iiinova — where your digital self keeps evolving

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dan @danabra.mov · 01/08/2026

man i can finally read papers now. what a time to be alive

$Berarducci 2000 — "Factorization in generalized power series" §1 — The problem The ring is the one from before: series in t with real exponents ≤ 0, well-ordered exponent sets. Call it R. What are the "units" — the things that divide everything, like ±1 in ℤ? Just the plain constants. So 2 is a unit here, and a series is irreducible if you can't write it as a product of two non-constants. Conway's candidate is 1 + t^{-1} + t^{-1/2} + t^{-1/3} + ⋯. Gonshor's is t^{-√2} + t^{-1} + 1. Nobody could prove either was irreducible, or even that irreducibles existed at all. And there was real reason to doubt: Wilkie had pointed out a very similar ring — series with only finitely many terms — with no irreducibles whatsoever. Everything there factors forever. So the question is genuinely open in both directions. §2 — The plan Here's the whole idea in one line: find a number attached to each series that multiplies. If you can attach v(a) to each series so that v(bc) = v(b)·v(c), and you find an a whose v(a) can't be written as a product of two smaller things, then a can't be factored either. Done. The obvious candidate: the shape of the exponent set (its order type). And the obvious problem: shapes don't multiply. Take any b and its inverse — b · b⁻¹ = 1, shape 1, no matter how complicated b was. But that's cheating: b⁻¹ has exponents going up, which isn't allowed in our ring. Berarducci's bet is that once you restrict to exponents ≤ 0, shapes really do multiply. That bet is the paper. §3 — Fixing ordinal arithmetic Before anything, he needs the arithmetic of shapes to behave. Ordinary ordinal addition is lopsided: 1 + ω = ω but ω + 1 ≠ ω. Sticking one item in front of an infinite list changes nothing; sticking one on the end changes everything.$

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Rado Kirov @radokirov.bsky.social · 01/08/2026

Golden age for the math hobbyist. I am excited about it too.

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