Proof. Clearly, \(L^{pi}\cap L^{sep}=K\) and there is a \(K\)-epimorphism \(\pi : L^{pi}\otimes L^{sep}\rightarrow L^{pi} L^{sep}\), \(a\otimes b\mapsto ab\). We haver to show that \({\rm ker } (\pi) = \{0\}\). Suppose that \({\rm ker } (\pi) \neq \{0\}\). We seek a contradiction. Then there is a nonzero element \(\alpha=\sum_{i=1}^na_i\otimes b_i \in {\rm ker } (\pi)\) where \(a_i\in L^{pi}\) and \(b_i\in L^{sep}\). Let \(A=K(a_1, \ldots , a_n)\) and \(B=K(b_1, \ldots , b_n)\). Then \(\alpha \in A\otimes B\), the extension \(A/K\) is a finite purely inseparable finite field extension and the extension \(B/K\) is a separable finite field extension. Since \(A\otimes B\subseteq L^{pi}\otimes L^{sep}\), we may assume that \[A =L^{pi}\;\; {\rm and}\;\; B=L^{sep}.\] Let \(L^{nor}\) be the normal closure of \(L^{sep}\) in \(\overline{K}\). Similarly, since \(L^{pi}\otimes L^{sep}\subseteq L^{pi}\otimes L^{nor}\), we may assume that \(L^{sep}=L^{nor}\), i.e. the finite field extension \(L^{sep}\) is a Galois field extension with Galois group \(G(L^{sep}/K)\). By the Primitive Element Theorem, \(L^{sep}=K(\theta)\) is a simple field extension where \(\theta\in L^{sep}\). Recall that if \(f(x)\in K[x]\) is the minimal polynomial of the element \(\theta\) over \(K\) then \[f(x)=\prod_{g\in G(L/K)}(x-g(\theta))\] and \(g(\theta)\in L^{sep}\) for all \(g\in G(L^{sep}/K)\). So, every automorphism \(h\in G(L^{sep}/K)\) permutes the roots \(\{ g(\theta)\, | \, g\in G(L^{sep}/K)\}\) of the polynomial \(f(x)\). Since the field extension \(L^{sep}/K\) is Galois, the field extension \(L^{pi}L^{sep}/L^{pi}=L^{pi}(\theta)/L^{pi}\) is also a Galois finite field extension such that the restriction map \[{\rm res}: G(L^{pi}L^{sep}/L^{pi})\rightarrow G(L^{sep}/K), \;\; \sigma\mapsto \sigma|_{L^{sep} }\] is a bijection (every automorphism of \(L^{sep}/K\) is necessarily uniquely extended to an automorphism of \(L^{pi}L^{sep}/L^{pi}\) by trivial action on the elements of \(L^{pi}\)). In particular, \[[L^{pi}L^{sep}:L^{pi}]=|G(L^{pi}L^{sep}/L^{pi})|=|G(L^{sep}/K)|=[L^{sep}:K].\] It follows from \(K\subseteq L^{pi}\subseteq L^{pi}L^{sep}=L^{pi}(\theta)\) that \[[L^{pi}L^{sep}:K]=[L^{pi}L^{sep}:L^{pi}][L^{pi}:K]=[L^{sep}:K][L^{pi}:K]=[L^{pi}\otimes L^{sep}:K],\] and so \(L^{pi} L^{sep}=L^{pi}\otimes L^{sep}\). ◻

Proof. 1. By the definition, the polynomial \(f^{sep}\in K[x]\) is a separable polynomial such that \(f^{sep}(\theta^{p^n})=f(\theta)=0\). Therefore, \(K(\theta^{p^n})\subseteq L^{sep}\). In particular, the field extension \(L^{sep}/K(\theta^{p^n})\) is a separable field extension. In fact, the equality holds, \[K(\theta^{p^n})= L^{sep}.\] This follows from the field inclusions \(K\subseteq K(\theta^{p^n})\subseteq L^{sep}\subseteq L=K(\theta)\) and the facts that field extension \(L/K(\theta^{p^n})=K(\theta)/K(\theta^{p^n})\) is a purely inseparable field extension and its subfield extension \(L^{sep}/K(\theta^{p^n})\) is a separable field extension.

