# Spec

In the introduction, we have explained the general idea of duality between spaces and rings of functions, and we have given two examples: one linear algebraic and the other topological. We present in what follows an algebro-geometric example. The functions considered will be algebraic. More precisely, the starting point will be to take as our ring of functions the ring of polynomials ${\displaystyle \mathbb {C} [x_{1},...,x_{n}]}$.

Algebraic geometry is concerned with the geometry of spaces whose functions are of this kind. Using different starting rings, it is possible to develop in a parallel way different theories, such as analytic geometry, starting from the ring ${\displaystyle \mathbb {C} \{x_{1},\ldots ,x_{n}\}}$ of power series converging on an open neighborhood of the origin, or formal geometry, starting from the ring ${\displaystyle \mathbb {C} [[x_{1},\ldots ,x_{n}]]}$ of all formal power series.

An affine (algebraic) variety over ${\displaystyle \mathbb {C} }$ is the vanishing locus in ${\displaystyle \mathbb {C} ^{n}}$ of finitely many polynomials ${\displaystyle p_{1},\ldots ,p_{k}\in \mathbb {C} }$:

${\displaystyle \{x\in \mathbb {C} ^{n}|p_{1}(x)=\ldots =p_{k}(x)=0\}.}$
Affine varieties define a notion of space in algebraic geometry.

We take as our notion of ring of functions the finitely generated unital commutative algebras over ${\displaystyle \mathbb {C} }$. Let ${\displaystyle R}$ be such an algebra. By hypothesis, there exists a finite number of generators ${\displaystyle x_{1},\ldots ,x_{n}}$ of ${\displaystyle R}$, i.e. ${\displaystyle R}$ is the quotient of the polynomial algebra ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]}$ by some ideal of relations. By the Hilbert basis theorem, i.e. the fact that the ring ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]}$ is Noetherian, the ideal of relations is itself finitely generated. Let ${\displaystyle p_{1}}$,...,${\displaystyle p_{k}\in \mathbb {C} }$ be some generators of the relations. We denote by ${\displaystyle (p_{1},...,p_{k})}$ the ideal generated by the ${\displaystyle p_{i}}$'s, i.e. the set of all elements of the form ${\displaystyle \sum _{i=1}^{k}f_{i}p_{i}}$, ${\displaystyle f_{i}\in \mathbb {C} }$, i.e. the ${\displaystyle \mathbb {C} }$-submodule of ${\displaystyle \mathbb {C} }$ generated by the ${\displaystyle p_{i}}$'s.

Then we have an isomorphism:

${\displaystyle R=\mathbb {C} /(p_{1},\ldots ,p_{k}).}$
We associate to such a ring ${\displaystyle R}$ the affine variety:
${\displaystyle X=\{x\in \mathbb {C} ^{n}\ |\ p_{1}(x)=\ldots =p_{k}(x)=0\}.}$
For example, we associate ${\displaystyle \mathbb {C} ^{n}}$ to the ring ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]}$. Each point ${\displaystyle x\in X}$ defines a ring homomorphism:
${\displaystyle {\begin{array}{ccccc}ev_{x}\colon &R&\longrightarrow &\mathbb {C} \\&f&\mapsto &f(x)\\\end{array}}}$
called the evaluation map at ${\displaystyle x}$. The spectrum of the ring ${\displaystyle R}$, denoted by ${\displaystyle \operatorname {Spec} \,R}$, is the set of all ring homomorphisms ${\displaystyle R\rightarrow \mathbb {C} }$ (since all our rings are in fact ${\displaystyle \mathbb {C} }$-algebras, ring homomorphism will always mean homomorphism of algebras). The evaluation map induces a map ${\displaystyle X\longrightarrow \operatorname {Spec} \,R}$ defined by ${\displaystyle x\mapsto ev_{x}}$. It is possible to show that this map is a bijection, i.e. every ring homorphism ${\displaystyle R\longrightarrow \mathbb {C} }$ is the evaluation ${\displaystyle ev_{x}}$ for some ${\displaystyle x\in X}$. In other words, we have
${\displaystyle \{p_{1}=\ldots =p_{k}=0{\text{ in }}\mathbb {C} ^{n}\}=\operatorname {Spec} \,\mathbb {C} /(p_{1},\ldots ,p_{k}).}$

