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In the dictionary in my head, I think of elements or $R$ as functions on $\mathop{\mathrm{Spec}} R$ -- i.e. as being analogous to scalar fields on manifolds. In my mind, this lessens the expectation that their properties should be directly interpretable in terms of subschemes of $\mathop{\mathrm{Spec}} R$.

That said, there is still correlation between the reducilibty of an element of $R$ and the reducibility of its zero set; the game just becomes cataloging the differences.

One of them is that we should be looking at the subscheme, not the subspace. For example, in the plane with coordinate ring $k[x,y]$, the origin $V(x,y)$ and the scheme $V(y, y-x^2)$ are different; the latter is the "double point" over $k$. The geometry remembers the difference between primary and prime ideals, even though the topology forgets.

That said, the real missing ingredient is the Picard group -- your reducibility criterion is not merely splitting $\mathop{\mathrm{Div}}(f)$ into a sum of nonzero effective divisors, but those divisors must also vanish into the Picard group.

Anyways, your analog of algebraic number theory for the circle is to consider the norm from $\mathbb{R}(x,y)$ down to $\mathbb{R}(x)$. If $(x,1-y)$ was principal with generator $f(x) + g(x) y$, then $$ \pm x = N(f(x) + g(x) y) = \frac{ (x^2 - 1) g(x)^2 + f(x)^2 }{g(x)^2} $$ From which you argue $g(x) = 1$, and $f(x)$ must be linear with leading coefficient $\mathbb{i}$, an impossibility.

Aside: I feel like there should be some sort of Galois descent argument to compare $\mathbb{C}[x,y]$ with $\mathbb{R}[x,y]$, but it's beyond my expertise to see how it should go.

user13113