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The following identity:

$\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots\right)=\left(1+(x+y)+\frac{(x+y)^2}{2!}+\frac{(x+y)^3}{3!}+\cdots\right)$

may be regarded as a "purely algebraic" identity in the ring of formal power series over $\mathbb{Q}$, without making any explicit references to the exponential function or the differential equation $f'=f$. Yet, it is by considering the latter two notions that the identity is most generally and most efficiently proved.

Having limited knowledge about analysis, topology and commutative algebra, I find most motivating examples of algebraic geometry (in its modern flavour) completely impenetrable. Is the above an example of the methods used in algebraic geometry: finding a "natural" geometric/topological setting for a purely algebraic question (in this case, the metric properties of $\mathbb{R}$ or $\mathbb{C}$ which allow us to naturally define a derivative) and exploiting the geometry/topology to answer the algebraic question? Is there another elementary example of such a technique?

I know a formal derivative (a purely algebraic operation) can be defined without speaking about limits and whatnot, but my question is not whether topology/analysis is necessary to answer a given algebraic question. Rather, it is about whether it is easier or more natural to do so. Historically, I'm sure, the formal derivative was defined after the limit-of-gradient derivative.

semisimpleton
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1 Answers1

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I don't think so, no.

I would say that this is an application of the permanence of identities. Here's the rough idea:

We want to prove a result about $\mathbb{Q}[[x,y]]$, namely that $\exp(x) \exp(y) = \exp(x+y)$, where we interpret $\exp$ as a formal power series.

Now, we embed $\mathbb{Q}[[x,y]] \hookrightarrow \mathbb{R}[[x,y]]$, and we see that $\exp(x)$ and $\exp(y)$ get sent to the subring $C^\omega \left (\mathbb{R}^2 \right ) \subseteq \mathbb{R}[[x,y]]$ of analytic functions. In summary, we get embeddings

$$ \require{AMScd} \begin{CD} \mathbb{Q}[\exp(x), \exp(y)] @>>> \mathbb{Q}[[x,y]]\\ @VVV \\ C^\omega \left (\mathbb{R}^2 \right) \end{CD} $$

Now since $\exp(x) \exp(y) = \exp(x+y)$ in $C^\omega \left ( \mathbb{R}^2 \right )$, as one can check via calculus, the same equation must hold in $\mathbb{Q}[\exp(x), \exp(y)]$ since the vertical arrow is an injection (if you like model theory, this is because embeddings reflect atomic truth). But since homomorphisms preserve atomic truth, once we know that $\exp(x) \exp(y) = \exp(x+y)$ in $\mathbb{Q}[\exp(x), \exp(y)]$, the same must be true of their image in $\mathbb{Q}[[x,y]]$ by moving along the horizontal arrow.

This is super convenient, since $\mathbb{Q}[[x,y]]$ has a certain universal property. So we get homomorphisms $\mathbb{Q}[[x,y]] \to R$ for many rings $R$, and since homomorphisms preserve atomic truth, this tells us that our equation must be true in any (commutative!) ring $R$ for which $\exp$ makes sense! See an old blog post of mine here for more information about this technique.

Now then, the obvious follow-up question is "if this isn't algebraic geometry, what is?". That's a much longer question, but thankfully there have been a number of good answers already posted to mse! See here for an overview and here for a wealth of information about getting started (obviously many of these introductory texts will also tell you what the subject is all about :P)

I hope this helps ^_^

HallaSurvivor
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