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The inverse different is the trace-dual latticeThe integral closure ${\mathcal{O}}_L$ is finite free over the complete discrete valuation ring ${\mathcal{O}}_K$. Choose an integral basis and use the nondegenerate trace pairing to form its dual basis over $K$. This exhibits ${\mathfrak{D}}^{-1}$ as a full, finite ${\mathcal{O}}_K$-lattice. It is stable under multiplication by ${\mathcal{O}}_L$, because $b{\mathcal{O}}_L\subseteq {\mathcal{O}}_L$ for $b\in {\mathcal{O}}_L$, and bounded denominators make it a fractional ideal of ${\mathcal{O}}_L$. Integral elements have integral field traces, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}^{-1}$. A nonzero fractional ideal of a discrete valuation ring is invertible; its inverse is consequently an integral ideal, the different ideal ${\mathfrak{D}}_{L/K}$.

Assume now $A={\mathcal{O}}_K[x]={\mathcal{O}}_L$, with monic separable minimal polynomial $g$ of degree $n$. Lagrange interpolation at its distinct roots givesIndeed, these are the leading coefficients in the interpolation formula for $X^j$. For any element of $A$, this trace is the coefficient of $X^{n-1}$ in its degree-less-than-$n$ representative modulo $g$. The resulting pairing on the basis $1,x,\dots ,x^{n-1}$ is integral and unimodular: reversing the order of one basis makes its matrix triangular with diagonal ones, since entries vanish when the exponent sum is less than $n-1$. Thus it identifies $A$ with its full ${\mathcal{O}}_K$-dual. Translating back to the trace pairing proves

For a totally ramified extension of degree $n$, any uniformizer ${\pi }_L$ is an Eisenstein generator of a totally ramified extension, and ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. To see the latter equality, use the common residue field to expand an integral element in powers of ${\pi }_L$ with digits from ${\mathcal{O}}_K$; reduce powers using its Eisenstein polynomial and take limits in the finite complete ${\mathcal{O}}_K$-module generated by $1,{\pi }_L,\dots ,{\pi }_L^{n-1}$. Write that polynomial as $X^n+{\sum }_{j<n}{a}_j{X}^j$, with $v_L(a_j)\ge n$ for $j<n$. When $p\nmid n$, the derivative's leading term has valuation $n-1$, while every other nonzero derivative term has valuation at least $n$. There can be no cancellation of the unique smallest term. Therefore

For the prime-power p-adic cyclotomic extension, put $\zeta ={\zeta }_{p^r}$ and $e=p^{r-1}(p-1)$. Modulo $p$, the shifted cyclotomic polynomial ${\Phi }_{p^r}(1+T)$ is $T^e$, while its constant term is $p$, not a multiple of $p^2$. It is therefore Eisenstein, so $\pi =\zeta -1$ is a uniformizer and ${\mathcal{O}}_{K_r}={\mathbb{Z}}_p[\zeta ]$. Fromwe obtain, by differentiating the numerator at $\zeta$,The denominator is a primitive-$p$ root of unity minus one and has $K_r$-valuation $p^{r-1}$. The different exponent is consequently $re-p^{r-1}$, givingThis includes $p=2,r=1$: the extension is trivial and its different exponent is zero.

If $f$ is an irreducible polynomial, any two roots $\alpha ,\beta$ have the same minimal polynomial. Thus the map $\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$ carrying $\alpha$ to $\beta$ is a field embedding. The extension theorem for field embeddings extends it to an embedding of the splitting field $L$ into an algebraic closure. Since $L/\mathbb{Q}$ is normal, its image is $L$, so it is an element of the Galois group.

Conversely, factor $f$ into irreducible factors over $\mathbb{Q}$. An automorphism preserves the vanishing of each factor, so it cannot send a root of one factor to a root of another. Since there are no repeated roots, different factors have disjoint root sets. Transitivity on all roots therefore forces a single irreducible factor. This proves the equivalence.

Write $J_r(\lambda )$ for a size-$r$ Jordan block. The characteristic polynomial fixes the total dimensions at $2$ and $-7$, while the minimal polynomial fixes the largest Jordan block at each eigenvalue. Consequently, the blocks for $2$ have total size four, largest size two, and at least one size-two block. The blocks for $-7$ both have size one. The complete list, up to similarity, isThese Jordan normal forms are inequivalent because their $2$-eigenspaces have different dimensions, two and three.

