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A field isomorphism is a bijective field homomorphism; its inverse is also a field homomorphism. For fields extending a specified field $K$, an isomorphism may be required to fix $K$ pointwise, a stronger requirement than an abstract field isomorphism.

For an unramified extension $L/K$ of non-Archimedean local fields and any finite extension $M/K$, reduction is a bijection between $K$-field homomorphisms $L\to M$ and $k_K$-field homomorphisms $k_L\to k_M$. Choose an integral generator with separable irreducible reduction. Each specified residue image lifts uniquely to a root in $M$ by Hensel lemma. Conversely a $K$-embedding preserves the unique extended valuation and therefore induces the residue embedding. The same argument applies when $M$ is any non-Archimedean local extension with the given compatible embedding of $K$.

For a finite field with $q$ elements, exactly $q+1$ functions from the field to itself preserve multiplication, without requiring additivity or preservation of the identity. They are the constant-zero map, the constant-one map, and the $q-1$ power maps $x\mapsto x^k$ for $1\le k\le q-1$.

To prove completeness, $f(1{)}^2=f(1)$ forces $f(1)\in \{0,1\}$ because a field has no zero divisors. The zero choice forces the constant-zero map. For $f(1)=1$, nonzero elements must map to nonzero elements, and the restriction is a group homomorphism of the cyclic multiplicative group of a finite field. Its generator has $q-1$ possible images, giving the $q-1$ power maps on nonzero elements. Setting $f(0)=0$ extends each. The only further possibility is $f(0)=1$, which forces every value to be one. For $q=5$, this gives six maps. These are generally not field homomorphisms, because preservation of addition was never required.

All the local fields here carry normalized discrete valuations. Write ${\mathcal{O}}_F$ for the valuation ring, ${\mathfrak{m}}_F$ for its maximal ideal, and $k_F$ for its residue field. We use the basic degree identity $[B:A]=e(B/A)f(B/A)$ for finite extensions of non-Archimedean local fields, with $e$ the ramification index and $f$ the residue degree.

First construct the maximal unramified subextension of a local field extension inside an arbitrary finite extension $L/K$. Choose a primitive element of a field extension $\bar{z}$ for the finite residue field extension $k_L/k_K$. Let $\bar{h}$ be its monic irreducible minimal polynomial of an algebraic element, of degree $f$, and lift its coefficients to obtain a monic $h\in {\mathcal{O}}_K[T]$ of the same degree. Finite finite fields are perfect fields, so ${\bar{h}}^{\prime }(\bar{z})\ne 0$. By Hensel lemma applied in $L$, there is $z\in {\mathcal{O}}_L$ reducing to $\bar{z}$ with $h(z)=0$. For $U=K(z)$ we haveEvery inequality is therefore an equality. In particular $U/K$ is unramified, $k_U=k_L$, and $h$ is the minimal polynomial of an algebraic element $z$.

If $L/K$ is unramified, then $[L:K]=f$, so this construction gives $U=L$. Taking $x=z$ supplies the required generator, and $h^{\prime }(x)$ is a unit. Conversely suppose $L=K(x)$ with $x$ integral and its monic minimal polynomial of an algebraic element $g$ satisfying $g^{\prime }(x)\notin {\mathfrak{m}}_L$. Its coefficients lie in ${\mathcal{O}}_K$, since all conjugates of an integral element are integral and $K$ is complete. In the constructed unramified extension $U$ the element $\bar{x}\in k_U=k_L$ is a simple root of $\bar{g}$. Hensel lemma in $U$ produces a root $x^{\prime }\in U$ with residue $\bar{x}$. In $L$ both $x$ and $x^{\prime }$ lift this same simple residue root, so Hensel uniqueness gives $x=x^{\prime }$. Thus $L\subseteq U\subseteq L$, proving

For lifting residue-field embeddings of unramified extensions, use the generator $z$ and polynomial $h$ just constructed when $L/K$ is unramified. Given a $k_K$-field homomorphism $\iota :k_L\to k_M$, its value $\iota (\bar{z})$ is a simple root of $\bar{h}$. Hensel lemma in $M$ produces the unique root $z^{\prime }\in {\mathcal{O}}_M$ with that residue. The assignment $z\mapsto z^{\prime }$ defines a $K$-field homomorphism $K[T]/(h)\to M$; it induces $\iota$ because $\bar{z}$ generates $k_L$. Conversely any such field homomorphism preserves the unique extended valuation, hence induces a residue field homomorphism, and its image of $z$ is uniquely forced by Hensel uniqueness. This proves the asserted bijection. The same proof only requires that $M$ be a non-Archimedean local field containing $K$ with its compatible valuation; its finiteness over $K$ is not essential.

Now set $M=L$. The finite residue field extension has exactly $f=[L:K]$ automorphisms, cyclically generated by $a\mapsto a^{|k_K|}$. Their unique lifts give $[L:K]$ automorphisms of $L$. Consequently $L/K$ is a Galois extension, and reduction identifies its Galois group with that cyclic residue Galois group. Its generator is the arithmetic Frobenius:

For the compositum, choose a uniformizer ${\pi }_M$ of the totally ramified extension $M/K$. The Eisenstein generator of a totally ramified extension characterization, proved in the next solution, says that $M=K({\pi }_M)$ and its minimal polynomial of an algebraic element is Eisenstein over ${\mathcal{O}}_K$. A uniformizer of $K$ remains a uniformizer in the unramified extension $L$. All its Eisenstein coefficient conditions therefore remain true over ${\mathcal{O}}_L$. Thus the same Eisenstein polynomial is irreducible over $L$ and generates $LM=L({\pi }_M)$ as a totally ramified extension. This proves $LM/L$ is totally ramified, of degree $[M:K]$.