Exercise: Check this is true by assuming that this is known for ${\displaystyle R=\mathbb {C} }$, i.e. that ${\displaystyle \operatorname {Spec} \,\mathbb {C} =\mathbb {C} ^{n}}$. The space ${\displaystyle \mathbb {C} ^{n}}$ seen as an affine variety is called the affine space of dimension ${\displaystyle n}$.

Conversely, starting from an affine variety ${\displaystyle X}$, it is possible to construct a ring of functions on ${\displaystyle X}$ by quotienting ${\displaystyle \mathbb {C} }$ by the ideal of functions vanishing on ${\displaystyle X}$. The resulting finitely generated unital algebra is called the coordinate ring of ${\displaystyle X}$ and is denoted by ${\displaystyle {\mathcal {O}}_{X}}$.

Example

We have an isomorphism ${\displaystyle \mathbb {C} [x,y]/(y)=\mathbb {C} }$. The ring ${\displaystyle \mathbb {C} /(y)}$ is the coordinate ring of the ${\displaystyle x}$-axis in ${\displaystyle \mathbb {C} ^{2}}$. The isomorphism with ${\displaystyle \mathbb {C} }$ corresponds to the obvious geometric fact that the ${\displaystyle x}$-axis in ${\displaystyle \mathbb {C} ^{2}}$ is a copy of ${\displaystyle \mathbb {C} }$.

Example

We have an isomorphism ${\displaystyle \mathbb {C} /(y-x^{2})=\mathbb {C} }$ given by ${\displaystyle y=x^{2}}$. The ring ${\displaystyle \mathbb {C} /(y-x^{2})}$ is the coordinate ring of the parabola ${\displaystyle y=x^{2}}$ in ${\displaystyle \mathbb {C} ^{2}}$. The isomorphism with ${\displaystyle \mathbb {C} }$ corresponds to the geometric fact that the projection of the parabola on the ${\displaystyle x}$-axis is an isomorphism of affine varieties.

Example

Let ${\displaystyle f\in \mathbb {C} }$ and ${\displaystyle R=\mathbb {C} /(1-yf(x))}$. The ring ${\displaystyle R}$ is the coordinate ring of the graph in ${\displaystyle \mathbb {C} ^{2}}$ of ${\displaystyle y=1/f(x)}$ for ${\displaystyle x}$ such that ${\displaystyle f(x)\neq 0}$. The projection on the ${\displaystyle x}$-axis gives an isomorphism between this affine variety and the ${\displaystyle x}$-axis minus the points where ${\displaystyle f=0}$. In other words, we have isomorphisms ${\displaystyle \operatorname {Spec} \,R=\mathbb {C} _{x}\setminus \{f=0\}}$ and ${\displaystyle R=\mathbb {C} }$.

The ring ${\displaystyle R=\mathbb {C} }$ is generally denoted by ${\displaystyle \mathbb {C} _{(f)}}$and is called the localization of ${\displaystyle \mathbb {C} }$ along the multiplicative set ${\displaystyle \{1,f,f^{2},\ldots \}}$ consisting of powers of ${\displaystyle f}$.

We have explained how to associate an affine variety to a finitely generated unital commutative algebra ${\displaystyle R}$, by taking its spectrum ${\displaystyle \operatorname {Spec} \,R}$, and how to associate a finitely generated unital commutative algebra to an affine variety ${\displaystyle X}$, by taking its ring of functions ${\displaystyle {\mathcal {O}}_{X}}$. But the constructions ${\displaystyle R\mapsto \operatorname {Spec} \,R}$ and ${\displaystyle X\mapsto {\mathcal {O}}_{X}}$ are not in general the inverse of each other.