Since \(f(x)=f^{sep}(x^{p^n})\in {\rm Irr}_m(K[x])\), we have that \(f^{sep}(x)\in {\rm Irr}_m(K[x])\). Now, \[L^{sep}=K(\theta^{p^n})\simeq K[x]/(f^{sep}(x)).\] Hence, \([L^{sep}:K]=\deg(f^{sep}(x))=s\) and \(f^{sep}(x)\in {\rm Irr}_m(K[x])\) is the minimal polynomial of the element \(\theta^{p^n}\) over the field \(K\).

2–4. By statement 1, \(\theta^{p^n}\in L^{sep}\). This implies that the field extension \(L/L^{pi}L^{sep}=K(\theta)/L^{pi}L^{sep}\) is a purely inseparable field extension. Now, the equality \[m^*:=\max\Big\{m'\in \mathbb{N}\, | \, \theta^{p^{n-m'}}\in L^{pi}L^{sep} \Big\}=\max\Big\{m'\in \mathbb{N}\, | \, L^{p^{n-m'}}\subseteq L^{pi}L^{sep} \Big\}\] follows from the equality \(L=K(\theta)\). By the definition of the number \(m^*\) and the equality \(L^{sep}=K(\theta^{p^n})\), \[L^{pi}L^{sep}=L^{pi}(\theta^{p^{n-m^*}}).\]

(i) \(L=L^{pi}\otimes L^{sep}(\theta)=L^{pi}\otimes L^{sep}[x]/\Big(x^{p^{n-m^*}}- \theta^{p^{n-m^*}}\Big)\), \([L:L^{pi}\otimes L^{sep}]= p^{n-m^*}\) and \(x^{p^{n-m^*}}- \theta^{p^{n-m^*}}\in {\rm Irr}_m(L^{pi}\otimes L^{sep}[x])\) is the minimal polynomial of the element \(\theta\) over the field \(L^{pi}\otimes L^{sep}\). The finite field extension \(L/L^{pi}\otimes L^{sep}\) is a simple purely inseparable field extension of exponent \(n-m^*\): The statement (i) follows from the definition of the number \(m^*\).

(ii) \([L:L^{sep}]=p^n\) and \([L^{pi}L^{sep}:L^{sep}]=p^{m^*}\): By statement 1, \([L^{sep}:K]=\deg (f^{sep}(x))=s\), and so \[[L:L^{sep}]=\frac{[L:K]}{[L^{sep}:K]}=\frac{\deg (f)}{s}=\frac{sp^n}{s}=p^n.\] Now, the second equality in the statement (ii) follows from the statement (i), \[[L^{pi}L^{sep}:L^{sep}]=\frac{[L:L^{sep}]}{[L:L^{pi}L^{sep}]}=\frac{p^n}{p^{n-m^*}}=p^{m^*}.\]

(iii) \([L^{pi}L^{sep}:K]=sp^{m*}\) and \([L^{pi}:K]=p^{m^*}\): \[[L^{pi}L^{sep}:K]= [L^{pi}L^{sep}:L^{sep}][L^{sep}:K]=p^{m*}s,\]

\[[L^{pi}:K] = \frac{[L^{pi}:K][L^{sep}:K]}{[L^{sep}:K]} = \frac{[L^{pi}\otimes L^{sep}:K]}{[L^{sep}:K]}=\frac{p^{m^*}s}{s}=p^{m^*}.\]

(iv) \([L^{pi}L^{sep}:L^{pi}]=s\):

\[[L^{pi}L^{sep}:L^{pi}]= \frac{[L^{pi}L^{sep}:K]}{[L^{pi}:K]}=\frac{sp^{m^*}}{p^{m^*}} =s.\]

Notice that \(f^{sep,\frac{1}{p^{m^*}}}(x):=\sum_{i=0}^s\lambda_i^\frac{1}{p^{m^*}}x^i\in \overline{K}^{pi}[x]\) where \(\overline{K}^{pi}/K\) is the maximal purely inseparable field extension in \(\overline{K}/K\) (\(\overline{K}\) is the algebraic closure of \(K\)). Then the polynomial \(f^{sep,\frac{1}{p^{m^*}}}(x)\) is a monic polynomial of degree \(\deg\Big(f^{sep,\frac{1}{p^{m^*}}}(x)\Big)=s\) and the element \(\theta^{p^{n-m^*}}\) is a root of it: \[f^{sep,\frac{1}{p^{m^*}}}(\theta^{p^{n-m^*}})= \sum_{i=0}^s\lambda_i^\frac{1}{p^{m^*}}\theta^{ip^{n-m^*}}=\bigg(\sum_{i=0}^s\lambda_i \theta^{ip^n} \bigg)^\frac{1}{p^{m^*}}=\Big( f(\theta)\Big)^\frac{1}{p^{m^*}}= 0^\frac{1}{p^{m^*}} =0.\]

