A Glimpse of Commutative Algebra and Algebraic Geometry

Solving a system of polynomial equations and studying an ideal in a polynomial ring are the same activity viewed from two sides. The equations $f_1=\cdots =f_m=0$ carve out a set of points; the polynomials $f_1,\dots ,f_m$ generate an ideal. Commutative algebra studies that correspondence; algebraic geometry is its geometric side. The dictionary between them rests on the ring theory of ideals and quotient rings and on the Gröbner-basis methods that make the computations effective. Throughout, $R$ denotes a commutative ring with $1\ne 0$, and $k$ denotes a field.

Noetherian rings

The polynomial rings $k[x_1,\dots ,x_n]$ are the ambient rings of algebraic geometry, and their central finiteness property is a condition on ascending chains of ideals.

This is the ring-theoretic special case of the module condition from the structure theorem over a PID, applied to $R$ as a module over itself, where the submodules are the ideals themselves.¹ Three conditions turn out to be equivalent, and each is used in practice.

The equivalence with finite generation is what connects A.C.C. to geometry: a Noetherian ring has no ideal that needs infinitely many generators. The class of Noetherian rings is broad and closed under the constructions that build polynomial rings.

• Every PID is Noetherian, since every ideal is generated by one element. In particular $\mathbb{Z}$, $k[x]$, and the Gaussian integers $\mathbb{Z}[i]$ are Noetherian; this recovers the Euclidean-domain hierarchy from the finiteness side.
• Quotients stay Noetherian. Any homomorphic image $R/I$ of a Noetherian ring is Noetherian, because a chain in $R/I$ pulls back to a chain in $R$.
• Polynomial rings stay Noetherian, by the next theorem.

Applied $n$ times, the Basis Theorem shows that $k[x_1,\dots ,x_n]$ is Noetherian, so every ideal of a polynomial ring over a field is finitely generated.² Not every ring is Noetherian. The polynomial ring $\mathbb{Z}[x_1,x_2,\dots ]$ in infinitely many variables fails A.C.C.: the ideal $(x_1,x_2,\dots )$ needs infinitely many generators. The ring of continuous real functions on $[0,1]$ also fails it. A Noetherian ring may still have infinite descending chains — in $\mathbb{Z}$,

so A.C.C. and the descending chain condition are genuinely different; rings with the descending condition are called Artinian and form a separate, smaller class.

Finitely generated algebras

A ring $R$ containing a field $k$ in its center is a $k$-algebra. It is finitely generated as a $k$-algebra if $R$ is generated as a ring by $k$ together with finitely many elements $r_1,\dots ,r_n$. This is equivalent to having a surjective $k$-algebra homomorphism

so every finitely generated $k$-algebra is a quotient of a polynomial ring, hence Noetherian.² These are the coordinate rings that the geometry attaches to its spaces.

Affine algebraic sets

Fix a field $k$ and let ${\mathbb{A}}^n$ denote affine $n$-space, the set of $n$-tuples of elements of $k$. Each polynomial $f\in k[x_1,\dots ,x_n]$ is a $k$-valued function on ${\mathbb{A}}^n$ by evaluation, and a set $S$ of polynomials picks out the points where all of them vanish.

Since $Z(S)=Z(I)$ where $I=(S)$ is the generated ideal, only ideals matter, and because $k[x_1,\dots ,x_n]$ is Noetherian every ideal is generated by finitely many $f_1,\dots ,f_q$. Consequently

so every affine algebraic set is the intersection of finitely many hypersurfaces. A purely geometric statement — a solution set needs only finitely many equations — falls out of an algebraic one, the Basis Theorem.³

Examples fix the picture:

• Points and finite sets. The single point $(a_1,\dots ,a_n)$ is $Z(x_1-a_1,\dots ,x_n-a_n)$, and any finite set is algebraic.
• Linear sets. Lines, planes, and coordinate subspaces are loci of degree-one polynomials; the $x$-axis in ${\mathbb{A}}^3$ is $Z(y,z)$.
• Hypersurfaces. The parabola $Z(y-x^2)$, the circle $Z(x^2+y^2-1)$, and the hyperbola $Z(xy-1)$ are algebraic sets in the plane.

The two maps

The correspondence runs in both directions. Going one way, an ideal determines its zero locus. Going the other, a set of points determines the ideal of everything that vanishes on it.

The maps $Z$ and $I$ are both inclusion-reversing (contravariant), and they satisfy

They are mutually inverse once restricted to the algebraic sets $V=Z(I)$ on one side and the ideals of the form $I(V)$ on the other.³ Determining exactly which ideals arise as $I(V)$ is the content of the Nullstellensatz below.

The coordinate ring

The functions on ${\mathbb{A}}^n$ that restrict to functions on a subset $V$ are the polynomials modulo those that vanish on $V$.