Example

Let ${\displaystyle R_{1}=\mathbb {C} /(x)=\mathbb {C} }$ and ${\displaystyle R_{2}=\mathbb {C} /(x^{2})=\mathbb {C} \oplus \mathbb {C} x}$.

The equations ${\displaystyle x=0}$ and ${\displaystyle x^{2}=0}$ define the same subset ${\displaystyle X}$ of ${\displaystyle \mathbb {C} }$, so we have ${\displaystyle X=\operatorname {Spec} \,R_{1}=\operatorname {Spec} \,R_{2}}$, but ${\displaystyle {\mathcal {O}}_{X}=\mathbb {C} =R_{1}\neq R_{2}}$.

To solve this difficulty and have a better correspondence between spaces and rings of functions, there are two possible approaches:

1) Restrict the notion of ring. Rather than to consider rings of the form ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]/I}$ for a general ideal ${\displaystyle I}$ of relations, one can consider only rings of the form ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]/I}$ for an ideal which is radical, i.e. such that ${\displaystyle {\sqrt {I}}=I}$ where ${\displaystyle {\sqrt {I}}=\{f\in \mathbb {C} \ \vert \ f^{N}\in I{\text{ for some }}N>0\}}$. With this notion of rings, one obtains a one-to-one correspondence bewteen affine varieties and rings. This results from the Nullstellensatz which states that ${\displaystyle {\mathcal {O}}_{\operatorname {Spec} \,\mathbb {C} /I}=\mathbb {C} [x_{1},\ldots ,x_{n}]/{\sqrt {I}}}$.

2) Enlarge the notion of space. Rather than considering spaces as subsets of ${\displaystyle \mathbb {C} ^{n}}$, one could remember more information. For example, one would like to say that the space associated to the ring ${\displaystyle \mathbb {C} /(x^{2})}$ is different from the one associated to the ring ${\displaystyle \mathbb {C} /(x)}$, the former being a first order infinitesimal thickening of the latter. The notion of scheme enlarges the notion of space to take into account the possibility of infinitesimal thickenings. There is a one-to-one correspondence between affine schemes and rings of the form ${\displaystyle \mathbb {C} [x_{1},\ldots ,x_{n}]/I}$.

We will describe the first approach in more detail. Given an ideal ${\displaystyle I\leftarrow \mathbb {C} }$ we have the associated affine variety ${\displaystyle X=\mathbb {V} (I)}$. To an affine variety ${\displaystyle X}$ we have associated the ideal of polynomials which vanish on it ${\displaystyle \mathbb {I} (X)=\{f\in C[x_{1},\ldots ,x_{n}]|f(x)=0{\text{ for all }}x\in \}}$ and have defined its ring of functions to be ${\displaystyle {\mathcal {O}}_{X}=\mathbb {C} /\mathbb {I} (X)}$. It is straightforward to check that ${\displaystyle X=\mathbb {V} (\mathbb {I} (X))}$. But in general ${\displaystyle \mathbb {I} (\mathbb {V} (I))\neq I}$. For example, ${\displaystyle \mathbb {I} (\mathbb {V} (x^{2}))=(x)}$. One of the many forms of the Nullstellensatz states that for an affine variety ${\displaystyle X=\mathbb {V} (I)}$ we have ${\displaystyle \mathbb {I} (\mathbb {V} (I))={\sqrt {I}}}$. So the ring of functions on ${\displaystyle X={\text{Spec}}\;\mathbb {C} /I}$ is just ${\displaystyle \mathbb {C} /{\sqrt {I}}}$, that is ${\displaystyle {\mathcal {O}}_{\operatorname {Spec} \,\mathbb {C} /I}=\mathbb {C} [x_{1},\ldots ,x_{n}]/{\sqrt {I}}}$. This gives a bijection between radical ideals and affine varieties which in turn sets up a contravariant equivalence between the category of affine varieties and the category of finitely-generated reduced algebras (i.e. nilpotent elements, it is easy to see that ${\displaystyle C[x_{1},\ldots ,x_{n}]/I}$ is reduced if and only if ${\displaystyle I}$ is radical).

