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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

14 votes
0 answers
549 views

Is the Zariski density proof of Cayley-Hamilton circular?

This old MO thread and its comments contains a discussion of the Zariski density proof of Cayley-Hamilton (I have also asked a separate question about the proof Victor gives in the comments here). Vic …
25 votes
7 answers
3k views

When can we prove constructively that a ring with unity has a maximal ideal?

Many commutative algebra textbooks establish that every ideal of a ring is contained in a maximal ideal by appealing to Zorn's lemma, which I dislike on grounds of non-constructivity. For Noetherian …
11 votes

Formal power series vs localization at non-constant polynomials

Here is a concrete example which may be useful to think about. It is not exactly the situation you describe but is similar. Let $k$ be a field of characteristic zero and let $R = k[x, y]/(y^2 - x^3 - …
Qiaochu Yuan's user avatar
36 votes

Do rings of smooth functions differ from rings of continuous functions?

Here is a different proof which maybe clarifies a different aspect of the situation. The ring $C(X)$ of continuous functions on a compact Hausdorff space, as an abstract ring, actually knows its $C^{\ …
Qiaochu Yuan's user avatar
16 votes

Does the rational power series ring $\mathbb{Q}[[X]]$ embed as a ring into the field of real...

Some commentary on David's very nice answer that might provide some useful context. If $D$ is a subring of $\mathbb{R}$ then it must be an integral domain and its field of fractions $\operatorname{Fra …
Qiaochu Yuan's user avatar
8 votes

How many solutions are there to the equation $x^2 + 3y^2 \equiv 1 \pmod{p}$?

Here is an alternate approach, more algebraic and less geometric. As in Noam's answer we'll consider the more general equation $x^2 - Dy^2 \equiv 1 \bmod p$. Consider the $\mathbb{F}_p$-algebra $A = \ …
Qiaochu Yuan's user avatar
17 votes
Accepted

The isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$

Writing $\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_I \mathbb{Q}$ where $I$ indexes a Hamel basis for $\mathbb{R}$ minus one element, we have $$\text{Ext}^1(\mathbb{R}/\mathbb{ …
Qiaochu Yuan's user avatar
15 votes
7 answers
4k views

Different definitions of the dimension of an algebra

I know of three ways to define the dimension of a finitely-generated commutative algebra A over a field F: The Gelfand-Kirillov (GK) dimension, based on the growth of the Hilbert function. The Krull …
7 votes
Accepted

Comultiplication on objects in an (abelian?) category

Sure, we can define such things. Let's work in the Morita 2-category $\text{Mor}(k)$ over a commutative ring $k$, which has objects $k$-algebras $A$, morphisms $k$-bimodules (with composition given …
Qiaochu Yuan's user avatar
8 votes

Is the multiplication between even numbers an associative algebra?

The set of even numbers is a non-unital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a non-unital algebra, cons …
Martin Sleziak's user avatar
25 votes
5 answers
2k views

Exotic principal ideal domains

Recently I realized that the only PIDs I know how to write down that aren't fields are $\mathbb{Z}, F[x]$ for $F$ a field, integral closures of these in finite extensions of their fraction fields that …
8 votes
Accepted

Classify commutative rings $R$ such that $A \otimes_{\Bbb Z} B = A \otimes _{R} B$

Fernando and Pierre-Yves in the comments are right; $R$ has this property (the version where the canonical map is an isomorphism, as YCor says in the comments) iff it is a solid ring, meaning the mult …
Qiaochu Yuan's user avatar
34 votes
2 answers
7k views

What is the geometric meaning of integral closure?

More precisely, how does one characterize integrally closed finitely generated domains (say, over C) based on geometric properties of their varieties? Given a finitely generated domain A and its inte …
6 votes
Accepted

Morita equivalence and isomorphisms in cohomology theories

The conceptual point is that all of these invariants are Morita invariant because they can be defined directly in terms of the category of modules. Explicitly: Starting from the category of modules …
Qiaochu Yuan's user avatar
50 votes
Accepted

Given a polynomial f, can there be more than one constant c such that every root of f(x)-c i...

This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero). We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and …
Neil Strickland's user avatar

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