0. What is a discrete valuation on a field. $v:K^{\times }\to \mathbb{Z}$ such that $v(a+b)\ge \min (v(a),v(b))$ and $v(ab)=v(a)+v(b)$.

1. This can be extented to whole of $K$ by defining $v(0)=\infty$. We follow the first definition in this lecture.

2. Example: P-adic valuation on Q v_p

3. The ring $A:=\{x\in K\mid v(x)\ge 0\}$ is called the valuation ring of $K$.

4. What is an absolute value or multiplicative valuation on a field. Archimedean, Non archimedean.

5. Example: P-adic abs val on Q.

6. Why is it called non archimedean?

7. Show examples as in the notes. Define p-adic absolute value on $\mathbb{Q}$.

8. Can go from a valuation to an absolute value and vice versa. $v(x)=-\log (|x|)$

9. Relation between absolute value and additive valuations: A discrete (additive) valuation $\mathrm{ord}:K^{\times }\to \mathbb{Z}$ determines an absolute value by $|x|=e^{-\mathrm{ord}(x)}$ An absolute value on $K$ determines an additive valuation (need not be discrete) on $K^{\times }$ by ${\log }_e|x|=-\mathrm{ord}(x)$

10. Lemma:

    Absolute value is non archimedean iff it takes bounded values on $\{m\cdot 1|m\in \mathbb{Z}\}$

    1. Corollary: If $\mathrm{char}K\ne 0$ then $K$ has only non archimedean abs values.

11. Say that the absolute value is discrete if the image of $K^{\times }$ is a discrete subgroup of ${\mathbb{R}}_{>0}$.

12. Let $v$ be a discrete valuation on $K$, then $$ A := { a \in K : v(a) \geq 0 }

is a PID with the unique maximal ideal $$ \mathfrak{m} := \{ a \in K : v(a) > 0 \}

Proof:

Take an ideal $I$ of $A$, $I$ does not contain any element with valuation 0, since they are units of A. So, take the element $a$ with least valuation, this generates $I$. any element with equal valuation as that of $a$, is an associate of $a$. Now if $v(b)=qv(a)+r$ with $0<r<b$, we get that $v(b{a}^{-q})=r>0$ hence $b{a}^{-q}\in A$, but has smaller valuation than $a$. Contradiction. So this is a PID.

Easy to see that $\mathfrak{m}$ is the unique maximal ideal.

Proposition 7.6 : | | discrete iff m is principal, which makes A a DVR (PID + local)

Proof:

If $\mid \cdot \mid$ is discrete, then take $\pi \in \mathfrak{m}$ such that $\mid \pi \mid$ is maximum. Then take any ${\pi }_1\in \mathfrak{m}$ and let $\mid \pi {\mid }^{q+1}<\mid {\pi }_1\mid <\mid \pi {\mid }^q$. and so, $|\pi |<\mid \frac{{\pi }_1}{{\pi }^q}\mid <1$ and so, $\frac{{\pi }_1}{{\pi }^q}\in \mathfrak{m}$ and has larger absolute value than $\pi$. Contradiction. So, $|{\pi }_1|=|\pi {|}^q$, this gives ${\pi }_1={\pi }^q\cdot u$ where $u$ is a unit in $A$, hence ${\pi }_1\in (\pi )⟹\mathfrak{m}\subset (\pi )⟹\mathfrak{m}=(\pi )$.

If $\mathfrak{m}=(\pi )$ then take any $\alpha \in K^{\times }$, let $|\pi {|}^{q+1}<|\alpha |<|\pi {|}^q$, then $\mid \frac{\alpha }{{\pi }^q}\mid <1⟹\frac{\alpha }{{\pi }^q}\in (\pi )⟹\alpha ={\pi }^{q+1}\beta$ where $\beta \in A$. This gives $|\alpha |\le |\pi {|}^{q+1}$ This is a contradiction, hence $|\alpha |=|\pi {|}^q$ for some $q\in \mathbb{Z}$. Hence $\mid K^{\times }\mid =(|\pi |)$.

Take an ideal $I$ of $A$, $I$ does not contain any element with abs value 1, since they are units of A. So, take the element $a$ with greatest abs val, this generates $I$. Why? Because any element with equal abs val as that of $a$, is an associate of $a$. Now if $|a{|}^{q+1}<|b|<|a{|}^q$ we get that $|b{a}^{-q}|<1$ hence $b{a}^{-q}\in A$, but has greater abs val than $a$. Contradiction.

So this is a PID.

9. Abs value induce metric induce topology
10. Abs value equiv if they induce same topology
11. Q with p adic abs value, what is it? what does closeness look like here? 1,p,p^2,… converges to 0
12. Equivalent abs values
13. Ostrowski
14. Completion of Q under p- adic abs value
15. Completions in non archimedean case
16. |K| = |$\hat{K}$|
17. $\hat{A}$ is the closure of A, $\hat{\mathfrak{m}}$ is the closure of $\mathfrak{m}$, ${\hat{\mathfrak{m}}}^n$ is the closure of ${\mathfrak{m}}^n$.
18. $\hat{\mathfrak{m}}=(\pi )=\{a\in \hat{K}\mid |a|\le |\pi |\}$
19. For every $n$, $\frac{A}{{\mathfrak{m}}^n}\to \frac{\hat{A}}{{\hat{\mathfrak{m}}}^n}$ Is an isomorphism.

Proof:

The function is $f(a+{\mathfrak{m}}^n)=a+{\hat{\mathfrak{m}}}^n$. Note that ${\mathfrak{m}}^n=\{a\in A\mid \mid a\mid \le \mid \pi {\mid }^n\}=\{a\in A\mid \mid a\mid <\mid \pi {\mid }^{n-1}\}$ is open and closed as a subset of $A$. Now $\ker (f)=\{a+{\mathfrak{m}}^n\mid a\in A,a\in {\hat{\mathfrak{m}}}^n\}$ Since $\mathfrak{m}$ is dense in $\hat{\mathfrak{m}}$, ${\mathfrak{m}}^n$ is dense in ${\hat{\mathfrak{m}}}^n$ (${\mathfrak{m}}^n=({\pi }^n)\subset A$, and ${\hat{\mathfrak{m}}}^n=({\pi }^n)\subset \hat{A}$). Since $a\in A$ and $a$ is a limit point of ${\mathfrak{m}}^n$, it must be in ${\mathfrak{m}}^n$ since it is closed in $A$. This gives $\ker (f)=\{0\}$.

Now, let $\hat{a}\in \hat{A}$, ${\lim }_{n\to \infty }{a}_n=\hat{a}$, This gives $a_k\in \hat{a}+{\hat{\mathfrak{m}}}^n$ for all large $k$. (Since ${\hat{\mathfrak{m}}}^n$ is an open set around 0). Hence, $f(a_k+{\mathfrak{m}}^n)=a_k+{\hat{\mathfrak{m}}}^n=\hat{a}+{\hat{\mathfrak{m}}}^n$. Hence, $f$ is surjective

20. Choose a set $S$ of representatives of $\frac{A}{\mathfrak{m}}$, and let $\pi$ generate $\mathfrak{m}$. Then each element of $\hat{K}$ is expressible as a cauchy series of the form $$ a_{-n}\pi^{-n} + \dots + a_{0} + a_{1}\pi + a_{2}\pi^{2}\dots, \ a_{i}\in S

and such a representation is unique. 21. Use this to show how elements of $\mathbb{Q}_{p}$ look like 22. Let $K = \mathbb{Q}$, then $\hat{K} = \mathbb{Q}_{p}$, $A = \left\{ \frac{a}{b} \in \mathbb{Q} \mid p \nmid b, (a,b) = 1 \right\}$, $\mathfrak{m} = \left\{ \frac{a}{b} \in \mathbb{Q} \mid p |a; (a,b) = 1 \right\}$ 23. Then the representatives for $A/\mathfrak{m}$ are just $0,1,\dots p-1$ 24. In other words, $A / \mathfrak{m} \cong \mathbb{Z}/p\mathbb{Z}$. 25. $A = \mathbb{Z}_{(p)}$, $\mathbb{Z}_{p} := \hat{A}$. So, $\mathbb{Z}_{p}$ is the elements of $\mathbb{Q}_{p}$ with series expansion starting from non negative integers. 27. Given a element of $\prod \limits_{ n=1}^{ \infty } \mathbb{Z} /p ^{n} \mathbb{Z}$, with $a_{n+1} = a_{n}(\mathrm{mod} \ p ^{n})$, can change it into a series representation. 28. Conversely, If then $\alpha = \sum\limits_{ i=0}^{ \infty }c_{i}p ^{i}$, $0 \le c < p-1$, then $c_{0},c_{1},\dots$ is the unique sequence of integers such that $\alpha \equiv \sum\limits_{ i=0}^{ n-1 }c_{i}p ^{i}\ \mathrm{mod} \ p^n$ - So basically, elements of $\mathbb{Z}_{p}$ have "compatibility" among the coefficients, and any truncation of the series gives mod p^n of that number 31. Any element $\alpha \in \mathbb{Q}_{p}$, can multiply it by high power of $p$ and bring it in $\mathbb{Z}_{p}$. hence, coefficients of $\mathbb{Q}_{p}$ elements also have compatibility. 32. Hensel's lemma 33. Nakayama lemma for local rings: Let $A$ be a local ring and $\mathfrak{a}$ is a proper ideal of $A$. Let $M$ be a f.g. module over $A$, then $\mathfrak{a}M = M$ implies $M = 0$.