For instance, given a morphism of varieties ${\displaystyle F:X\rightarrow Y}$ we have an induced homomorphism on the rings of functions ${\displaystyle F^{*}:\mathbb {C} /J\rightarrow \mathbb {C} /I}$ given by ${\displaystyle y_{i}\mapsto y_{i}\circ F}$.

Going the other way, given a homomorphism between finitely-generated reduced ${\displaystyle \mathbb {C} }$-algebras, ${\displaystyle \phi :S\rightarrow R}$, we can choose representations

${\displaystyle R=\mathbb {C} /I,\,\,S=\mathbb {C} /J}$
and define an induced morphism between the associated affine varieties ${\displaystyle X=\mathbb {V} (I)}$ and ${\displaystyle Y=\mathbb {V} (J)}$ to be the restriction of the polynomial map ${\displaystyle \phi _{*}:\mathbb {C} ^{n}\rightarrow \mathbb {C} ^{m}}$, ${\displaystyle x\mapsto (f_{1}(x),\ldots ,f_{n}(x))}$ where ${\displaystyle f_{i}\in \mathbb {C} }$ represents ${\displaystyle \phi (y_{i})}$. This gives a well defined morphism when restricted to ${\displaystyle X}$. Then, if ${\displaystyle x\in X}$ and ${\displaystyle p\in J}$ we have ${\displaystyle p(\phi _{*}(x))=p(f_{1}(x),\ldots ,f_{n}(x))}$ but ${\displaystyle p(f_{1},\ldots ,f_{n})\equiv p(\phi (y_{1}),\ldots ,\phi (y_{m}))\equiv \phi (p)\equiv 0\mod {I}}$ so ${\displaystyle p(\phi _{*}(x))=0}$ and ${\displaystyle \phi _{*}(x)\in Y}$. It is straight forward to check that ${\displaystyle (F\circ G)^{*}=G^{*}\circ F^{*}}$ and ${\displaystyle (\phi \circ \psi )_{*}=\psi _{*}\circ \phi _{*}}$ so that isomorphisms in one category induce isomorphisms in the other and vice versa.

Example

Let ${\displaystyle X=\mathbb {C} }$, ${\displaystyle Y=\mathbb {V} (y-x^{2})\subset \mathbb {C} ^{2}}$. The morphism ${\displaystyle F:X\rightarrow Y}$, ${\displaystyle t\mapsto (t,t^{2})}$ has inverse morphism ${\displaystyle (x,y)\mapsto x}$. This corresponds to the isomorphisms of rings of functions ${\displaystyle \mathbb {C} /(y-x^{2})\rightarrow \mathbb {C} }$,

${\displaystyle x\mapsto x\circ F(t)=t,y\mapsto y\circ F(t)=t^{2}}$

Example
Let ${\displaystyle X=\mathbb {C} }$, ${\displaystyle Y=\mathbb {V} (y^{2}-x^{3})\subset \mathbb {C} ^{2}}$. The morphism ${\displaystyle F:X\rightarrow Y}$, ${\displaystyle t\mapsto (t^{2},t^{3})}$ induces the following homomorphism on the rings of functions ${\displaystyle \mathbb {C} /(y^{2}-x^{3})\rightarrow \mathbb {C} }$,
${\displaystyle x\mapsto x\circ F(t)=t^{2},\,\,y\mapsto y\circ F(t)=t^{3}.}$
This is clearly not onto and so this is not an isomorphism of varieties. (In fact, there is no surjective homomorphism ${\displaystyle \mathbb {C} /(y^{2}-x^{3})\rightarrow \mathbb {C} }$. Suppose ${\displaystyle x\mapsto p(t)}$ and ${\displaystyle y\mapsto q(t)}$ with ${\displaystyle f(p(t),q(t))=t}$ for some ${\displaystyle f\in \mathbb {C} }$. Then, differentiating, ${\displaystyle f_{x}(p,q)p'+f_{y}(p,q)q'=1}$ so ${\displaystyle p'}$ and ${\displaystyle q'}$ are coprime and in particular ${\displaystyle p}$ and ${\displaystyle q}$ cannot have any multiple zeros in common. But we also have ${\displaystyle p^{3}=q^{2}}$ so ${\displaystyle p}$ and ${\displaystyle q}$ must have the same roots but neither can have any single roots. This is a contradiction.)

