Search Results

Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram. $$ \begin{array}{ccc} R&\stackrel{f}{\rightarrow}&S\\\\ \downarrow\scriptstyle{}&&\downarrow\scriptstyle{} …

Take any compact Hausdorff space $X$ and let $R$ consist of the locally constant functions $X\to k$. Then, $R$ is a $k$-algebra. Its prime ideals are all of the form $\mathfrak{p}=\left\{f\in R\colon …

No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, …

Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense. As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in …

To answer Question 1: Yes, there do exist integrally closed abstract number rings with infinite class group. By factorization of ideals, for $R$ to be an abstract number ring it is enough that it is …

I can show that this is true for your "simple" case. If g(x,y) ∈ C∞(ℝ2) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C∞(ℝ2). This …

I can get quite close to proving this. That doesn't mean that the result is true but it does at least seem to be very nearly true. We can also see what any counterexamples must look like if it does fa …