By the statement (iv) and the equality \(L^{pi}(\theta^{p^{n-m^*}})=L^{pi}L^{sep}\), \[[L^{pi}(\theta^{p^{n-m^*}}):L^{pi}]=[L^{pi}L^{sep}:L^{pi}]=s=\deg(f^{sep,\frac{1}{p^{m^*}}}).\] This implies that the polynomial \(f^{sep,\frac{1}{p^{m^*}}}(x)=\sum_{i=0}^s\lambda_i^\frac{1}{p^{m^*}}x^i\in L^{pi}[x]\) is the minimal polynomial of the element \(\theta^{p^{n-m^*}}\in L^{pi}L^{sep}\) over the field \(L^{pi}\). In particular, all its coefficients belong to the field \(L^{pi}\), i.e. \[\lambda_i^\frac{1}{p^{m^*}}\in L^{pi}\;\; \text{ for all}\;\;i=1, \ldots , s-1.\] Therefore, \(f^{sep,\frac{1}{p^{m^*}}}(x)\in {\rm Irr}_m(M^*[x])\) where \(M^*:=K\bigg(\lambda_0^\frac{1}{p^{m^*}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m^*}}\bigg)\subseteq L^{pi}\).

The polynomial \(f^{sep,\frac{1}{p^{m^*}}}(x)=\sum_{i=0}^s\lambda_i^\frac{1}{p^{m^*}}x^i\in \in {\rm Irr}_m(M^*[x])\) is a separable polynomial over the field \(M^*\) (since it’s derivative is a nonzero polynomial). So, we have proven the statement (v).

(v) \(f^{sep,\frac{1}{p^{m^*}}}(x)\in {\rm Irr}_m(M^*[x])\) and the field extension \(M^*(\theta^{p^{n-m^*}})/M^*\) is a separable field extension of degree \([M^*(\theta^{p^{n-m^*}}):M^*]=\deg\Big(f^{sep,\frac{1}{p^{m^*}}}(x)\Big)=s\).

(vi) \(L^{pi}(\theta^{p^{n-m^*}})=M^*(\theta^{p^{n-m^*}})\): There is a chain of fields \[L^{sep}=K(\theta^{p^n})\subseteq M^*(\theta^{p^{n-m^*}})\subseteq L^{pi}(\theta^{p^{n-m^*}})=L^{pi}L^{sep}\subseteq L=K(\theta).\] Since \(p^{n-m^*}\stackrel{{\rm (i)}}{=}[L:L^{pi}L^{sep}]=[L:L^{pi}(\theta^{p^{n-m^*}})]\leq [L:M^*(\theta^{p^{n-m^*}})]\leq p^{n-m^*}\) , we must have \[[L:L^{pi}(\theta^{p^{n-m^*}})]= [L:M^*(\theta^{p^{n-m^*}})].\] Now, the inclusion \(L^{pi}(\theta^{p^{n-m^*}})\supseteq M^*(\theta^{p^{n-m^*}})\) implies the equality \(L^{pi}(\theta^{p^{n-m^*}})=M^*(\theta^{p^{n-m^*}})\).

(vii) \([M^*:K]=p^{m^*}\): By the statement (vi), there is a diagram of fields where the numbers at the edges are the degrees of the corresponding field extensions:

\[\begin{tikzpicture}[scale=1.8, every node/.style={font=\normalsize}] \node (A) at (0,0) {K}; \node (B) at (-1.2,0.9) {M^*}; \node (D) at (1.2,0.9) {L^{sep}=K(\theta^{p^n})}; \node (C) at (0,1.7) {L^{pi}(\theta^{p^{n-m^*}})=M^*(\theta^{p^{n-m^*}})}; \draw (C) -- node[right] {p^{m^*}} (D); \draw (D) -- node[right] {s} (A); \draw (A) -- node[left] {} (B); \draw (B) -- node[left] {s} (C); \end{tikzpicture}\]

The equalities \([L^{sep}:K]=s\), \([M^*(\theta^{p^{n-m}}):M^*]=s\) and \[[M^*(\theta^{p^{n-m}}):L^{sep}]= [L^{pi}(\theta^{p^{n-m^*}}):L^{sep}]= [L^{pi}L^{sep}:L^{sep}] =p^{m^*}\] follow from statement 1 and the statements (v) and (ii), respectively. Now, the diagram yields the equality \([M^*:K]= p^{m^*}\): \[[M^*:K]=\frac{[M^*(\theta^{p^{n-m}}):L^{sep}][L^{sep}:K]}{[M^*(\theta^{p^{n-m}}):M^*]}= \frac{p^{m^*}s}{s} =p^{m^*}.\]

