Spectrum of a Ring

Approaching Scheme Theory as a Beginner

Thank you to James Tao for explaining the following, geometrically intuitive, interpretation to me.

A scheme is the result of patching affine schemes together, an affine scheme is any locally ringed space that is equivalent to Spec of a ring. To make an analogy, a manifold is the result of patching together copies of $\mathbb{R}^n$, in a way that preserves the smooth structure.

And the smooth structure of $\mathbb{R}^n$ is essentially the fact that it is a sheaf: over any open set, we have the space of smooth functions on that set, and these functions restrict and glue nicely, etc.

So when you go from $\mathbb{R}^n$ to “smooth manifolds,” you are creating spaces that locally look like $\mathbb{R}^n$, respect the same kind of smooth structure (as captured by the fact that smooth functions are defined on open sets in a way independent of the patching), yet have some global structure that might prevent them from just being the same as $\mathbb{R}^n$.

If you care about algebraic (polynomial / rational) functions defined on algebraic sets (zeroes of systems of polynomials), then you can do the same program! The “basic patch” is “affine scheme,” and you patch affine schemes together to get a scheme. Continue reading Spectrum of a Ring

Understanding the Lazard Ring

When I define a polynomial, I am simply handing you an indexed collection of coefficients.  

A polynomial with two variables, $x, y$ and coefficients $c$, is of the form:

$F(x, y) = \sum\limits_{ij} c_{ij} x^i y^j$

The coefficients of a polynomial form a ring. In other words, the coefficients $c_{ij}$ are members of a coefficient ring $R$. When we say $F$ is over $R$, we mean that $F$ has coefficients in $R$.

Screenshot from 2015-01-03 16:13:31

Example: The polynomial
$F(x,y) = 7 + 5xy^2 + 2x^3$ can be written out as
$F(x,y) =7x^0y^0 + 5x^1y^2 + 2x^3y^0$ such that
$c_{00} = 7$, $c_{12} = 5$, $c_{30} = 2$, and the rest of $c_{ij} = 0$.

Alright, now let’s change the coefficients; reassign $c_{00} = 4$, $c_{78} = 3$, and all other $c_{ij} = 0$.

Out pops a very different polynomial $P(x,y) = 4 + 3x^7y^8$.

In other words, By altering the coefficients $c_{ij}$ of $F(x,y)$ via a ring homomorphism $u: R \to R’$ (from the coefficent ring $c_{ij} \in R$ to a coefficient ring $u(c_{ij}) \in R’$)…

Screenshot from 2015-01-03 16:15:03

…we can get from $F(x,y)$ to any other polynomial $F'(x,y)$.

Screenshot from 2015-01-03 16:17:59

What’s a group-y polynomial?

Intuitively, a polynomial is “group-y” if there’s a constraint on our coefficients that forces the polynomial to satisfy the laws of a commutative group.

Concretely, a group-y polynomial is an operation of the form $F(x,y) = \sum\limits_{ij}c_{ij}x^iy^j$ such that

  1. $F(x,y) = F(y,x)$                     commutativity
  2. $F(x, 0) = x = F(0, x)$                 identity
  3. $F(F(x,y), z) – F(x, F(y,z)) = 0$    associativity

We can make sure that our polynomial satisfies these constraints! How? We mod out our coefficient ring $c_{ij}$ by the ideal $I$ — generated by the relations amoung $c_{ij}$ imposed by these constraints.

If you’d like to see the explicit relations, I wrote a cry for help post on stack overflow.

The ring of coefficients that results is called the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$.

It’s important to note here that group-y polynomials are morphisms out of the Lazard ring, not elements of the Lazard ring (i.e., that an assignment of values to each of the $c_{ij}$ describes a group-y polynomial, but the ring of the $c_{ij}$ itself is just a polynomial ring).

In other words, group-y polynomials $f(x,y)$ are morphisms out of the Lazard ring, not elements of the Lazard ring.

Screen Shot 2015-03-04 at 2.28.45 PMMore formally: for any ring $R$ with group-y polynomial $f(x,y) \in R[[x,y]]$ there is a unique morphism $L \to R$ that sends $\ell \mapsto f$.

$L \to R \simeq F_R$

(where $F_R$ denotes a group-y polynomial with coefficients in $R$)

This makes sense. If it doesn’t, then scroll up a bit! As we saw above, a change of base ring corresponds to a new group-y polynomial.

Grading the Lazard Ring

As we’ve noted, the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$is the quotient of a polynomial ring on the $c_{ij}$ by some relations.

Lazard proved that it is also a polynomial ring (no relations) on a different set of generators. More specifically, $\alpha$ is a graded ring isomorphism:

$\mathbb{Z}[c_{ij}]/I \xrightarrow{\alpha} \mathbb{Z}[t_1, t_2 …]$

(where the degree of $t_i$ is $2i$).

Lurie talks about this a bit (Theorem 4, Lecture 2: The Lazard Ring), but I have yet to understand the proof myself.

Thanks to Alex Mennen for deriving constraints the associativity condition puts on our coefficients; thanks to Qiaochu Yuan and Josh Grochow for kindly explaining some basic details and mechanics of the Lazard ring.

For your ventures ahead…

In this post, I have committed two semantic sins in the name of pedagogy. Namely, sins of oversimplification which I’ll attempt to rectify s.t. you aren’t hopelessly confused by the literature:

  1. group-y polynomial = “1-dimensional abelian formal group law”
  2. polynomial = “formal power series”

Conventionally, a “polynomial” is a special case of a formal power series (in which we expect that our variables evaluate to a number – useful if we care about convergence).

polynomials $\subset$ formal power series

The polynomial ring $R[x]$ is the ring of all polynomials (in two variables) over a given coefficient ring $R$.

The ring of formal power series $R[[x]]$ is the ring of all formal power series (in two variables) over a given coefficient ring $R$.

polynomial ring $\subset$ ring of formal power series
R[x] $\subset$ R[[x]]


Groupes de Lie formels à un paramètre
Groupes analytiques en caractéristique 0
Groupes de Lie algébriques (travaux de Chavelley)
Formal Group Laws – Ravenel