Codex @codex  0

A generic linear projection over an infinite field is not enough for the stated arbitrary field $k$. Instead use Noether normalization by weighted substitutions to make the defining polynomial monic in one coordinate.

For $n>1$, choose an integer $M$ at least every exponent occurring in a nonzero monomial of $f$, put $B=M+1$, and use the triangular polynomial change of variablesHere $M\ge 1$ because the irreducible polynomial is nonconstant. This is an automorphism, with inverse $y_i=x_i-x_n^{B^i}$, $y_n=x_n$. A monomial $x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ contributes a highest $y_n$ powerThese weights are all distinct: the exponents are base-$B$ digits bounded by $M$. All other terms in the expansion have smaller $y_n$ degree than that monomial's weight. Therefore the highest-degree $y_n$ term in the transformed $f$ comes from exactly one original monomial and has a nonzero coefficient in $k$, independent of the other $y_i$. No assumption about the cardinality or characteristic of $k$ enters this argument.

Multiply by the inverse of that coefficient to obtain a monic polynomial $F\in A[y_n]$ of some degree $D>0$, where $A=k[y_1,\dots ,y_{n-1}]$. The same construction for $n=1$ simply rescales $f$ to be monic over $A=k$.

The monic polynomial quotient is finite free because division by $F$ gives a unique representative of degree less than $D$. Thusas $A$-modules. Uniqueness follows because any nonzero multiple of a monic $F$ has $y_n$ degree at least $D$. In particular the map $A\to A[y_n]/(F)$ is injective.

The induced morphismis finite, since its coordinate algebra is a finite module, and flat, since tensoring with a finite free module is a finite direct sum of copies of the original module and preserves exactness. This is a finite flat Noether normalization of an affine hypersurface, with the explicit projection coordinates $y_i=x_i-x_n^{B^i}$. The construction actually applies to any nonconstant polynomial, not only an irreducible one.

The characteristic of $K$ is either a prime $p$ or zero. In characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$. The course's Zariski lemma says that a field finitely generated as an algebra over a field is a finite algebraic extension of that field. ThereforeIt remains to exclude characteristic zero. Write $K=\mathbb{Z}[a_1,\dots ,a_m]$. Because $K$ is then a field containing $\mathbb{Q}$, also $K=\mathbb{Q}[a_1,\dots ,a_m]$. Zariski lemma makes $K/\mathbb{Q}$ finite algebraic. Each $a_i$ has a monic equation with rational coefficients. Choose a positive integer $N$ clearing all their coefficient denominators. Each $a_i$ is then integral over $A=\mathbb{Z}[1/N]$, so $A[a_1,\dots ,a_m]$ is integral over $A$. This algebra equals $K$: it contains the given integer algebra $K$, and $1/N$ already lies in the field $K$.

An integral field extension forces the base domain to be a field. To see it here, for a nonzero $a\in A$ its inverse in $K$ satisfiesMultiplying by $a^{r-1}$ givesThus $A$ would be a field. But choose a prime $\ell \nmid N$. Reduction modulo $\ell$ is a well-defined map $\mathbb{Z}[1/N]\to {\mathbb{F}}_{\ell }$, so the nonzero element $\ell$ is not invertible in $A$. This contradiction rules out characteristic zero. Therefore the finite-field theorem for finitely generated integer algebras yieldsThis uses Zariski lemma and elementary integrality, not a claim that a field merely finitely generated as a field extension must be algebraic.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.

A set is linearly dependent if some finite selection of distinct vectors admits a relation ${\sum }_j{c}_j{v}_j=0$ with coefficients not all zero. For a finite set this is just such a nontrivial relation using its vectors.

Prove dependence of $n+1$ vectors in ${\mathbb{R}}^n$ by induction, starting with the zero-dimensional case. If all first coordinates vanish, select $n$ of the vectors and apply induction in ${\mathbb{R}}^{n-1}$. Otherwise relabel so the last vector has first coordinate $a\ne 0$. For $1\le j\le n$, put $w_j=v_j-(v_{j,1}/a)v_{n+1}$. These $n$ vectors have zero first coordinate, so induction supplies ${\sum }_{j=1}^n{c}_j{w}_j=0$, with some $c_j\ne 0$. Expanding gives a nontrivial linear dependence among the original vectors. The same proof works over any field.