(viii) \(L^{pi}=M^*\) and \(\Big( L^{pi}\Big)^{p^{m^*}}\subseteq K\): By the statements (iii) and (vii), \([L^{pi}:K]=p^{m^*}=[M^*:K]\). Then the equality \(L^{pi}=M^*\) follows from the inclusion \(L^{pi}\supseteq M^*\). The equality \(L^{pi}=M^*\) yields the inclusion \[\Big( L^{pi}\Big)^{p^{m^*}}=\Big( M^*\Big)^{p^{m^*}}=\bigg( K\bigg(\lambda_0^\frac{1}{p^{m^*}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m^*}}\bigg)\bigg)^{p^{m^*}}\subseteq K.\] (ix) \((L/L^{pi})^{sep}=L^{pi}L^{sep}/L^{pi}\): Since the field extension \(L^{pi}L^{sep}/L^{pi}\) is a separable field extension, we have the inclusion \[(L/L^{pi})^{sep}\supseteq L^{pi}L^{sep}/L^{pi}.\] Since the field extension \(L/L^{pi}L^{sep}=K(\theta)/L^{pi}(\theta^{p^{n-m^*}})\) is purely inseparable, we must have the equality the equality \((L/L^{pi})^{sep}=L^{pi}L^{sep}/L^{pi}\).

(x) \(m^*=m\): By the definition of the number \(m^*\), \[\theta^\frac{1}{p^{n-m'}}\not\in L^{pi}\otimes L^{sep}\;\; \text{for all m' such thatm^*<m'\leq n. }\] By the statements (v) and (viii), \(f^{sep,\frac{1}{p^{m^*}}}(x)\in {\rm Irr}_m(M^*[x])={\rm Irr}_m(L^{pi}[x])\). By the definition of the number \(m=m_f\), \[f^{sep,\frac{1}{p^{m}}}(x):=\sum_{i=0}^s\lambda_i^\frac{1}{p^m} x^i\in L^{pi}[x].\] Therefore, \(m^*\leq m\) (by the maximality of \(m\)). Since \(f(x)\in {\rm Irr}_m(K[x])\), \(f(x)=\Big(f^{sep,\frac{1}{p^{m}}}(x^{p^{n-m}}) \Big)^{p^m}\) and \(\Big( L^{pi}\Big)^{p^{m^*}}\subseteq K\) (the statement (viii)), the polynomial \(f^{sep,\frac{1}{p^{m}}}(x)\in L^{pi}[x]\) is an irreducible polynomial over the field \(L^{pi}\). Otherwise, \(f^{sep,\frac{1}{p^{m}}}(x)=a(x)b(x)\) for some non-scalar polynomials \(a(x), b(x)\in L^{pi}[x]\), and so \[f(x)=\bigg(f^{sep,\frac{1}{p^{m}}}(x^{p^{n-m}}) \bigg)^{p^m}=\bigg(a(x^{p^{n-m}})b(x^{p^{n-m}}) \bigg)^{p^m}=a(x^{p^{n-m}})^{p^m}b(x^{p^{n-m}})^{p^m}\] where \(a(x^{p^{n-m}})^{p^m}, b(x^{p^{n-m}})^{p^m}\in K[x]\backslash K\), a contradiction.

The polynomial \(f^{sep,\frac{1}{p^{m}}}(x)\in L^{pi}[x]\) is a separable polynomial over the field \(L^{pi}\) (since it is an irreducible polynomial over \(L^{pi}\) and its derivative is a nonzero polynomial). The equality \[0=f(\theta)=\Big(f^{sep,\frac{1}{p^{m}}}(\theta^{p^{n-m}}) \Big)^{p^m}\] implies the equality \(f^{sep,\frac{1}{p^{m}}}(\theta^{p^{n-m}})=0\). This means that \[\theta^{p^{n-m}}\in (L/L^{pi})^{sep}\stackrel{{\rm (ix)}}{=}L^{pi}L^{sep}/L^{pi}.\] Now, we must have \(m^*=m\) (since otherwise, \(m^*<m\) and \(\theta^{n-m}\not\in L^{pi}L^{sep}\), a contradiction).

