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This is a list of short stand-alone results, (counter) examples, and tricks that I have picked up here and there.August 5, 2023 (algebra)

Suppose we have a ring \(A\) with a finitely generated module \(M\), and sections \(m_1,...,m_n\in M\) whose images span the fiber \(M_p\otimes \kappa(p)\) at some prime ideal \(p\subset A\). The usual form of the Nakayama lemma asserts that the sections \(m_1,...,m_n\) also generate the stalk \(M_p\) as an \(A_p\)-module.

With little work this stalk-local version of the lemma can be upgraded to an affine-local form, i.e. we show there is an \(a\in A\setminus p\) such that the sections \(m_1,...,m_n\) generate the localisation \(M_a\) as an \(A_a\)-module. Indeed write \(M'=M/\langle m_1,...,m_n \rangle\) for the cokernel of the surjection \(A^n\to M\) determined by the sections, and note that the stalk \(M'_p\) vanishes by stalk-local Nakayama lemma. Thus \(p\) is not in the support of \(M'\), and since finitely generated modules have closed support in \(\text{Spec}\,A\), we can find a distinguished affine neighbourhood of \(p\) (defined by some element \(a\in A\setminus p\)) contained outside \(\text{Supp}\,M'\). For all points \(q\) in this neighbourhood, \(M'_q=0\) so that \(M_q\) is generated by \(m_1,...,m_n\). It follows that \(M_a\) is generated by \(m_1,...,m_n\) by examining the corresponding map \(A_a^n\to M_a\).

March 3, 2022 (algebraic geometry)

We know that the affine spectrum of a local ring is a *connected doubleton*, i.e. topologically it is the set \(\{p,q\}\) with open sets \(\{\emptyset, \{p\}, \{p,q\}\}\). Now consider the topological space with underlying set \(\{p,q,r\}\) with open sets \(\{\emptyset, \{p\}, \{p,q\}, \{p,q,r\}\}\). Is there a way to realise this as an affine scheme?

If we think about \(\mathbb{C}[X,Y]_{(X,Y)}\), this has the right number of primes at heights \(0\) and \(2\) but way too many at height \(1\), so we need to remove those. However, we can’t do this with the usual operations of localization and quotients– something more exotic is needed. The trick is to consider the subring \[R:=\{f(X,Y)\in \mathbb{C}[X,Y^\pm]\;|\; f(0,Y)\in \mathbb{C}[X,Y]\}\] of \(\mathbb{C}[X,Y^\pm].\) In other words, we allow inverting \(Y\) but only if it appears with an \(X\).

The ideal \(\mathfrak{m}=(X,Y)\subset R\) is maximal, and \(0\) is prime. Suppose we have \(0\subsetneq \mathfrak{p}\subsetneq \mathfrak{m}\) for some prime ideal \(\mathfrak{p}\), then it must have terms with no positive power of \(Y\) (by multiplying with \(XY^{-n}\) for \(n\) sufficiently large). Choose \(f\in \mathfrak{p}\) such that \(f\in \mathbb{C}[X,Y^{-1}]\) and \(\text{deg}_Xf\) is minimal among these. Since \(f\in (X,Y)\cap \mathbb{C}[X,Y^{-1}]\), we must have \(f=X\cdot g\) for some \(g\in \mathbb{C}[X,Y^{-1}]\). But then \(\text{deg}_Xg<\text{deg}_Xf\) hence \(g\notin \mathfrak{p}\), so \(X\in \mathfrak{p}\) i.e. \(\mathfrak{p}=(X)\). Thus the localisation \(R_\mathfrak{m}\) has the required spectrum, with exactly one prime at heights \(0\),\(1\) and \(2\) each.

There is a way to state and extend this example using valuation rings, on totally ordered groups other than \(\mathbb{Z}\)– see Hahn series rings. The moral to me is that there are more exotic rings than I know of!

September 12, 2021 (algebra)

*The image of an ideal is an ideal of the image*.

If \(\phi: A\rightarrow B\) is a ring homomorphism and \(J\subset A\) is an ideal then \(\phi(A)\) is a subring of \(B\) and \(\phi(J)\) is an ideal of this subring: for any \(a,b\in J\) and \(t\in A\), we have \(\phi(a)+\phi(t)\cdot \phi(b)=\phi(a+tb)\in \phi(J)\). In particular, surjectivity is sufficient for the image of an ideal to be an ideal.

May 5, 2021 (analysis)

Let \((E,\mathscr{E}, \mu)\) be a measure space. If \(A,B\in \mathscr{E}\) are such that \(\mu(A)+\mu(B)> \mu(E)\) then \(A\cap B \neq \emptyset\). This is essentially a measure-theoretic consequence of De-Morgan’s theorem (which for two sets \(A\) and \(B\) says \(|A\cap B| = |A| + |B| - |A\cup B|\).)

This has a neat application. Suppose \(A\subset [0,1]\) is measurable with \(\mu(A)= \frac{1+\alpha}{2}\). Then for any \(\alpha'<\alpha\), the set \(A' =(\alpha' + A)\cap [0,1]\) has measure \(\geq \mu(A)-\alpha'\), hence \(\mu(A)+\mu(A')>1\). It follows that there is an \(x\) with \(\{x, x-\alpha'\}\in A\). Conclude that \((-\alpha, \alpha)\subset A-A\).

March 17, 2021 (logic)

This is a neat way to think about recursive languages (i.e. the languages recognised by a Deterministic Finite Automaton) that I learnt from Tom Forster. Suppose you are sitting in a cabin in the middle of the forest, and are being sent an infinite stream of symbols. Every time you see a new symbol, you are supposed to flip a switch saying whether or not the string received *so far* is in the language \(L\).

Tired, you wish to take a break and let your assistant take over. How much information about the string you have seen so far do you need to convey to the assistant, so that they can continue operating the switch? The least amount of information you could get away with passing on gives the number of states in the smallest DFA which recognises \(L\).

March 16, 2021 (algebra)

Watch out for analytical tricks to deduce irreducibility of a polynomial:

Suppose \(a_1,...,a_n\) are integers and \(p\) is a prime such that \(\sum_{i=1}^n|a_i|< p\). Then the polynomial \(a_nX^n+...+ a_1X+p\) is irreducible over \(\mathbb{Z}\): if not, the polynomial can be written as \(f\cdot g\) for non-constant polynomials \(f,g\). Then \(f(0)\cdot g(0)= p\), so wlog \(f(0)=\pm 1\). This implies the product of the (complex) roots of \(f\) is \(1\), in particular there is a root \(\alpha\) with \(|\alpha|\leq 1\). But then \(|p| = |a_n\alpha^n+...+a_1\alpha|\leq \sum_{i=1}^n |\alpha_i|\), a contradiction.