An algebraic integer is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$. It is irreducible over $\mathbb{Q}$. If $H\in \mathbb{Z}[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb{Q}$. Gauss lemma for polynomials implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in \mathbb{Z}[x]$, and its irreducibility over $\mathbb{Q}$ implies irreducibility in $\mathbb{Z}[x]$.

For $h\in \mathbb{Z}[x]$ with $h(\alpha )=0$, division by the monic $f$ works in $\mathbb{Z}[x]$, giving $h=qf+r$ with $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.

The polynomial $f(x)=x^5+2x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb{Q}$ and is the minimal polynomial of every one of its roots. Part (a) givesButIn the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an integral domain, so $(5)$ is not a prime ideal.

For ring homomorphisms preserving the identity, an image $\beta$ of $\alpha$ in the Gaussian integers would satisfy $f(\beta )=0$. The minimal polynomial of $\beta$ over $\mathbb{Q}$ would then be $f$, of degree five. This is impossible because $\beta \in \mathbb{Q}(i)$ has degree at most two. Thus there is no unital homomorphism $\mathbb{Z}[\alpha ]\to \mathbb{Z}[i]$. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

The denominator is ${\pi }^{n}X$; dividing by $X$ removes the zero at $X=0$. A uniformizer of a degree-$n$ totally ramified extension generates the field: its valuation over $K$ is $1/n$, so $K(\pi )/K$ already has ramification index at least $n$. Hence $K(\pi )=L$, its minimal polynomial has degree $n$, and its conjugates are the distinct $\sigma (\pi )$ for $\sigma \in G$.

Factor the minimal polynomial over $L$. The reduced ramification polynomial isEvery displayed root $r_{\sigma }=(\sigma (\pi )-\pi )/\pi$ is nonzero and integral, since two uniformizers have a difference of valuation at least one. Thus $g$ is monic of degree $n-1$, has nonzero constant coefficient, and belongs to ${\mathcal{O}}_L[X]$. For $n=1$ it is the constant polynomial $1$, with no slopes and no ramification jumps.

Here is also a justification of the uniformizer criterion for lower ramification groups in this setting. The powers $1,\pi ,\dots ,{\pi }^{n-1}$ are a $K$-basis. In an expansion $x=\sum a_j{\pi }^j$, the nonzero terms have valuations $n{v}_K(a_j)+j$, distinct modulo $n$; therefore there is no cancellation at the least valuation. If $x$ is integral, all those values are nonnegative, which forces $a_j\in {\mathcal{O}}_K$. Thus ${\mathcal{O}}_L={\mathcal{O}}_K[\pi ]$. For $j\ge 1$, $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$, with the remaining factor integral. Consequently every $\sigma (x)-x$ has valuation at least $v_L(\sigma (\pi )-\pi )$, and equality is attained at $x=\pi$.

For a nonidentity automorphism, write $i(\sigma )=v_L(\sigma (\pi )-\pi )$. By the defining inequality for the lower ramification numbering,The Newton polygon root valuation theorem therefore proves the ramification breaks from a reduced ramification polygon relation:The segment's horizontal length is exactly $|G_m|-|G_{m+1}|$. Equivalently, in the reflected Newton polygon convention required for a positive-slope formulation,All occurring root valuations are nonnegative integers. For negative indices, extend $G_m=G$; total ramification means that there are no negative-index jumps, so the reflected formulation remains true for all integers.

The sign distinction is substantive. Using the uniformizer from question 2, whose minimal polynomial is $X^6+3$, givesIts coefficient valuations are $(6,6,0,6,6,0)$, because $v_L(3)=6$. The usual lower hull has vertices $(0,6),(2,0),(5,0)$ and slopes $-3,0$, of lengths $2,3$. These correspond exactly to the drops $|G_3|-|G_4|=2$ and $|G_0|-|G_1|=3$. With the reflected axis the slopes are $0,3$, as in the printed positive-$m$ wording. If the coefficient-exponent definition is used throughout, the printed assertion needs the minus sign shown above.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

A splitting field of a nonzero polynomial $f$ over $K$ is an extension in which $f$ splits into linear factors and which is generated over $K$ by all its roots. For a constant nonzero polynomial the splitting field is $K$; the zero polynomial does not have a root-generated splitting field in this standard definition.