(xi) \(\max\{m'\in \mathbb{N}\, | \, \theta^{p^{n-m'}}\in L^{pi}L^{sep} \}=\max\{m'\in \mathbb{N}\, | \, L^{p^{n-m'}}\subseteq L^{pi}L^{sep} \}\): By statement 1, the field extension \(L/L^{pi}L^{sep}=K(\theta)/L^{pi}L^{sep}\) is a purely inseparable field extension since \(\theta^{p^n}\in L^{sep}\). Now, the equality in the statement (xi) follows from the equality \(L=K(\theta)\).

(xii) The number \(m\) is an isomorphism invariant of the field extension \(L/K\): By the definition, the number \(m^*=\max\{m'\in \mathbb{N}\, | \, L^{p^{n-m'}}\subseteq L^{pi}L^{sep} \}\) (the statement (xi)) is an isomorphism invariant of the field extension \(L/K\). Hence, so is the number \(m=m^*\) (the statement (x)). ◻

Proof. 1. By the definition, \(m_f=\max\bigg\{ m=0,1,\ldots , n\, \bigg| \, \lambda_i^\frac{1}{p^{m}}\in L\) for all \(i=0, \ldots , s-1\bigg\}\). Now, statement 1 follows from the equalities \(L=\bigoplus_{i=0}^{sp^n-1}K\theta^i\) and \(L^{pi}=K\Big(\lambda_0^\frac{1}{p^{m_f}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m_f}}\Big)\) (Theorem 2.(2)).

2. By Theorem 2.(1,3) and Theorem 2.(2), \[\begin{align} L^{pi}L^{sep}&=&L^{pi}\otimes L^{sep}=L^{pi}\otimes K(\theta^{p^n})=\bigoplus_{i=0}^{s-1}L^{pi}\theta^{ip^n},\\ L^{pi}&=&K\Big(\lambda_0^\frac{1}{p^{m_f}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m_f}}\Big)=\sum_{\alpha\in \mathbb{N}_{<p^{m_f}}}K\lambda^\frac{\alpha}{p^{m_f}}, \end{align}\] respectively. By Theorem 2.(4), the number \(m_f\) is the maximal number \(m\in \{ 0,1,\ldots , n\}\) such that \(\theta^{p^{n-m}}\in L^{pi}L^{sep}\). Now, statement 2 follows. ◻

Proof. By Theorem 2 or diagram (6 ), \(L^{pi}=K\) iff \(m_f=0\) iff either \(n=0\) or \(n\geq 1\) and \(\lambda_i^\frac{1}{p}\not\in L\) for some index \(i\in \{ 0,1,\ldots, s-1\}\) and \(m_f=0\) iff \([L:L^{sep}]=n\). ◻

Proof. \((1\Leftrightarrow 2)\) By Theorem 2.(4), \([L:L^{pi}\otimes L^{sep}]=p^{n-m_f}\) and the result follows.

\((2\Leftrightarrow 3)\) By Theorem 2.(2), the equality \(m_f=n\) is equivalent to the equality \(L^{pi}=K\Big( \lambda_0^\frac{1}{p^n}, \ldots , \lambda_{s-1}^\frac{1}{p^n} \Big)\). ◻

Proof. 1. The equality \(f_= f_N^{p^{m_{f,N}}}\) is obvious.

Suppose that \(f_N\not\in {\rm Irr}_m(N[x])\) and we seek a contradiction. Then \(f_N(x)=ab\) for some non-scalar polynomials \(a,b\in N[x]\). Then \(a^{p^{m_{f,N}}}, b^{p^{m_{f,N}}}\in K[x]\) (since \(N^{p^{m_{f,N}}}\subseteq K\)) and \[f= f_N^{p^{m_{f,N}}}=\Big(ab \Big)^{p^{m_{f,N}}}=a^{p^{m_{f,N}}}b^{p^{m_{f,N}}}.\] Therefore, \(f\not\in {\rm Irr}_m(K[x])\), a contradiction. Thus, \(f_N\in {\rm Irr}_m(N[x])\), and so \(f_N\in {\rm Irr}_m\bigg(K\bigg(\lambda_0^\frac{1}{p^{m_{f,N}}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m_{f,N}}}\bigg)\bigg)\) (since \(K\bigg(\lambda_0^\frac{1}{p^{m_{f,N}}}, \ldots , \lambda_{s-1}^\frac{1}{p^{m_{f,N}}}\bigg)\subseteq N\)).