Here is an explicit example showing that the converse decomposition of an arbitrary extension can fail. Take $K={\mathbb{Q}}_3$ and $L=K(i)$ with $i^2=-1$. Since $T^2+1$ is irreducible and separable modulo three, the degree and residue-degree argument above shows that $L/K$ is unramified quadratic. PutThe element $1+i$ is a unit in ${\mathcal{O}}_L$, so $T^2-3(1+i)$ is Eisenstein over $L$. Hence $[N:L]=2$, $e(N/K)=2$, and $f(N/K)=2$. Moreover $i={\alpha }^2/3-1$ gives $N=K(\alpha )$, withWe verify its Galois group, since that is what rules out a totally ramified complement. By Hensel lemma there is $r\in K$ with $r^2=-2$: the residue root $1$ of $T^2+2$ is simple modulo three. Set $\beta =(1+i)/r$. ThenThe assignments $\sigma (i)=-i$ and $\sigma (\alpha )=\beta \alpha$ respect the defining relation, because $(\beta \alpha {)}^2=-3i(1+i)=3(1-i)$. They therefore define a $K$-automorphism of $N$. Its square satisfies ${\sigma }^2(i)=i$ and ${\sigma }^2(\alpha )=\bar{\beta }\beta \alpha =-\alpha$, so $\sigma$ has order four. There are thus four automorphisms of the degree-four extension, andThis is the cyclic quartic local extension without a totally ramified complement. Its unique quadratic subfield is $L$, fixed by ${\sigma }^2$, and is unramified. If $N=L^{\prime }{M}^{\prime }$ with $L^{\prime }/K$ unramified and $M^{\prime }/K$ totally ramified, then $[L^{\prime }:K]\le f(N/K)=2$ and $[M^{\prime }:K]\le e(N/K)=2$. To obtain the full degree four both would have degree two, forcing $M^{\prime }$ to be a totally ramified quadratic subfield. There is no such subfield. Thus $N/{\mathbb{Q}}_3$ is the required counterexample.

The multiplicative Hilbert theorem 90 states that, for a finite Galois extension $L/K$ with Galois group $G$, every multiplicative cocycle $c:G\to L^{\times }$ satisfying $c_{\sigma \tau }=c_{\sigma }\sigma (c_{\tau })$ has the form $c_{\sigma }=\sigma (y)/y$ for some $y\in L^{\times }$. In the cyclic case $G=⟨\sigma ⟩$, it says

Here is the full existence argument. Distinct field homomorphisms $\sigma :L\to L$ are linearly independent as functions over $L$. To see this Artin independence theorem directly, assume a nonzero identically vanishing relation with the fewest nonzero coefficients. There are at least two terms. Choose two participating embeddings ${\sigma }_1,{\sigma }_j$ and an element $a$ with ${\sigma }_1(a)\ne {\sigma }_j(a)$. Evaluate the relation at $at$ and subtract ${\sigma }_j(a)$ times its value at $t$. The $j$th term disappears while the first remains nonzero, giving a shorter vanishing relation, a contradiction.

Apply this independence toThe coefficients are nonzero, so $A$ is not identically zero; choose $t$ with $y=A(t)\ne 0$. The cocycle identity gives $\tau (c_{\sigma }^{-1})=c_{\tau \sigma }^{-1}{c}_{\tau }$, and henceThis proves the general statement. In the cyclic case, given $N(x)=1$, set $c_1=1$ and $c_{{\sigma }^j}=x\sigma (x)\cdots {\sigma }^{j-1}(x)$ for $1\le j<n$. The norm-one identity makes this definition compatible with ${\sigma }^n=1$, and the cocycle identity follows by concatenating the products. The argument above gives $\sigma (y)/y=x$. Conversely that quotient has norm one because its conjugate product telescopes.

Finally let $L/K$ be unramified, and take its arithmetic Frobenius $\varphi$. For a unit $x$ of norm one, the cyclic theorem gives $b\in L^{\times }$ with $\varphi (b)/b=x$. Write $r=v_L(b)$ and choose the common uniformizer $\pi \in K$. Then $y={\pi }^{-r}b$ has valuation zero, and $\varphi (\pi )=\pi$ implies $\varphi (y)/y=\varphi (b)/b=x$. Thus the unit form of Hilbert theorem 90 isThe unramified hypothesis is what permits shifting an arbitrary integer valuation of $b$ by a power of a uniformizer belonging to $K$.

A field homomorphism fixing $K$ is injective, since its kernel is an ideal of a field and cannot contain one. It is also $K$-linear. Equal finite field extension degrees then make it a bijection, hence a $K$-isomorphism.

Equal degrees alone do not ensure that any such homomorphism exists. For example $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ both have degree two, but a $\mathbb{Q}$-isomorphism would require a square root of two in $\mathbb{Q}(\sqrt{3})$. Squaring $a+b\sqrt{3}$ with rational $a,b$ would give $2ab=0$ and $a^2+3b^2=2$. Neither $a^2=2$ nor $b^2=2/3$ has a rational solution. Thus the two degree-two extensions are not isomorphic over $\mathbb{Q}$.