ has $\deg g_i=[k_i:k_{i-1}]$ distinct roots in $M$. By item 3 of proposition above all elements $\sigma \in \Hom_k(k_{i-1}, M)$ admit $[k_i:k_{i-1}]$ embeddings $\tau\in\Hom_k(k_i,M)$. Hence, $$ \begin{align*} |\Hom_k(k_i,M)|&=[k_i:k_{i-1}]\cdot|\Hom_k(k_{i-1},M)|\text{ for }1\leq i\leq n\\ |\Hom_k(L,M)|&=\left(\prod_{i=1}^n[k_i:k_{i-1}]\right)|\Hom_k(k,M)|=[L:k] \end{align*} $$
  3. Let $M_0'=k', M_i'$ a splitting field of $f_i$ on $M_{i-1}'(1\leq i\leq n), M=M_n'$. Then all elements $\sigma\in\Hom_k(k_{i-1},M)$ admit a lifting $\tau\in\Hom_k(k_i,M)$ by item 2 of proposition above or $$ |\Hom_k(L,M)|\geq |\Hom_k(k,M)|=1$$

Corollary

Proof
  1. Let $M$ be a splitting field of $f\in k[X]$ with root $\alpha$, then by item 2 above and item 3 in proposition above $$|\Hom_k(k(\alpha),M)|=\deg f=[k(\alpha):k] $$
  2. Inductively apply the above part to all elements in the extension.

Primitive element theorem

Proof If $|k|<\infty$ (number of elements in $k$ is finite), then $L=k(\alpha)$ where $\alpha$ is a generator of the finite cyclic group $L^\times$. Suppose that $|k|=\infty$ (number of elements of $k$ is infinite). Then by item 2 of this proposition, there exists an extension $M/k$ and $n=[L:k]$ distinct homomorphisms $\sigma_1,\ldots,\sigma_n\in \Hom_k(L,M)$. For all $1\leq i<j\leq n,\Ker(\sigma_i-\sigma_j)\subsetneq L$ is a strict vector $k$ sub space of $L$. The lemma below shows that there exists $\alpha\in L$ such that $i\neq j$ implies $\sigma_i(\alpha)\neq \sigma_j(\alpha)$, hence, $$n=[L:k]\geq [k(\alpha):k]\geq |\Root_f\in M|\geq|\{\sigma_1(\alpha),\ldots,\sigma_n(\alpha)\}|=n $$ where $f\in k[X]$ is the minimal polynomial of $\alpha$ on $k$. The above inequalities give that $[L:k]= [k(\alpha):k]=n$ or $[L : k(α)] = 1$, hence $L = k(α)$.

Lemma


Proof 1 Induction on $m$. The case $m=1$ is trivial, thus suppose that $m>1$ and $V=V_1\cup\cdots V_m$. There exists $u\in V, u\notin V_m$ which implies $u\in V_1\cup\cdots\cup V_{m-1} $, and by inductive hypothesis $v\in V, v\notin V_1\cup\cdots\cup V_{m-1}$ implies $v\in V_m$. The set $\{u+av\text{ such that }a\in k\}$ is infinite, hence there exists $1\leq j\leq m$ and $a,b\in k, a\neq b$, such that $u+av,u+bv\in V_j$ which implies that $(a-b)v\in V_j$ which implies $v\in V_j$ consequently $u\in V_j$, an impossibility.
Proof 2 As in the above proof assume $m>1$, and also that $V_i$ are of codimension one, that is space given by the solution of a single homogeneous equation $$a_1x_1+a_2x_2+\ldots+a_nx_n=0,$$ where $x_1,\ldots,x_n$ is a basis of $V$. The polynomial $a_1+a_2t+a_3t^2+\ldots+a_nt^{n-1}$ has atmost $n-1$ many roots, hence the space $$S=\{(1,t,t^2,\ldots, t^{n-1})\text{ such that }t\in k\} $$ will intersect $V_i$ in finitely many points. But, $S$ is an infinite set contained within $V$, and it is impossible to cover this infinite set with finitely many points coming from finitely many subspaces. Consequently, it is impossible to cover $V$.



  1. Consider $L/k$ where $L=\F_p(X,Y)$ and $k=\F_p(X^p,Y^p)$, then for all $\alpha \in L$, $ [k(\alpha):k]\leq p\leq p^2=[L:k]$, hence $L\neq k(\alpha)$.
  2. Lemma: For $n\geq 1$ an integer, let $\zeta_n=\exp(2\pi\ci /n)$. Then for all $n,m\geq 1$ we have $\Q(\zeta_n,\zeta_m)=\Q(\zeta_j)$ where $j$ is least common multiple of $n$ and $m$.
    Proof Since $\zeta_j^{j/n}=\zeta_n$ we get $\zeta_n\in\Q(\zeta_j)$ (same for $m$) and thus $\Q(\zeta_n,\zeta_m)\subset \Q(\zeta_j)$. Inversely if $j/m,j/n$ are prime to each other the Bezout indentity gives $uj/m+vj/n=1$ for $u,v$ as integers. Multiply this by $2\pi\ci/j$ and taking the exponential gives $\zeta_j=\zeta_n^u\zeta_m^v$ giving the reverse inclusion $\Q(\zeta_n,\zeta_m)\supset \Q(\zeta_j).$
  3. Corollary: Let $K/k$ be a finite extension, then $|{\Hom_k(K,\Omega)}|=[K:k]$.
    Proof By the primitive element theorem $K=k[x]$ where $x$ is the primitive element.