2. Statement 2 follows from statement 1.

3. Statement 3 follows from statement 1. ◻

Proof. By (9 ), \(m_{f,N_i}=i\) for all \(i=0,1,\ldots, n\). Now, by (10 ), the corollary follows from Proposition 1. ◻

Proof. Suppose that the polynomial \(\phi\) is a reducible polynomial over the field \(N\). Then \(\phi = ab\) for some nonscalar polynomials \(a,b\in N[x]\). By the assumption the polynomial \(\phi \in K[x]\) is a separable polynomial over \(K\). Therefore, \(\phi = \prod_{i=1}^d(x-\theta_i)\) where \(d=\deg (\phi)\) and \(\theta_1, \ldots , \theta_d\in \overline{K}^{sep}\) are distinct roots of the polynomial \(\phi\) (by the separability of \(\phi\)). Then, up to order of the roots of \(\phi\), \[a=\prod_{i=1}^n (x-\theta_i)=\sum_{i=0}^n (-1)^is_i(\theta_1, \ldots ,\theta_n)x^{n-i} \;\] where \(s_i (x_1, \ldots , x_n)\) is the elementary symmetric polynomial/function of degree \(i\) in the variables \(x_1, \ldots , x_n\) for \(i=1,\ldots, n\) and \(s_0 (x_1, \ldots , x_n):=1\). Since \(\theta_1, \ldots , \theta_d\in \overline{K}^{sep}\) and \(a\in N[x]\), the coefficients of the polynomial \(a\) belong to the intersection \(\overline{K}^{sep}\cap N=K\), i.e. \(a\in K[x]\). By symmetry, \(b\in K[x]\). Therefore, the polynomial \(f=ab\) is a reducible polynomial over \(K\), a contradiction. ◻

Proof. 1. Clearly, \(g:=\sum_{i=0}^s\lambda_i^\frac{1}{p^n}x^i\in N_{f,n}[x]\).

(i) The polynomial \(g\in {\rm Irr}_m(N_{f,n}[x])\) is a monic irreducible separable polynomial over the field \(N_{f,n}\): By Corollary 2.(1), \[g=f_{N_{f,n}}\in {\rm Irr}_m(N_{f,n}[x]).\] Therefore, the polynomial \(g\) is a separable polynomial over the field \(N_{f,n}\) (since the polynomial \(g\) is an irreducible polynomial over \(N_{f,n}\) and \(\frac{dg}{dx}\neq 0\)).

(ii) For all purely inseparable field extensions \(N/K\) such that \(N_{f,n}\subseteq N\), the polynomial \(g\in {\rm Irr}_m(N[x])\) is a monic irreducible separable polynomial over the field \(N\): The statement (ii) follows from the statement (i) and Lemma 2.

(iii) For each \(j=0,1,\ldots , n\), \(K({\rm coef} (g^{p^j}))=N_{f,n-j}\): The statement (iii) follows from the equality \[g^{p^j}=\sum_{i=0}^s\lambda_i^\frac{1}{p^{n-j}}x^{ip^{n-j}}.\] (iv) \(K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1}=N_{f,n-\delta}\) for all natural numbers \(i=i_\delta p^\delta\) where \(i_\delta=1,\ldots , p-1\) and \(\delta=0,1,\ldots , n\): By the statement (iii) and the inclusion \(N_{f,n-\delta}\supseteq N_{f,n-\delta-1}\), \[N_{f,n-\delta}=K\Big({\rm coef} \Big(g^{p^\delta}\Big)\Big) \supseteq K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1}.\] Since \({\rm gcd } (i_\delta, p)=1\), \(\alpha i_\delta-\beta p=1\) for some integers \(\alpha,\beta \geq 1\) (take \(\alpha\in \{ 0,1, \ldots, p-1\}\) such that \(\alpha =i_\delta^{-1}\in \mathbb{F}_p\)). It follows from the equalities and the inclusions, \[g^{p^\delta}=g^{1\cdot p^\delta}=g^{(\alpha i_\delta-\beta p)p^\delta}=\frac{g^{\alpha i_\delta p^\delta}}{g^{\beta p^{\delta+1}}}, \;\; {\rm coef}\Big(g^{\alpha i_\delta p^\delta}\Big)\subseteq {\rm coef}\Big(g^{ i_\delta p^\delta}\Big) \;\; {\rm and}\;\; {\rm coef}\Big(g^{\beta p^{\delta+1}}\Big)\subseteq N_{n-\delta-1},\] that \(N_{f,n-\delta}=K({\rm coef} \Big(g^{p^\delta}\Big)\Big) \subseteq K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1},\) and the statement (iv) follows.