Given two splitting fields $L,L^{\prime }$, enumerate the roots generating $L$ and extend a $K$-embedding one root at a time. If an embedding is defined on $K({\alpha }_1,\dots ,{\alpha }_j)$, the minimal polynomial of the next root divides $f$ over that field. Its image under the embedding therefore divides $f$ over the corresponding subfield of $L^{\prime }$, and has a root in $L^{\prime }$. The quotient construction $K(\alpha )\cong K[t]/(m_{\alpha })$ extends the embedding to that root. The polynomial need not be a separable polynomial.

At the end this gives a $K$-embedding $L\to L^{\prime }$. It sends the distinct roots of $f$ injectively to its roots, hence permutes the finite set of distinct roots and contains every root in its image. Since these generate $L^{\prime }$, the embedding is onto. Thus the splitting field is unique up to a (K)-isomorphism.

Choose a nonzero $\alpha \in \mathfrak{a}$. It is an algebraic integer, with monic minimal polynomialIts constant coefficient is nonzero, since otherwise the polynomial would have the root zero and be divisible by its variable, contradicting irreducibility and $\alpha \ne 0$. Rearranging gives $c_0=-\alpha ({\alpha }^{d-1}+\cdots +c_1)$. The expression in parentheses lies in the ring of integers of a number field, so the ideal property gives $c_0\in \mathfrak{a}$. Hence

A finite field extension $L/K$ is a separable field extension if every element of $L$ has a minimal polynomial over $K$ with distinct roots in an algebraic closure. The primitive element theorem says that a finite separable field extension $L/K$ is $K(\alpha )$ for some $\alpha \in L$.

If $K$ is finite, then so is $L$. The multiplicative group of a finite field is cyclic, so a generator of $L^{\times }$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a finite abelian group its exponent is attained as an element order, while the polynomial root bound forces its size to be at most that exponent.

If $K$ is infinite, first consider $L=K(a,b)$. The minimal polynomials of $a$ and $b$ have distinct roots $a_i,b_j$ in an algebraic closure. Choose $c\in K$, $c\ne 0$, avoiding the finitely many values for which $a_i+c{b}_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha =a+cb$. Over $K(\alpha )$, the polynomials $f(T)$ and $g((\alpha -T)/c)$ have precisely one common root, $a$. Their monic greatest common divisor is therefore $T-a$, so $a\in K(\alpha )$ and $b=(\alpha -a)/c\in K(\alpha )$. Thus $K(a,b)=K(\alpha )$. Induct over finitely many generators of the finite separable field extension. This proves every finite separable extension is simple.

Put $a=\sqrt{p}$, $b=\sqrt{-q}$ and $\alpha =a+b$. Since $a^2-b^2=p+q$, the identity $(a+b)(a-b)=p+q$ givesThus $\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt{p},\sqrt{-q})$. The latter has degree $4$: $p$ is not a square and $\mathbb{Q}(\sqrt{p})$ is real, whereas $\sqrt{-q}$ is not real. Squaring ${\alpha }^2=p-q+2\sqrt{-pq}$ gives its monic degree-$4$ minimal polynomial

In characteristic $p$, let $K={\mathbb{F}}_p(s,t)$ and $L={\mathbb{F}}_p(s^{1/p},t^{1/p})$, with $s,t$ having algebraic independence. The monomials $s^{i/p}{t}^{j/p}$, $0\le i,j<p$, form a $K$-basis, so $[L:K]=p^2$. Their independence follows by grouping polynomial monomials according to their exponents modulo $p$, after clearing rational-function denominators.

Every $\alpha \in L$ has ${\alpha }^p\in K$, so its minimal polynomial divides $T^p-{\alpha }^p$ and $[K(\alpha ):K]\le p$. Thus this finite purely inseparable extension has no primitive element. It demonstrates why the separable field extension hypothesis cannot simply be removed.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

The characteristic polynomial is $(t-\frac{1}{2}{)}^3$. For $N=A-\frac{1}{2}I$, one finds $N\ne 0$ but $N^2=0$, so the minimal polynomial is $(t-\frac{1}{2}{)}^2$. Hence the invariant factors are $t-\frac{1}{2}$ and $(t-\frac{1}{2}{)}^2$, and the rational canonical form is

Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.