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Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

Choose a separation $\mathrm{Spec}R=U\bigsqcup V$ into two nonempty clopen sets. Both are closed in the Zariski topology, so write $U=V(I)$ and $V=V(J)$ for ideals $I,J$. Their disjointness gives $V(I+J)=\varnothing$. A proper ideal lies in a maximal ideal, so this forces $I+J=R$.

Their union is all of the spectrum of a commutative ring, and $V(I)\cup V(J)=V(IJ)$. Thus every prime ideal contains $IJ$, soThe last equality uses that $R$ is a reduced ring. For comaximal ideals, $I\cap J=IJ$: if $i+j=1$ and $r\in I\cap J$, then $r=ri+rj\in IJ$. Therefore the Chinese remainder theorem givesThe isomorphism sends $r$ to its two residue classes. Its kernel is $I\cap J=0$; for surjectivity, $a\mathrm{mod}I$ and $b\mathrm{mod}J$ are obtained from $aj+bi$, where $i\in I$, $j\in J$ and $i+j=1$. The two factors are nonzero because $U,V$ are nonempty, so $I,J$ are proper.

This is a disconnected reduced spectrum product decomposition. The elements $i,j$ also satisfy $ij=0$, $i^2=i$, $j^2=j$, exhibiting the ring product decomposition by an idempotent. No Noetherian hypothesis is required.

Factor $X^3+X=X(X-i)(X+i)$. These factors generate pairwise comaximal ideals, so the Chinese remainder theorem givesThe three copies of $\mathbb{C}$ have different module structures: $X$ acts by $0,i,-i$, respectively. Thus $q(X)$ acts on $(a,b,c)$ as $(q(0)a,q(i)b,q(-i)c)$, making $\Phi$ a module homomorphism.

The inverse, obtained by Lagrange interpolation, isEvaluating at $0,i,-i$ recovers $(a,b,c)$. Conversely, a polynomial whose three evaluations vanish is divisible by $X(X-i)(X+i)$, so these maps are mutual inverses on the quotient module. Three copies with identical trivial $X$-action would not give the required isomorphism.

Let $R$ be a Euclidean domain and $I$ a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing the Euclidean size. For any $a\in I$, Euclidean division gives $a=qd+r$, where either $r=0$ or $r$ has smaller size than $d$. But $r=a-qd\in I$, so minimality forces $r=0$. Consequently $I=(d)$. The zero ideal is $(0)$, proving that every Euclidean domain is a principal ideal domain.

A commutative ring is Noetherian if every ideal is finitely generated; equivalently, every ascending chain of ideals stabilizes. To see the equivalence, the union of an ascending chain is an ideal. Finitely many generators for that union already belong to one member of the chain, after which the chain is constant. Conversely, an ideal which is not finitely generated permits successive choices outside the ideals generated by previous choices, producing a strictly ascending chain.

Every principal ideal domain is therefore Noetherian. The given Euclidean domain property of the Gaussian integers yields

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.