(v) For all natural numbers \(i=\sum_{\nu=0}^\mu i_\nu p^\nu\) where \(i_\nu \in \{ 0,1,\ldots , p-1\}\) and \(\mu \leq n\), \[K\Big( {\rm coef}(g^i)\Big)N_{f,n-\delta-1}=K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big)N_{f,n-\delta-1}=N_{f,n-\delta}\;\; {\it where} \;\; \delta:=\min \{\nu \, | \, i_\nu\neq 0\}:\] It suffices to show that the first equality holds since the second one is the statement (iv). In view of the statement (iv), we may assume that \(\delta<\mu\). By the statement (iii), \[g^i=\prod_{\nu=\delta}^\mu \Big(g^{p^\nu}\Big)^{i_\nu}\in g^{i_\delta p^\delta}\prod_{\nu>\delta}^\mu N_{f,n-\nu}[x] \subseteq g^{i_\delta p^\delta} N_{f,n-\delta-1}[x]\] since \(N_{f,n-\delta}\supset N_{f,n-\delta-1}\supset \cdots \supset N_{f,-1}\supset N_{f,0}=K\). Therefore, \(K\Big({\rm coef}(g^i)\Big)\subseteq K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1}\). Hence, \[K\Big({\rm coef}(g^i)\Big)N_{f,n-\delta-1}\subseteq K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1}.\] The equality \(g^{i_\delta p^\delta}=\frac{g^i}{\prod_{\nu>\delta}^\mu \Big(g^{p^\nu}\Big)^{i_\nu}}\) implies the inclusion \(K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big)\subseteq K\Big({\rm coef}(g^i)\Big)N_{f,n-\delta-1}\). Hence, \[K\Big({\rm coef}\Big(g^{i_\delta p^\delta}\Big)\Big) N_{f,n-\delta-1}\subseteq K\Big({\rm coef}(g^i)\Big)N_{f,n-\delta-1},\] and the statement (v) follows.

Let \(N/K\) be a purely inseparable field extension. By the definition of the natural number \(m:=m_{f,N}\) and the field \(M_{f,N}\), \[N_{f,m}=M_{f,N}\subseteq N.\] (vi) \(f_N=\sum_{i=0}^s\lambda_i^\frac{1}{p^m}x^{ip^{n-m}}=f_{N_{f,m}}\in {\rm Irr}_m(N_{f,m}[x])\): The statement (vi) follows from Corollary 2.(1).
(vii) \(f_N\in {\rm Irr}_m(N[x])\): The polynomial \(f_N\) is a monic polynomial. Suppose that the polynomial \(f_N\) is a reducible polynomial over the field \(N\), i.e. \(f=ab\) for some non-scalar polynomials \(a,b\in N[x]\). We seek a contradiction. Notice that \[f_N=g^{p^{n-m}}\;\; {\rm and }\;\; g\in {\rm Irr}_m\Big(\overline{K}^{pi}[x]\Big)\;\; \text{(by the statement (ii)). }\] Therefore, \(a=g^i\) and \(b=g^j\) for some natural numbers \(i\geq 1\) and \(j\geq 1\) such that \(i+j=p^{n-m}\). Therefore, \(i=\sum_{\nu=0}^\mu i_\nu p^\nu\) where \(i_\nu \in \{ 0,1,\ldots , p-1\}\) and \(\mu <n-m\). Let \(\delta:=\min \{\nu \, | \, i_\nu\neq 0\}\). Then \(\delta< n-m\) or, equivalently, \(m<n-\delta\). By the statement (v), \[K\Big( {\rm coef}(g^i)\Big)N_{f, n-\delta-1}=N_{f, n-\delta}\supset N_{f, n-\delta}\supseteq N_{f, m}.\] Since \(g^i\in N[x]\), we must have \(K\Big( {\rm coef}(g^i)\Big)\subseteq N_{f, m}\). Therefore, \[N_{f, n-\delta}=K\Big( {\rm coef}(g^i)\Big)N_{f, n-\delta-1}\subseteq N_{f, m}N_{f, n-\delta-1}=N_{f, n-\delta-1},\] a contradiction.

