Provide construction details for ring 148 (Ram's Ore extension ring)

[rschwiebert]

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.

[dyunov]

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).

Properties

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 $K$.

Some properties

- :x: I_0. $x_0 T \not \subset J(T)$, as $1 + x_0$ is not invertible (the inverse would also have to lie in $R$, and then be equal to $\sum (-1)^n x_0^n$, which is not available there). - ✔️ Abelian. - ✔️ anti-automorphic. $\sigma$ seems to be an anti-automorphism if I'm not lost in the definitions. - :x: clean. $1+x_0$ and $x_0$ are not units. - ❔ countable. Whenever $K$ is. - :x: finite. The set $\{x_0^n\}$ is infinite. - :x: fully prime. $(x_0 x_1)$ is not prime. - ✔️ lift/rad. - :x: local. $(1+x_0)$ and $(-x_0)$ are not invertible and have an invertible sum. - :x: L/R ACC/DCC annihilator. $l(x_1) \subset l(x_1 x_2) \subset l(x_1 x_2 x_4) \supset \ldots$ and symmetrically. - :x: L/R Bezout. $(x_0, x_1)$ is not cyclic. - :x: L/R DCC annihilator. See above. - :x: right Ikeda-Nakayama. $x_0 x_1 T \cap x_1 x_2 T = 0$ (no nonzero monomial contains all three variables); their left annihilators clearly do not contain elements of $K$. - :x: L/R Rickart. Too few idempotents for (7.45) in Lam II to hold. - :x: L/R cohopfian. $1 + x_0$ is neither invertible nor a zero divisor. - :x: right dual. $l(x_0 x_2 R) = R x_{-2} + R x_1 + R x_4 = I$; $r(I)$ seems to be trivial. - :x: PLI/PRI. Take $(x_0, x_1)$.

Additional comments

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.

[rschwiebert]

@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.

[rschwiebert]

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.

[dyunov]

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:

• $I_0$: Every right ideal not contained in $J(R)$ contains a nonzero idempotent. Also called "semipotent"
• $\pi$-regular: For all $a$ in $R$ there exists an $x\in R$ and a natural number $n$ such that $(a^n)x(a^n)=a^n$
• $\ldots$ Then it should probably be linked to at the top of the full property list.

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!

[rschwiebert]

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.

[dyunov]

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 $R_\sigma[x]$.

[rschwiebert]

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.

Lemma: The idempotents of a commutative $\mathbb N$-graded ring lie in the degree zero part.

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 $0$.

Claim 1: $T$ is an $\mathbb N$-graded ring, where the degree of a monomial is the total of its exponents. (This is the grading on the polynomial ring passed down to the quotient by a homogenous ideal.) 1) The idempotents are trivial, and also 2) $T$ is reduced.

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.

Claim 2: $T[x;\sigma]$ has only trivial idempotents.

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 $1$.

[rschwiebert]

(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.)

[dyunov]

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 $\sigma$.

[rschwiebert]

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!