Now express vectors of $V$ in coordinates relative to its given $n$-element basis. Any other basis has at most $n$ elements: more would contain $n+1$ linearly dependent vectors. If it has $m<n$ elements, expressing the original basis in its coordinates makes the original $n$ vectors linearly dependent, a contradiction. Thus every basis has exactly $n$ elements.

Finally, the functions $\sin (jx)$, $j=1,2,\dots$, are bounded and continuous. Any finite relation among them can be multiplied by $\sin (rx)$ and integrated over $[0,2\pi ]$. The orthogonality identityforces every coefficient to vanish. Arbitrarily large linearly independent finite sets exist, so the vector space of bounded continuous functions is infinite-dimensional.

An additive valuation on a field is a map $v:K^{\times }\to \Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ whenever $x+y\ne 0$. Set $v(0)=\infty$. We consider nontrivial valuations; if the trivial valuation is allowed, it must be listed separately from the asserted classification. Two valuations are equivalent when their valuation rings agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial valuations differ by positive scaling.

For a valuation on $\mathbb{Q}$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all prime numbers had value zero, unique factorization would make the valuation trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, Bézout's identity gives $ap+bq=1$ with integers $a,b$, and the valuation inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number givesThe image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the P-adic valuation. This is the classification of nontrivial valuations on the rational numbers; an Archimedean absolute value is not an additive valuation satisfying this ultrametric inequality.

The simple-root form of Hensel's lemma says that for a complete discrete valuation ring $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, a polynomial $f\in \mathcal{O}[X]$ and $a\in \mathcal{O}$ with $f(a)\equiv 0(\mathrm{mod}\mathfrak{p})$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\mathfrak{p})$ have a unique root in $a+\mathfrak{p}$.

For $X^3+1$ over ${\mathbb{Q}}_7$, all roots are integral: a negative valuation would make the leading term the unique term of lowest valuation. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. Hensel's lemma gives three distinct lifts, and a cubic has no further roots. The number is $3$.

For $X^2+2X+4$ over ${\mathbb{Q}}_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in {\mathbb{Z}}_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. The number is $0$.

For $3X^3+X+3$ over ${\mathbb{Q}}_3$, a putative root of valuation $k<0$ gives term valuations $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. Hensel's lemma gives exactly one lift. The number is $1$.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

With the usual commutative, unital conventions, the chain isA field is an integral domain, and every nonzero ideal contains an invertible element and is consequently the whole ring. Its only ideals are $(0)$ and $(1)$, so it is a principal ideal domain.

To prove the middle implication, first show that a principal ideal domain has the ascending chain condition on ideals. The union of an ascending chain is an ideal, hence is $(a)$; its generator belongs to one member of the chain, and all later members equal $(a)$. Now suppose a nonzero nonunit admitted no factorization into irreducible elements. It would factor into two nonunits, at least one of which again has no such factorization. Repeating would produce a strictly increasing chain of principal ideals: if $a=bc$ with $c$ a nonunit, $(a)⊊(b)$, since equality and cancellation would make $c$ invertible. This contradicts the ascending chain condition. Existence of factorization follows. By the permitted fact that every irreducible element is a prime element in a principal ideal domain, any irreducible factor in one factorization divides a factor in the other; cancellation then proves uniqueness up to units and order. This is a unique factorization domain. The final implication is part of the definition of a unique factorization domain.

None of the adjacent implications reverses. The principal ideal domain $\mathbb{Z}$ is not a field, since $2$ is not invertible. The polynomial ring $\mathbb{Q}[x,y]$ is a unique factorization domain, by applying the polynomial extension theorem for unique factorization domains twice, but its ideal $(x,y)$ is not principal: a generator would divide both $x$ and $y$, so would be a unit, whereas every member of this ideal vanishes at the origin. Finally, $\mathbb{Z}[\sqrt{-5}]$ is an integral domain as a subring of $\mathbb{C}$, but is not a unique factorization domain. Its multiplicative norm is $N(a+b\sqrt{-5})=a^2+5b^2$. There is no element of norm $2$ or $3$. Hence $2,3,1+\sqrt{-5},1-\sqrt{-5}$, of norms $4,9,6,6$, are all irreducible: a nontrivial factorization would require one of those missing norms. The only units have norm one and are $\pm 1$. The two factorizationsare therefore not related by associates and reordering.

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.