2. Statement 2 follows from statement 1. ◻

Proof. The theorem is Theorem 2 but for the field extension \(N(\theta)/N\) rather than \(K(\theta)/K\). It is obtained in a straightforward manner from Theorem 2 using Theorem 4. ◻

Proof. By Theorem 5 or diagram (12 ), \((NL)^{pi}=N\) iff \(m_{f, N(\theta)}=0\) iff either \(n=m_{f,N}\) or \(n>m_{f,N}\) and \(\lambda_i^\frac{1}{p^{m_{f,N}+1}}\not\in NL\) for some index \(i\in \{ 0,1,\ldots, s-1\}\). Clearly, \(m_{f, N(\theta)}=0\) iff \([NL:(NL)^{sep}]=p^{n-m_{f,N}}\). ◻

Proof. \((1\Leftrightarrow 2)\) By Theorem 5.(4), \([NL:(NL)^{pi}\otimes_N (NL)^{sep}]=p^{n-m_{f,L}-m_{f,NL}}\) and the result follows.

\((2\Leftrightarrow 3)\) By Theorem 5.(2), the equality \(n=m_{f,N}+m_{f, N(\theta)}\) is equivalent to the equality \((NL)^{pi}=N\Big( \lambda_0^\frac{1}{p^n}, \ldots , \lambda_{s-1}^\frac{1}{p^n} \Big)\). ◻

Proof. \((1\Leftrightarrow 2)\) Since \(L \supseteq L^{pi}L^{sep}=L^{pi}\otimes L^{sep}\), the equality \(L = L^{pi}\otimes L^{sep}\) holds iff \([L:K] = [L^{pi}:K] \cdot [L^{sep}:K]\).

\((1\Leftrightarrow 3)\) By diagram (2), \(L = L^{pi}\otimes L^{sep}\) iff \(L/L^{pi}\) is a separable field extension.

\((3\Leftrightarrow 4)\) The equivalence is obvious. ◻

Proof. \((1\Leftrightarrow 2)\) By Theorem 7.(3), the equality \(L = L^{pi}L^{sep}\) holds iff the field extension \(L/L^{pi}\) is a separable field extension iff for each \(i=1, \ldots , \nu\), the polynomial \(f_{i, L^{pi}}(x)=\sum_{j=0}^{s_i}\lambda_{ij}^\frac{1}{p^{n_i}} x^j\) is the minimal polynomial of the element \(\theta_i\) over the field \(L^{pi}\), by Theorem 4.(1) (since \(L=L^{pi}(\theta_1,\ldots , \theta_\nu)\)).

\((3\Rightarrow 2\Leftrightarrow 4)\) Clear.

\((2\Rightarrow 3)\) Let \(N_i:=K\bigg(\lambda_{i,0}^\frac{1}{p^{n_i}}, \ldots, \lambda_{i,s_i-1}^\frac{1}{p^{n_i}} \bigg)\). Suppose that statement 2 holds. Then, by Theorem 3, \(L_i=L_i^{pi}L_i^{sep}\) where \(L_i^{pi}=N_i\) for \(i=1, \ldots , \nu\). Therefore, \[\begin{align} L&=&\prod_{i=1}^\nu L_i=\prod_{i=1}^\nu L_i^{pi}L_i^{sep}=\prod_{i=1}^\nu L_i^{pi} \prod_{i=1}^\nu L_i^{sep}\stackrel{{\rm Thm.}\, \ref{A27Apr26}}{=}\prod_{i=1}^\nu L_i^{pi}\otimes\prod_{i=1}^\nu L_i^{sep} \end{align}\] since \(\prod_{i=1}^\nu L^{pi}_i\subseteq L^{pi}\) and \(\prod_{i=1}^\nu L_i^{sep} \subseteq L^{sep}\) and \(L^{pi}L^{sep}=L^{pi}\otimes L^{sep}\) (Theorem 1). Now, two inclusions above and the inclusions \[L^{pi}\otimes L^{sep}=L^{pi}L^{sep}\subseteq L= \prod_{i=1}^\nu L_i^{pi}\otimes\prod_{i=1}^\nu L_i^{sep} \subseteq L^{pi}\otimes L^{sep} \subseteq L\] imply that \(L=L^{pi} L^{sep}=L^{pi}\otimes L^{sep}\), \[L^{pi}=\prod_{i=1}^\nu L_i^{pi}=\prod_{i=1}^\nu N_i=K\bigg(\lambda_{ij}^\frac{1}{p^{n_i}}\bigg| i=1, \ldots , \nu; j=0,1, \ldots , s_i-1\bigg)\] and \(L^{sep}= \prod_{i=1}^\nu L_i^{sep}\). ◻