Closed rschwiebert closed 7 months ago
New description: Let $K$ be an infinite field, $S = K[\ldots, x_{-1}, x_0, x_1, \ldots]$, $I = ( { xk x{k + a} x_{k + 2a}: k \in \Bbb Z, a \in \Bbb Z \setminus {0} } )$ an ideal of $S$, and $\sigma$ the $K$-automorphism of $S$ such that $\sigma(xi) = x{i + 1}$ for all $i$. Then, $\sigma$ induces a $K$-automorphism of $R = S/I$. Ram's Ore extension ring $T$ is the skew polynomial ring $R[x; \sigma]$.
Would you like to specialize $K$ to say $\Bbb Q$, or any other one? Many properties don't depend on $K$ here. Marks and Lam (p167) don't impose the cardinality condition on $K$; Ram needed it to be infinite to have $R$ semisimple semiprimitive (for primeness of $T$ it is not necessary).
Suppose $e = \sum \alpha_i x^i \in T$ is an idempotent. Take the lowest $i$ with $\alpha_i \not = 0$. But then, if $i > 0$, each of the terms in the expansion of $e^2$ is either zero or is a multiple of $x^{2i}$. Thus all idempotents of $T$ lie in $R$. For similar reasons one deduces that 1) an idempotent in $R$ has constant term $0$ or $1$ (you can't get a number by multiplying terms that contain letters), and finally 2) $R$ is strongly connected.
The right zero divisors (=elements with nonzero left annihilator) of $T$ are exactly the elements with zero constant term (that contains neither $x$ nor $x_i$). If ${x_i: i \in I}$ is the set of variables that are involved in $r$, where $|I|<\infty$, then for a suitable fast-growing sequence ${a_i}$ of positive integers the product $\prod\limitsi x{i + ai} x{i + 2 a_i}$ is nonzero and left-annihilates $r$. On the other hand, nothing could cancel with the term that results from multiplication by a constant from $K$.
GitHub's Markdown won't show \{
and \}
in my formulae.
I wonder what one can say about $J(T)$.
The ring might be Armendariz and unlikely to have stable range 1. Likely also neither left Ikeda-Nakayama not L/R distributive or left dual.
Maybe add the right Ore condition to the property page? Are Ore extensions related to the Ore condition?
@rschwiebert Have you considered adding a search page for the text content? While compiling the list I stumbled upon a familiar-sounding property that was defined as "Noetherian and self-injective" and decided to put it aside until I get to the finiteness conditions. A Google search over the site gave abysmally unrelated results, even though the phrase in italics appears exactly in this form on the QF page.
@dyunov Regarding the text content search page: I think that's a great idea, but I don't have good ideas how to do it. You are thinking mainly of the properties, yes? What if, as a first approximation, there were a page that just listed all the definitions, so one could Ctrl-F stuff? I would greatly appreciate if you opened an issue as a feature request for something like this.
Maybe add the right Ore condition to the property page? Are Ore extensions related to the Ore condition?
Those both are already there: Ore domain Ore ring
I don't know the detailed history of Ore extensions but I surmise Ore must have made use of them. You can build Ore domains using Ore extension that are not right Ore, though.
You are thinking mainly of the properties, yes?
Currently even the obscure ring entry names are way more descriptive than some property names, so for now, yes. An automatically-generated full list of properties with definitions would be a good start, maybe something like:
Those both are already there
I know, I meant add the "formula" as stated in Lam II (10.3, pp300–303): $aS \cap sR \not = \varnothing$, where $S$ is the set of regular elements; or the "classical" per-element variant like at the end of Application section on Wikipedia to the property page. Sorry for poor wording and thanks for the example!
Is example 3.2 supposed to show necessity of the annihilator condition mentioned in Theorem 2.1? Theorem 2.1 says if $R$ is semiprime then $R_\sigma[x]$ is semiprimitive. But example 3.2 is not semiprimitive (it has nonzero Levitzki radical, and I think that's contained in the Jacobson radical) but R is claimed to be semiprimitive and so it should also be semiprime. So i guess one can only conclude that $R$ does not have the ACC on left annihilator ideals.
Yes, the purpose of Ex3.2 seems to be exactly that. $R$, which is commutative, doesn't satisfy ACC on annihilators of elements (for some reason I inverted the $\subset$'s in that part of my comment), and therefore neither does $R_\sigma[x]$.
