Study Plan, tentatively, + Algebraic Geometry Exercises

So I think it’s probably best to have a rotating study plan schedule that allows me to do certain topics on certain days. So far, I’m thinking of having a rotation that looks something like:

Differential Geometry -> Algebra -> Clifford Stuff -> Algebraic Topology (optional),

and since yesterday was (unofficially) differential geometry day, I’m going to spend today doing algebra.

First order of business: Eisenbud and Harris. And, since I’ve been meaning to write down some of the solutions to exercises I’ve passed, I guess I’ll do that here.

I-1. Find $\text{Spec} R$ when $R$ is:

1. $\mathbb{Z}$
2. $\mathbb{Z}/(3)$
3. $\mathbb{Z}/(6)$
4. $\mathbb{Z}_{(3)}$
5. $\mathbb{C}[x]$
6. $\mathbb{C}[x]/(x^2)$.

Proof. Recall that the elements of the set $\text{Spec}R$ are the prime ideals of $R$ and that the 0 ideal is included (I’ll not write 0 hereon except when absolutely necessary). Therefore, for part 1, $\text{Spec}R$ is the set of all $\mathbb{Z}/(p)$ where $p$ is prime. For (2), note that $\mathbb{Z}/(3)$ is a field, that every field is a PID, and that in a PID, maximal ideals and prime ideals are equivalent; moreover, note that in $\mathbb{Z}/(n)$, $(d)=R$ for all integers $d$ relatively prime to $n$, and because all integers are relatively prime to the prime number 3, $\text{Spec}R = 0$.

On the other hand, $\mathbb{Z}/(6)$ isn’t a field – or even an integral domain – and so the possible elements of $\text{Spec}R$ are of the form $(d)$ for $d = 2,3,4$ (0 is already mentioned above, and $(1)=(5)=\mathbb{Z}/(6)$ due to 1,5 being relatively prime to 6). Note that $(4)$ isn’t prime since $4=2\cdot 2$ for $2\not\in(4)$. Simple computation verifies that $(2),(3)$ are prime in $\mathbb{Z}/(6)$, however, so $\text{Spec}R=\{(2),(3)\}$. Because $\mathbb{Z}_{(3)}\cong\mathbb{Z}/(3)$, part 4 is equivalent to part 2.

For part 5, $\mathbb{C}$ being a field implies that $\mathbb{C}[x]$ is a PID, so it suffices to find all maximal ideals there, and because $K[x]/(f(x))$ is a field (for $K[x]$ a PID) precisely when $f(x)$ is $K[x]$-irreducible, it follows that

$\text{Spec}R=\{(f(x))\,:\,f(x)\text{ is irreducible in }\mathbb{C}[x]\}$.

Of course, $\mathbb{C}$ is algebraically closed, whereby it follows that the collection of such ideal-generating polynomials is precisely the linear ones, namely those of the form $f(x)=x+\alpha$, $\alpha\in\mathbb{C}$. Finally, note that $\mathbb{C}[x]/(x^2)$ consists of elements of the form $ax+b$ for $a,b\in\mathbb{C}$. Ignoring units, this ring $R$ therefore consists precisely of the linear polynomials $x+\alpha$, $\alpha\in\mathbb{C}$, and because of part 5, it follows that $R=\text{Spec}R$.  $\square$

I-2. What is the value of the “function” 15 at the point $(7)\in\text{Spec }\mathbb{Z}$? At the point $(5)$?
Proof. The notation here is as follows. Elements $f\in R$ are considered as functions on the space $\text{Spec} R$ whose valuations are computed as: If $x=[\mathfrak{p}]\in\text{Spec}R$, the quotient field $\kappa(x)=\kappa(\mathfrak{p})$ of the integral domain $R/\mathfrak{p}$ consists of values $f(x)$ defined to be the image of $f$ via the canonical maps $R\to R/\mathfrak{p}\to\kappa(x)$. So then, because the quotient field of $R_1=\mathbb{Z}/(7)$ is $R_1$ itself (since $R_1$ is a field), the value of the “function” 15 in $R_1$ is simply $15\mod 7 = 1$. Similarly, $R_2=\mathbb{Z}/(7)$ is again a field, whereby the value of 15 there is given by $15\mod 5 = 0$.  $\square$

I-3. (a) Consider the ring of polynomials $\mathbb{C}[x]$, and let $p(x)$ be a polynomial. Show that if $\alpha\in\mathbb{C}$ is a number, then $(x-\alpha)$ is a prime of $\mathbb{C}[x]$, and there is a natural identification of $\kappa((x-\alpha))$ with $\mathbb{C}$ such that the value of $p(x)$ at the point $(x-\alpha)\in\text{Spec }\mathbb{C}[x]$ is the number $p(\alpha)$. (b) Show that this holds in the general case as well.
Proof. As demonstrated above, the fact that $\mathbb{C}[x]$ is a PID and that $\mathbb{C}[x]/(x-\alpha)$ is a field for all $\alpha\in\mathbb{C}$ shows that $(x-\alpha)$ is prime. Moreover, because the image of $p(x)$ in $\kappa$ is defined to be the image of $p$ in the quotient field of $\mathbb{C}[x]/(x-\alpha)$ – that is, $\mathbb{C}[x]/(x-\alpha)$ itself, since $(x-\alpha)$ is maximal – the identification given is derived equivalently to the derivation of the fact that $\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$ by way of the correspondence $a+bi\in\mathbb{C}\sim ax+b\in\mathbb{R}/(x^2+1)$.

In the general case, $R$ is the coordinate ring of an affine variety $V$ over an algebraically closed field $K$ and $\mathfrak{p}$ is the maximal ideal corresponding to a point $x\in V$ in the usual sense, whereby the conclusion is that $\kappa(x)=K$ and $f(x)$ is the value of $f$ at $x$ in the usual sense. This result follows from a construction identical to that in case (a) due to the fact that $\mathfrak{p}$ being maximal implies that $V/\mathfrak{p}$ is a field that’s therefore equal to $K$. Hence, because $x\in V\sim\mathfrak{p}\in\text{Spec}R$, $\kappa(x)$ corresponds to $K$ and $f(x)\in\kappa(x)$ corresponds to $f(x)\in K$.  $\square$