There is a table of dualities:

{\displaystyle {\begin{aligned}\mathbb {C} ^{n}&\longleftrightarrow &\mathbb {C} \\X=\{x\in \mathbb {C} ^{n}|p_{1}(x)=\ldots =p_{k}(x)=0\}&\longleftrightarrow &R=\mathbb {C} /(p_{1},\ldots ,p_{k})\\&\longleftrightarrow &I=(p_{1},\ldots ,p_{k})\\X_{1}\cap X_{2}&\longleftrightarrow &I_{1}+I_{2}\\X_{1}\cup X_{2}&\longleftrightarrow &I_{1}\cap I_{2}\\X{\text{ irreducible}}&\longleftrightarrow &I{\text{ prime ideal}}\\{\text{Points of }}X&\longleftrightarrow &{\text{Maximal ideals of }}R\end{aligned}}}

There is an alternative description of ${\displaystyle \operatorname {Spec} \,R}$ as the set of maximal ideals of ${\displaystyle R}$; we will now show that this is equivalent to our original definition. Firstly, if ${\displaystyle {\mathfrak {m}}}$ is a maximal ideal of ${\displaystyle R}$ then we let ${\displaystyle f_{\mathfrak {m}}}$ be the quotient morphism ${\displaystyle R\rightarrow R/{\mathfrak {m}}}$. It is a basic fact (see the accompanying exercises) that ${\displaystyle R/{\mathfrak {m}}\cong \mathbb {C} }$, so we have a morphism ${\displaystyle f_{\mathfrak {m}}:R\rightarrow \mathbb {C} }$. In the reverse direction, given a ring homomorphism ${\displaystyle f:R\rightarrow \mathbb {C} }$, we let ${\displaystyle {\mathfrak {m_{f}}}=\ker {f}}$. Then ${\displaystyle m_{f}}$ is an ideal of ${\displaystyle R}$, and it is maximal because ${\displaystyle \operatorname {Im} =\mathbb {C} }$ (since ${\displaystyle f}$ is an algebra morphism and ${\displaystyle \mathbb {C} \hookrightarrow R}$), and so ${\displaystyle R/{\mathfrak {m_{f}}}}$ is a field. It is straightforward to check that these constructions are mutually inverse.

Exercise: Show that ${\displaystyle \operatorname {Spec} \,\mathbb {C} [x_{1},\ldots ,x_{n}]=\mathbb {C} ^{n}}$, i.e. that for every maximal ideal ${\displaystyle I}$ of ${\displaystyle \mathbb {C} }$ is of the form ${\displaystyle (x_{1}-a_{1},\ldots ,x_{n}-a_{n})}$ for some (unique) ${\displaystyle a\in \mathbb {C} ^{n}}$.

Let ${\displaystyle a\in \mathbb {C} ^{n}}$ and consider the evaluation map ${\displaystyle ev_{a}:\mathbb {C} \rightarrow \mathbb {C} }$ defined by ${\displaystyle ev_{a}(f)=f(a)}$. It is clear that the ideal ${\displaystyle (x_{1}-a_{1},\ldots ,x_{n}-a_{n})}$ is contained in the kernel of ${\displaystyle ev_{a}}$. Now, given ${\displaystyle g\in \mathbb {C} }$ we can write it in the form

