Intro

I haven’t really studied any Math seriously since I graduated with my undergrad in 2015. It seems I kinda miss it, so I’ll be going over Introduction to Algebraic Geometry by Serge Lang today.

It starts out with the theory of Places. The idea being that homomorphisms between fields are necessarily trivial. Before we prove that, we define a field homomorphism as a function $\varphi:F\to K$ such that:

  1. $\forall a,b \in F, \varphi(a + b) = \varphi(a) + \varphi(b)$
  2. $\forall a,b \in F, \varphi(a \cdot b) = \varphi(a) \cdot \varphi(b)$
  3. $\varphi(0) = 0$, and $\varphi(1) = 1$

Axiom 3 only guarantees that the homomorphism doesn’t send every element to 0.

Field homomorphisms are injective

Computational proof

It’s been a long time since I’ve done any math, so let’s warm up a bit with a direct, computational proof.

Proof:

First we prove $\forall x \neq 0, \varphi(x) \neq 0$. To that end, let’s assume toward a contradiction that there exists an $x \neq 0$ such that $\varphi(x) = 0$. Then…

\[\begin{align*} 0 =& \varphi(x) \\\\ 0 \cdot \varphi(x^{-1}) =& \varphi(x) \cdot \varphi(x^{-1}) \\\\ 0 =& \varphi(x \cdot x^{-1}) \\\\ 0 =& \varphi(1) \\\\ 0 =& 1 \end{align*}\]

Next, inverses necessarily get sent to inverses, although you may have noticed that we’ve already proven it above.

\[\begin{align*} 1 =& \varphi(1) \\\\ =& \varphi(x \cdot x^{-1}) \\\\ =& \varphi(x) \cdot \varphi(x^{-1}) \end{align*}\]

Okay, now let’s prove it’s injective:

\[\begin{align*} \varphi(x_0) =& \varphi(x_1) \\\\ \varphi(x_0) \cdot \varphi(x_1)^{-1} =& \varphi(x_1) \cdot \varphi(x_1)^{-1} \\\\ \varphi(x_0) \cdot \varphi(x_1^{-1}) =& \varphi(x_1) \cdot \varphi(x_1^{-1}) \\\\ \varphi(x_0 \cdot x_1^{-1}) =& \varphi(x_1 \cdot x_1^{-1}) \\\\ 1 - \varphi(x_0 \cdot x_1^{-1}) =& 1 - \varphi(1) \\\\ \varphi(1 - x_0 \cdot x_1^{-1}) =& 0 \\\\ 1 - x_0 \cdot x_1^{-1} =& 0 \\\\ 1 =& x_0 \cdot x_1^{-1} \\\\ x_1 =& x_0 \end{align*}\] \[\blacksquare\]

A more mature approach

Okay, so now that our mathematical juices are starting to flow, let’s try a more mature mathematical approach… The proof is built upon the theory of rings and ideals. So, let’s define them.

A ring is a set $R$ along with a commutative operation $+$, a $0$ element, inverses for all non-zero elements. There is also a $\cdot$ operation where inverses are not necessarily defined.

An ideal $I \lt R$ is a subset of $R$ such that

  1. $\forall a, b \in I$, $a + b \in I$
  2. $\forall \lambda \in R, \forall a \in I$, $\lambda \cdot a \in I$

Proof:

The kernel of a ring homomorphism1 is an ideal2, and the only ideals of a field are $\{0\}$ and $F$ itself. Since $\varphi(1) = 1$, we know the kernel is not $F$, and hence $\varphi(x) = 0 \iff x = 0$.

Yay! We’re real mathematicians now! Let’s be happy and all drink.

Places

Mathematicians are really running out of names for things… But places are the topic of chapter one, so let’s do it…

Let $F$ be a field. We can extend $F$ to include the notion of $\infty$ by adding it to the field with the following definitions for the different operations. Let $a\in F_{\neq 0}$.

\[\begin{align*} a \pm \infty =& \infty \\\\ a \cdot \infty =& \infty \\\\ 1 / \infty =& 0 \\\\ 1 / 0 =& \infty \\\\ \infty \cdot \infty =&\infty \end{align*}\]

Note: $\infty \pm \infty$, $0\cdot \infty$, $\infty / \infty$, $0 / 0$ are undefined.

Let, $K,F$ be fields, then an $F$-valued place is mapping $\varphi:K\to F\cup\{\infty\}$ that satisfies the field homomorphism axioms (whenever defined).

Finite ring

The set, $\nu \subset K$, of elements that don’t get mapped to $\infty$ (called the finite elements of $\varphi$) form a ring. This can be seen because $+$ and $\cdot$ are extended to include $\infty$, but not overridden for finite elements.

The kernel is an ideal of $\nu$, and in fact, $\nu/\text{ker} \cong \varphi(\nu)$ is a field. This can be seen because if $\varphi(x)$ is finite, then $\varphi(x^{-1})$ is finite too (by axiom 2 of homomorphisms). It is also a consequence of the kernel being a maximal ideal of $\nu$3.

A valuation ring of $K$ is a ring $R\subseteq K$ such that $\forall a \in K, a\notin R \implies a^{-1}\in R$. $\nu$ is a valuation ring of $K$. It turns out, there is a 1-to-1 correspondence between valuation rings and places. To prove this, we only need to show that the non-units of $R$ form a maximal ideal, then we’ll be able to construct a map between $K$ and $(K/$ker$(\varphi))\cup\{\infty\}$ that makes this a place.

Valuation rings are places

Let $a, b\in R$ be non-units, and let $\lambda \in R$ be arbitrary.

  1. Addition: Since this is a valuation ring, if $ab^{-1}\notin R$, then $a^{-1}b\in R$. Without loss of generality, assume $ab^{-1}\in R$. Then, $1 + ab^{-1} = (a + b)b^{-1}\in R$. If $(a + b)^{-1}\in R$, then $b^{-1}\in R$. Since $b^{-1}\notin R$ ($b$ is a non-unit of $R$), $a+b$ is a non-unit of $R$.
  2. Multiplication: If $(\lambda a)^{-1}$ is in $R$, then necessarily $\lambda(\lambda a)^{-1}$ is as well, and hence, $a^{-1}$. But $a$ is not a unit in $R$, so $\lambda a$ is not either.

Summary

So valuation rings and places are somewhat interchangeable (at least up to equivalence of places4). And now we’re caught up to half-way down page 3.

  1. Ring homomorphisms follow the same axioms as field homomorphisms. ↩

  2. if $\varphi(a) = 0$ and $\varphi(b) = 0$, then axiom 1 says, $\varphi(a + b) = 0$, so axiom 1 of ideals is satisfied. Similar reasoning gives you axiom 2 of ideals. ↩

  3. A ring modded by a maximal ideal is a field. ↩

  4. Two places $\varphi_0:K\to F_0\cup\{\infty\}$ and $\varphi_1:K\to F_1\cup\{\infty\}$ are equivalent if there exists an isomorphism $\lambda:F_0 \to F_1$ such that $\varphi_1 = \lambda\varphi_0$. We define $\varphi(\infty) = \infty$. ↩