5
$\begingroup$

Let $k$ be a valued field. Is there a special term for a commutative (Banach) $k$-algebra $A$ such that for any maximal ideal $m$ we have $A/m=k$? Is there an easy to check criterion that would imply this property?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ I don't understand the question. Actually, I think it's difficult to understand it as it is currently phrased. $\endgroup$
    – Fernando Muro
    Commented Apr 3, 2012 at 13:49
  • 1
    $\begingroup$ To give an example for the 2nd question: In case $k = \mathbb R$, the Banach algebra $E := C_b(X)$ of bounded continuous real-valued functions over a completely regular space X has this property. If $m$ is a max. ideal, then one defines a total ordering on $E/m$ that extends the order of $\mathbb R \le E/m$ and has no infinitely large elements (that's because the functions are bounded). Now a general theorem on ordered fields implies $E/m = \mathbb R$. $\endgroup$
    – Ralph
    Commented Apr 3, 2012 at 21:53
  • 1
    $\begingroup$ @Fernando: Please tell me which part of the question is unclear or needs more explanation and I will edit it. \\ @Ralph: Another example is the algebra of continuous functions on a compact space. So, the actual question is a bit broad - how one can check (algebraically) if an algebra is the algebra of continuous k-valued functions on some "good" space? $\endgroup$
    – Anton Lyubinin
    Commented Apr 5, 2012 at 2:13
  • 1
    $\begingroup$ One way to construct such an algebra is to start with any algebra and then invert every $x$ in the algebra and every polynomial $f$ with no roots in $k$, invert $f(x)$. It is easy to see that this kills all the non-$k$ points and keeps all the $k$-points. I don't think there is any reasonable classification of such spaces. $\endgroup$
    – Will Sawin
    Commented Apr 29, 2015 at 2:52

1 Answer 1

Reset to default
3
$\begingroup$

I don't think that there is a standard term for what you are looking for, but I would be inclined to call such an algebra "compact". My reason is the following. Suppose that $X$ is a manifold (resp. locally compact Hausdorff space). Then the maximal spectrum of the ring $C^\infty(X)$ of smooth real-valued functions (resp. $C^0(X)$ of continuous real-valued functions) with the Zariski topology is homeomorphic to the Stone--Cech compactification $\beta X$ of $X$. If $X$ is not astronomically large*, then the points $\mathfrak m \in \beta X$ for which the quotient field $C^\infty(X)/\mathfrak m$ (resp. $C^0(X)/\mathfrak m$) is isomorphic to $\mathbb R$ are precisely the points in $X \hookrightarrow \beta X$. In particular, $X$ is compact iff $\beta X = X$ iff every maximal ideal comes from a point of $X$ iff every maximal ideal has quotient field $\mathbb R$.

I said this over $\mathbb R$, but really any infinite field $\mathbb K$ will do. (Then I need to take "isomorphic" to mean "isomorphic as $\mathbb K$-algebras", so that if something is isomorphic to $\mathbb K$, then it is so uniquely. Over $\mathbb R$, I can in fact take "isomorphic" to mean "isomorphic as rings", since $\mathbb R$ has no nontrivial ring automorphisms.)

*Footnote: The claim fails if there is a measurable cardinal $\lambda > |\mathbb K|$ with $|X| \geq \lambda$. Note that in many universes there are no uncountable measurable cardinals at all, and if uncountable measurable cardinals exist, then they are strongly inaccessible, and hence very very large.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .