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I'm trying to understand what DVRs are. I have seen two formulations, one in terms group homomorphisms and a discrete valuation function satisfying an axiom of the sort:

$$ v(x+y) \geq \text{min} \left( v(x) , v(y) \right)$$

and another one saying it is a ring generated by one element and is a maximal ideal.

Neither of these definition help me understand what sort of thing the DVRs are. Could someone please explain through a concrete intuitive example what DVRs are, why they are important and how these two different types of definitions fit together?

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A problem with learning about DVRs in algebra is that such rings are completely unlike the usual rings you see in (beginning) algebra courses. They are not like the integers or like polynomials. The intuition for them comes from geometry or analysis. A DVR should be thought as the functions on a one-dimensional space (a "curve") that make sense at a specific point. The functions that make sense at different points lead to different DVRs. The maximal ideal in the DVR is the functions that vanish at the point. The reason DVRs are important is for geometric and analytic reasons more than for algebraic reasons. If you want to understand a space, then one way to do it is through understanding the functions on that space, and the functions that are defined at a point on a curve is what a DVR is trying to model.

Example 1. On the Riemann sphere, its field of meromorphic functions is the field of rational functions $\mathbf C(z)$. The rational functions that are defined at a point on the Riemann sphere form a DVR. For example, at $z = 0$, the rational functions that are defined there are those with denominator (in reduced form) not vanishing at $0$, meaning the denominator is not divisible by $z$. Such rational functions, except for $0$, have the form $z^ka(z)/b(z)$ where $k \geq 0$ and $a(z)$ and $b(z)$ are polynomials nonvanishing at $0$ (not divisible by $z$). Such functions (including $0$) form a ring $\mathcal O_0$, with its units being those functions where $k = 0$ and its unique maximal ideal $\mathfrak m$ being those functions with $k \geq 1$ (they vanish at the point $0$) together with the zero function: this ideal is principal with generator $z$. Every nonzero rational function $f(z)$ can be written as $z^ka(z)/b(z)$ where $k \in \mathbf Z$ and $a(z)$ and $b(z)$ are polynomials that don't vanish at $0$: just factor out the highest power of $z$ from the numerator and denominator and that gives you $z^k$. The exponent $k$ will be negative if $f(z)$ has a pole at $0$. The exponent $k$ is the valuation of $z^ka(z)/b(z)$. It tells us the order of vanishing of $f$ at $0$.

Example 2. A variant on the first example is the ring $H_0$ of holomorphic functions at the origin in $\mathbf C$. (Don't think of $\mathbf C$ as a two-dimensional real plane, but as a one-dimensional "complex line"). Each element of $H_0$ is a function holomorphic on some neighborhood of $0$ (having a positive radius of convergence), where the radius is allowed to vary with the function. These functions form a ring and its units are the holomorphic functions $f(z)$ on a disc around $0$ where $f(0) \not= 0$: the power series at $0$ has a nonzero constant term. The nonunits of $H_0$ have power series at $0$ with constant term $0$, and such a function is divisible by $z$. So the nonunits form an ideal $zH_0$ and this is a maximal ideal. It's the only maximal ideal since all elements of $H_0 - zH_0$ are units in $H_0$. Each function in $H_0$ except for $0$ has a first nonzero term in its power series, and factoring that out lets use write the function as $z^kg(z)$ where $k \geq 0$ and $g \in H_0$ with $g(0) \not= 0$. The ring $H_0$ is an integral domain with fraction field $M_0$ being the meromorphic functions on a disc around $0$ (the disc varies with the function). Every nonzero meromorphic function $f(z)$ on a disc around $0$ can be written as $z^kg(z)$ where $k \in \mathbf Z$ and $g(z)$ is holomorphic and nonvanishing at $0$: just factor out the highest power of $z$ from the Laurent expansion of $f$ at $0$ and that gives you $z^k$. The exponent $k$ will be negative if $f$ has a pole at $0$. The exponent $k$ is the valuation of $f$ at $0$. It tells us the order of vanishing of $f$ at $0$ (positive for a zero at $0$, zero for finite and nonvanishing holomorphic functions at $0$, and negative for meromorphic functions with a pole at $0$).

Example 3. Think of rational numbers as "functions" on the primes, where the value of a rational number $r$ at the prime $p$ is $r \bmod p$: reduce the numerator and denominator (when $r$ is in reduced form) and as long as the denominator is not divisible by $p$, $r \bmod p$ makes sense. For example, at the prime (point) $5$, we have $7/3 \equiv 2/3 \equiv (2)(2) \equiv 4 \bmod 5$ and $10/3\equiv 0 \bmod 5$. The rational numbers that makes sense mod $5$, except for $0$, have the form $5^ka/b$ where $k \geq 0$ and $a$ and $b$ are integers not divisible by $5$. Such fractions (including $0$) form a ring $\mathbf Z_{(5)}$, with its units being those fractions where $k = 0$ and its unique maximal ideal $\mathfrak m$ being those fractions with $k \geq 1$ (they vanish mod $5$) together with the number $0$: this ideal is principal with generator $5$. Every nonzero rational number $r$ can be written as $5^ka/b$ where $k \in \mathbf Z$ and $a$ and $b$ are integers that are nonzero mod $5$ (they don't vanish at $5$): just factor out the highest power of $5$ from the numerator and denominator and that gives you $5^k$. The exponent $k$ will be negative if $r$ has $5$ in its denominator in reduced form (it blows up mod $5$). The exponent $k$ is the valuation of $5^ka/b$. It tells us the "order of vanishing" of $r$ at $5$.

Complete DVRs are like power series rings at a point on a curve, such as $F[[x]]$ for a field $F$ (formal power series at $0$) or the $p$-adic integers (not a power series ring, but having many properties reminiscent of such rings except lacking a "field" of coefficients). The DVR $H_0$ in Example 2 is almost the completion of the DVR $\mathcal O_0$ in Example 1, except the completion at $\mathcal O_0$ is actually bigger: it is the ring $\mathbf C[[z]]$ of all formal power series at $0$

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  • $\begingroup$ Woah the second last paragraph felt like I opened the third eye or something $\endgroup$ Jun 26 at 1:09
  • $\begingroup$ What's the idea behind restricting the study of functions defined on curves $\endgroup$ Jun 26 at 12:30
  • $\begingroup$ Do you mean why curves? On higher-dimensional spaces, there is not a single order-of-vanishing at a point. On an $n$-dimensional space with $n>1$, the functions vanishing at a point will be an ideal with $n$ generators and not one generator. Look at the ring of smooth functions near a point $(a_1,\ldots,a_n)$ in $\mathbf R^n$: the functions vanishing there form an ideal with generators $x_1-a_1,\ldots,x_n-a_n$. It is not a principal ideal when $n \geq 2$. $\endgroup$
    – KCd
    Jun 26 at 12:39
  • $\begingroup$ I don't understand the last message, could you explain in a simpler way? $\endgroup$ Jun 26 at 13:05
  • $\begingroup$ Please start telling us something about your own background. Have you studied differential geometry or algebraic geometry? $\endgroup$
    – KCd
    Jun 26 at 13:06

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