${\displaystyle g(x_{1},\ldots ,x_{n})=g(a_{1},\ldots ,a_{n})+\sum c_{i}(x_{i}-a_{i})+\sum c_{ij}(x_{i}-a_{i})(x_{j}-a_{j})+\cdots }$
and see that if ${\displaystyle g(a)=0}$ then ${\displaystyle g\in {\text{Ker}}(ev_{a}).}$ Hence, ${\displaystyle {\text{kernel}}(ev_{a})=(x_{1},\ldots ,x_{n}).}$

Conversely, given a maximal ideal ${\displaystyle M\leftarrow \mathbb {C} }$ we know by exercise 1 that the quotient is isomorphic to ${\displaystyle \mathbb {C} }$. If ${\displaystyle a_{i}}$ is identified with ${\displaystyle x_{i}+M}$ under this isomorphism then ${\displaystyle (x_{1}-a_{1},\ldots ,x_{n}-a_{n})}$ is contained in the natural projection ${\displaystyle \mathbb {C} \rightarrow \mathbb {C} }$ since ${\displaystyle x_{i}-a_{i}\mapsto a_{i}-a_{i}=0}$. By maximality, we conclude that it must be equal to ${\displaystyle M}$.

Exercise: Using that ${\displaystyle \operatorname {Spec} \,\mathbb {C} [x_{1},\ldots ,x_{n}]=\mathbb {C} ^{n}}$, prove the Nullstellensatz, i.e. for ${\displaystyle I\leftarrow \mathbb {C} [x_{1},\ldots ,x_{n}]}$ we have ${\displaystyle \mathbb {I} (\mathbb {V} (I))={\sqrt {I}}.}$

Checking that ${\displaystyle {\sqrt {I}}\subset \mathbb {I} (\mathbb {V} (I))}$ is straightforward.

Let ${\displaystyle f\in \mathbb {I} (\mathbb {V} (I))}$. Now introduce a new variable ${\displaystyle t}$ and consider the ideal ${\displaystyle I'=(I,ft-1)\in \mathbb {C} }$. The corresponding variety ${\displaystyle \mathbb {V} (I')}$ must be empty. For if not, there exists some ${\displaystyle (a_{1},\ldots ,a_{n},b)\in \mathbb {C} ^{n+1}}$ such that ${\displaystyle f(a)b=1}$ and yet ${\displaystyle f(a)=0}$ since ${\displaystyle a\in \mathbb {V} (I)}$ and ${\displaystyle f\in \mathbb {I} (\mathbb {V} (I))}$. Now it must be the case that ${\displaystyle 1\in I'}$. Again, supose not, then ${\displaystyle I'}$ lies in some maximal ideal ${\displaystyle M\leftarrow \mathbb {C} }$ but then we would have ${\displaystyle \emptyset \neq \mathbb {V} (M)\subset \mathbb {V} (I')=\emptyset }$. So write

${\displaystyle 1=\sum _{i=1}^{n}c_{i}f_{i}+c_{0}(ft-1)}$
for some generators ${\displaystyle f_{i}}$ of ${\displaystyle I}$ and ${\displaystyle c_{i}\in \mathbb {C} }$. Formally substituting ${\displaystyle t=1/f}$ and then multiplying through by a sufficiently higher power of ${\displaystyle f}$ we get an expression of the form
${\displaystyle f^{N}=\sum _{i=1}^{n}c'_{i}f_{i}}$
and conclude that ${\displaystyle f\in {\sqrt {I}}.}$

Remark: What we call spectrum is what is generally called the maximal spectrum, since it is the set of maximal ideals. What is generally called the spectrum is the set of prime ideals. If we only consider varieties over ${\displaystyle \mathbb {C} }$, the maximal spectrum is enough.

Exercise: Why the name spectrum? Given ${\displaystyle A\in M_{n\times n}(\mathbb {C} )}$ consider the (commutative!) ${\displaystyle \mathbb {C} }$-algebra generated by ${\displaystyle A}$. What is its Spec?