Since much relies on the establishments of the idempotents in Ring 148, the following is a brief justification of that based on ideas from dyunov and josebrox.
Below, $T$ is the quotient ring of $\mathbb Q[\{x_i\mid i\in \mathbb Z\}]$ by the arithmetic-progression-triple monomials, and $R$ is the skew polynomial ring $T[x;\sigma]$ where $\sigma$ is the right shift operator on the variables.
Proof: If an element $p$ has a nonzero homogenous component in grade $n$, call it $p_n$, with $n>0$ minimal, then $p^2=p$ entails that $p_0^2=p_0$ and $p_n=2p_0p_n$. Multiplying the latter one b $p_0$ on both sides one obtains $p_n=p_0p_n=2p_0p_n$. The latter equality implies $p_0p_n=0 whence $p_n=0$, a contradiction. Hence an idempotent must have degree $0$.
Proof: Using the well-known lemma above, the idempotents must lie in $T_0$ which is $\mathbb Q$ in this case, hence the idempotents are trivial.
For a nonzero element $b$, et the largest (nonzero) homogenous component of highest degree $d$ be $b_d$. Each of the monomials in $b_d$ has grade $d$. Let $m_d$ be one such monomial that appears with a nonzero coefficient. Then $(m_d)^2$ must appear in the homogenous component of highest degree $2d$ in $b^2$, so $b^2\neq 0$. This implies that any nilpotent elements would have to lie in $T_0=\mathbb Q$, so $T$ is reduced.
Proof: This ring is graded but the Lemma as proven above doesn't work without commutativity. But fortunately the same idea works because the only options for $p_0$ are $0$ and $1$, so they are central. So, using the same argument, one finds that if an idempotent $p$ had a degree $d>0$ part, there would be the same contradiction. Hence the only idempotents are $0$ and $1$.
(With $T$ the commutative quotient ring and $R=T[x;\sigma]$)
@dyunov I feel confident about all the properties you mentioned except these: R IS anti-automorphic ($\sigma$ is not defined on all of $R$.) R IS lift-rad (why? I'm not sure what J(R) is, although it seems it must be contained within $(x)$.) R IS NOT Ikeda-nakayama: (From where you left off, you'd have to show ann(x_0x_1R)+ann(x_1x_2R)\neq R, which isn't clear to me. The left annihilators of course do not contain units, but their sum possibly could?) R IS NOT cohopfian: (It's not clear to me why 1+x_0 is regular and not a unit. You might have a trick that helps.)
In the IN construction I: (1) mixed up sides (however for disproving right Ikeda-Nakayama one could use the "alternative convention" of writing $x$'s on the left); (2) by the paragraph just before the properties spoiler, implied that $I = r(T x_0 x_1)$ — actually even $r(x_0 x_1)$ — does not contain any elements with nonzero constant term, since all elements of $I$ are right zero divisors. (Isn't it just $x2 T + x{-1} T$, by the way?)
$1 + x_0$ is regular because it has nonzero constant term. Let $t \in T$ be its inverse. Claim 3: $t \in R$. Again for grading reasons $t$ cannot contain a term $\tau$ with $x^k$, $k \ge 1$ (else in the product $(1+x_0) t$, there would have been a nonzero term $1 \cdot \tau$ with $R$-grade less than all other terms $R$-proportional to $x^k$). Claim 4: $t \in \Bbb Q[x_0]$. Suppose otherwise. Define a $\Bbb Q$-algebra endomorphism $\phi$ of $R$ by $xk \mapsto x{2k}$. Then $\phi$ fixes $1$ and $1 + x_0$, but not $t$. Now use the description of units in polynomial rings.
As for anti-automorphic and lift/rad, these should indeed be disregarded, not sure what I was thinking about when writing them, especially about $\sigma$.
Your observation that we can say $x_0R\cap x_1x_2R={0}$ enables me to say the ring is not right uniform, and with all the other properties it implies $R$ is not dual or Ikeda-Nakayama on the right.
You've convinced me about the cohopfian thing as well: very nice!
https://github.com/rschwiebert/dart_data/blob/90f76e76f8c44e2f14c632379de8e7a7c92deedb/db/ringapp/ring/RING_000148/data.yaml#L5
Visit the ring page for sources to draw information from.