Two polynomials define the same function on $V$ exactly when their difference lies in $I(V)$, so $k[V]$ is genuinely the ring of polynomial functions on $V$. It is a finitely generated $k$-algebra, generated by the restricted coordinate functions. Maps between algebraic sets translate perfectly into maps between coordinate rings: a morphism $\phi :V\to W$ (given by polynomials in each coordinate) induces a $k$-algebra homomorphism ${\phi }^{\ast }:k[W]\to k[V]$ by $f\mapsto f\circ \phi$, and every $k$-algebra homomorphism arises this way from a unique morphism.³ The correspondence is contravariant, and it is an isomorphism of algebraic sets exactly when the induced map of coordinate rings is an isomorphism. Gröbner bases make this computational: the reduced remainder after division by a Gröbner basis for $I(V)$ gives a canonical representative for each coset in $k[V]$, which is how one computes in coordinate rings, tests membership in an ideal, and finds kernels and images of $k$-algebra maps in practice.³

Radicals

An algebraic set does not determine its defining ideal uniquely: the zeros of $f$ are the zeros of $f^2$, so $Z(f)=Z(f^k)$ for all $k\ge 1$, and $(x),(x^2),(x^3),\dots$ all cut out the same $y$-axis. The redundancy is captured by the radical.

The radical is again an ideal containing $I$, and $(\mathrm{rad}I)/I$ is the nilradical of $R/I$; so $R/I$ has no nonzero nilpotents exactly when $I$ is radical.⁴

• Radical as an intersection of primes. For a proper ideal $I$, $\mathrm{rad}I$ equals the intersection of all prime ideals containing $I$. In particular the nilradical is the intersection of all primes of $R$, and prime (hence maximal) ideals are radical.
• In a UFD, radicals kill exponents. If $a=p_1^{e_1}\cdots p_r^{e_r}$ is the factorization into distinct primes, then $\mathrm{rad}(a)=(p_1\cdots p_r)$ is the squarefree part. For instance $\mathrm{rad}(180)=(30)$ in $\mathbb{Z}$.

Because $k$ has no nilpotents, the coordinate ring $k[V]=k[{\mathbb{A}}^n]/I(V)$ has none either, so $I(V)$ is always a radical ideal.⁴ Over an arbitrary field the converse fails: $(x^2+1)$ is maximal, hence radical, in $\mathbb{R}[x]$, but $x^2+1$ has no real zeros, so it is not $I(V)$ for any $V$. The obstruction comes down to polynomials with no roots, and it disappears over an algebraically closed field.

The Zariski topology and varieties

The algebraic sets are closed under finite union and arbitrary intersection, and they include $\varnothing =Z(1)$ and ${\mathbb{A}}^n=Z(0)$. Those are precisely the axioms for the closed sets of a topology.

The topology is coarse. On ${\mathbb{A}}^1$ the only closed sets are the finite sets, the empty set, and the whole line, so any two nonempty open sets over an infinite field meet. Points are closed, but distinct points cannot be separated by disjoint opens, so the Zariski topology is not Hausdorff. Over $\mathbb{R}$ it is strictly coarser than the Euclidean topology.⁵ Within it, the natural building blocks are the sets that do not split.

Irreducibility is a ring-theoretic condition on the ideal: $V=Z(I)$ is irreducible if and only if $I$ is a prime ideal, equivalently the coordinate ring $k[V]$ is an integral domain.⁵ Since $k[x_1,\dots ,x_n]$ is Noetherian, every algebraic set decomposes into finitely many irreducible pieces.

The decomposition mirrors the algebra exactly: over an algebraically closed field a radical ideal is a finite intersection of prime ideals, and the union of the corresponding varieties is the decomposition of the algebraic set into irreducible components.⁵

Hilbert's Nullstellensatz

Over an algebraically closed field the correspondence closes into a clean bijection. The theorem comes in a weak form, about points and maximal ideals, and a strong form, about arbitrary ideals and their radicals.

The last clause is the origin of the name — Nullstellensatz, zero-locus theorem: any proper system of polynomial equations over an algebraically closed field has a common solution. The proof runs through Noether's Normalization Lemma, which presents any finitely generated $k$-algebra as a finite (integral) extension of a polynomial subalgebra.⁶ An element $s$ of a ring extension $S\supseteq R$ is integral over $R$ if it satisfies a monic polynomial with coefficients in $R$; the integral elements form a subring, and integrality is transitive. This is the ring-theoretic sibling of an algebraic field extension, and it supplies the finiteness that drives the maximal-ideal statement.

Passing to the zero set and back to the ideal recovers the radical of $I$ rather than $I$ itself, the largest ideal with the same zero set. Over $\mathbb{C}$, then, the geometrically defined ideals $I(V)$ are precisely the radical ideals, a condition that is purely algebraic.⁶ A computational corollary makes radical membership decidable: for a proper ideal $I=(f_1,\dots ,f_s)$ in $k[x_1,\dots ,x_n]$,

the Rabinowitsch trick, which reduces a radical-membership question to whether an ideal in one extra variable is the whole ring — a Gröbner-basis computation.⁶

The dictionary

Everything above collapses into a single translation table. Over an algebraically closed field, geometry on the left becomes commutative algebra on the right.

The table is the reason commutative algebra and algebraic geometry are studied as one subject. A question about the geometry of solution sets becomes a question about ideals and quotient rings, and the Nullstellensatz guarantees that nothing is lost in translation over an algebraically closed field. Dropping algebraically closed leads to arithmetic geometry and, through the integral extensions above, to the ring of integers of a number field and algebraic number theory. Allowing the base to vary leads to schemes, where the prime spectrum $\mathrm{Spec}R$ replaces ${\mathbb{A}}^n$ and every commutative ring becomes a geometric space. The Chinese Remainder Theorem reappears as the statement that a disjoint union of points corresponds to a product of coordinate rings, and Gröbner bases make all of it computable. The representation and character theory capstone applies the same modules-over-a-ring viewpoint to the group ring in place of the polynomial ring.