Perfect

Proof Suppose $k$ is imperfect, then there exists an irreducible polynomial $f\in k[X]$ such that $f'=0$. Since, $f$ is non constant, $k$ must be of characteristic $p$ and $f$ be of form $\sum_na_{np}X^{np}$. If the Frobenius of $k$ is surjective, we could find $b_{np}\in k$ such that $b^p_{np}=a_{np}$, and we would then have $f=(\sum_n b_{np}X^n)^p $ contradicting the irreducibility of $f$.

 Conversely, suppose $k$ has characteristic $p$, and the Frobenius is non surjective. Pick $\alpha\in k\backslash k^p$ and consider the polynomial $f:=X^p-\alpha$, then $f'=0$. Let us show that $f$ is irreducible, thus it will have repeated roots (since $f'=0$) implying not separable. Let $f=f_1f_2$ a non trivial factorization of $f\in k[X]$. If $\beta$ denotes a $p$th root of $\alpha$ in a algebraic closure $\wbr{k}$ of $k$, then $f=(X-\beta)^p$ in $\wbr{k}[X]$, and so $f_i=(X-\beta)^{r_i}$ with $0<r_i<p$. Note that $r_1$ and $r_2$ are coprime and thus there exist $u,v$ such that $ur_1+vr_2=1$. Consequently, we have the equality $f_1^uf_2^v=(X-\beta)$ in $\wbr{k}(X)$ (since, $u,v$ could be negative). Since $k(X)\cap \wbr{k}[X]=k[X]$, we can deduce that $X-\beta\in k[X]$ and so that $\beta\in k$, which contradicts the hypothesis on $\alpha.$


  1. All finite fields are perfect, since an injective map from a finite set to itself is also bijective.
  2. All algebraically closed fields $k$ are perfect, since the equation $X^p-a$ has a solution for all $a\in k$.
  3. The field $\F_p(T)$ is never perfect as $T$ does not have a $p$th root, so is not in the image of Frobenius. But, the field $\F_p(T,T^{1/p},T^{1/p^2},\ldots)$ is perfect.

Proposition

Proof The question poses a problem only in $\Char p>0$. Let $F$ be the frobenius morphism of $K$, that is bijective on $k$ (since $k$ is perfect and has $\Char p$, see theorem above), thus, there is an inverse $F^{-1}:k\ra k$.
 $F(K)$ is a $F(k)$ vector subspace of $K$, thus, a $k$ vector subspace since $F(k)=k$. Similarly, if $x_i\in K$ are free on $k$, the $F(x_i)$ are free on $k$ in $F(K)$. In fact, if $\sum a_i F(x_i)=0$ with $a_i\in k$, then rewriting gives $F(\sum F^{-1}(a_i)x_i)=0$ and so $\sum F^{-1}(a_i)x_i=0$. The family of $x_i$ being free implies $F^{-1}(a_i)=0$ for all $i$, and thus $a_i=0$. Notice, the inclusion $F(K)\subset K$ and the inequality $[K:k]\leq [F(K):k]$; (on the base $k$) which implies $K=F(K)$, making $K$ perfect.
  The opposite is also true. Let $k$ be a subfield of an algebraically closed field $\Omega$ with $\Char p>0$. As the Frobenius morphism is bijective on $\Omega$ (since, $\Omega$ is perfect), the $p$th roots $x^{1/p}=F^{-1}(x)$ of all elements of $\Omega$ are well defined. This is also true for $p^n$th roots in a similar manner.

Lemma


Proof Let $\tau$ be the $p^n$th root of $t$. Since, $t^{1/p}\notin k$ gives $t^{1/p}=\tau^{p^{n-1}}\notin k$ or $\tau\notin k$. Let $P\in k[X]$ be the minimal polynomial of $\tau$ which divides $Q=X^{p^n}-t$ since, $Q(\tau)=0$. Decomposing $Q$ in $k[X]$ we could write $$ Q=P^mR \text{ with } R\in k[X] \text{ and }\lr{P,R}=1 $$ where $\lr{P,R}$ is the ideal generated by $P+R$. This leads to Bezout identity $PA+RB=1$ with $A,B\in k[X]$. Consequently, $P$ and $R$ don't have common roots in $\Omega$. Thus, $Q(X)=(X-\tau)^{p^n}$ so that $R$ has no root at all and thus $Q=P^m$. Comparing the degrees we obtain $m=p^v, 0\leq v\leq n$. Evaluation at zero gives $$Q(0)=-t=(P(0))^{p^v}. $$ Since, $t$ does not have a $p$th power in $k$, we get $v=0$. Hence, $Q=P$ and thus $Q$ is irreducible.

Corollary


Proof Indeed, if $t\in k$ is not a $p$th power, then $t^{p^{-n}}\in K$ has degree $p^n$ on $K$ which tends to infinity with $n$. This is a contradiction since $K/k$ being a finite extension, the degree of elements of $K$ on $k$ is maximum of $[K:k]$.


The reference for this page is INTRODUCTION À LA THÉORIE DE GALOIS by Yves Lazlo.