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Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector $z$ has squared norm ${\sum }_i{a}_i{z}_i^2$. These four numbers are linearly independent over $\mathbb{Q}$: the four sign changes of $\sqrt{2},\sqrt{3}$ in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.

LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives $S^T=-S$, $S^2=-3I$, andIn particular both $L_{\pm }$ are full-rank Euclidean lattices. Denote their four generator columns by $v_i^{\pm }$. Modulo $3$ the two matrices have the same columns, and the nonzero words of the given ternary tetracode $T$ are exactlyThese are the reductions of $\pm v_i^{\pm }$, respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo $3$ maps either Euclidean lattice onto precisely $T$.

To construct a length-preserving correspondence, find the kernels $M_{\pm }=L_{\pm }\cap 3{\mathbb{Z}}^4$ of this reduction. The inverse matrix is $A_{\pm }^{-1}=A_{\pm }^T/12$. Hence $3w\in L_{\pm }$ exactly when $A_{\pm }^{T}w\equiv 0(\mathrm{mod}4)$. These congruences are equivalent toFor clarity, the first $+$ congruence is $-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; subtracting the others gives evenness of each required pair sum. For the $-$ matrix the first congruence is $2w_1-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.

Let $R_i$ be the orthogonal reflection changing only coordinate $i$. If the coordinates of $w$ are even, this changes their sum by a multiple of $4$; if all are odd, it changes the sum by $2$ modulo $4$. ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets $\pm v_i^{\pm }+M_{\pm }$, since $T$ has nine words. Define a map $\Phi :L_{+}\to L_{-}$ by using $R_1$ on $M_{+}$ and $R_i$ on each of $v_i^{+}+M_{+}$ and $-v_i^{+}+M_{+}$. Each piece maps bijectively onto the corresponding $-$ coset. Thus $\Phi$ is a bijection satisfyingIt also commutes with $z\mapsto -z$. The closed geodesics of a flat torus have lengths $\Vert z\Vert$ for nonzero translation vectors $z$ in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at $\ell$ is the sum of primitive multiplicities at $\ell /k$, and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.

It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry $L_{+}$ onto $L_{-}$. In integer coordinates write this part as $B$. For every $z\in L_{+}$, both $z$ and $Bz$ have integer coordinates and have equal squared norms. Rational independence of the $a_i$ impliesApply this to $z=A_{+}n$ for all $n\in {\mathbb{Z}}^4$. Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since $A_{+}$ is invertible, $B$ is consequently a diagonal sign change $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_4)$.

Modulo $3$, such a $D$ must preserve $T$. Each nonzero word of $T$ has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4$. Thus only $D=I$ and $D=-I$ are possible. Neither changes $L_{+}$.

Finally $v_1^{+}\notin L_{-}$: it has the same ternary word as $v_1^{-}$, but their difference is $6e_1=3(2,0,0,0)$, which fails the congruence defining $M_{-}$. Hence $L_{+}\ne L_{-}$ and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.

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.

Because $f$ is irreducible, the ideal $(f)$ is maximal in the polynomial ring $k[X]$. Henceis a field extension of $k$. Let $\alpha =X+(f)\in F$. Evaluation in the quotient gives $f(\alpha )=0$, and the factor theorem in $F[X]$ givesfor some $g\in F[X]$. This is the standard adjoining a root of an irreducible polynomial construction.

The tower law for field extensions states that for finite field extensionsone hasTo prove it, let $(u_i{)}_{i=1}^m$ be a $K$-basis of $E$ and $(v_j{)}_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as ${\sum }_j{a}_j{v}_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_i{v}_j$ span $L$ over $K$.

Ifthen independence of the $v_j$ over $E$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_i{v}_j)$ is a $K$-basis of $L$, proving the formula.

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.

True. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

For a ring homomorphism $A\to B$, the Module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.

If $\alpha \ne 0$ has degree $d$, then at every Archimedean embeddingIndeed, the factor of the defining product belonging to that Archimedean place shows $\max (1,|\alpha |{)}^{1/d}\le H(\alpha )$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to ${\alpha }^{-1}$ and use $H({\alpha }^{-1})=H(\alpha )$ to obtain the lower bound. This is the Liouville height inequality.

If $L,L^{\prime }$ are field extensions of $K$, choose a $K$-basis of $L$ containing $1$. After tensoring with $L^{\prime }$, that basis expresses $L{\otimes }_K{L}^{\prime }$ as a nonzero direct sum of copies of $L^{\prime }$, and $1\otimes 1$ is nonzero. A prime ideal of this nonzero ring supplies a point used in the point-lifting property of a scheme fibre product.

A transcendence basis of a field extension $L/k$ is an algebraically independent set $T$ such that $L/k(T)$ is algebraic. Its cardinality is the transcendence degree. For finitely generated extensions it is finite, and the remaining algebraic extension is finite.

The transcendence degree is the cardinality of a transcendence basis. It counts independent transcendental parameters. In a finitely generated separably generated field extension, it equals the dimension of the Kähler differentials as a vector space over the extension field.