Timeline for $\mathbb{Z}$-module structure of the subring generated by an algebraic number
Current License: CC BY-SA 3.0
24 events
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| Apr 25, 2017 at 23:58 | comment | added | YCor | @GerryMyerson yes because it's not a technical term, it's a bit informal. Example "the group structure of the ring $\mathbf{Z}[1/n]$ remembers the set of prime divisors of $n$, but does not remember $n$ (does not distinguish $n=2$ and $n=4$". Whenever from a mathematical object $X$ you define another object $F(X)$ (an "invariant" of $X$), you can wonder how $F(X)$ "remembers" $X$. I found a few others examples with Google, e.g. about how the von Neumann algebra $L\Gamma$ of a group $\Gamma$ "remembers" $\Gamma$. | |
| Apr 25, 2017 at 23:41 | comment | added | Gerry Myerson | @YCor, OK, but what does it mean for a ring to "remember" another ring? I have never seen "remember" used as a technical term in Mathematics. | |
| Apr 25, 2017 at 23:31 | comment | added | YCor | @GerryMyerson I meant "remember the ring $\mathbf{Z}[a]$" (up to isomorphism); it's deliberately vague because I was (and am still) unsure, expecting that in some cases the ring $A$ of endomorphisms of the additive group would be reduced to $\mathbf{Z}[a]$. Or not much bigger. For instance, if $A$ is reduced to $\mathbf{Z}[a]$, then the minimal polynomial over $\mathbf{Z}$ of $a$ is determined by the additive group structure modulo multiplication by an invertible element of this ring. | |
| Apr 25, 2017 at 23:18 | comment | added | Gerry Myerson | "but in some others it might remember ${\bf Z}[a]$." @YCor, what was this meant to say? | |
| Apr 25, 2017 at 18:42 | comment | added | YCor | Just to summarize the "local" information more explicitly, it gives the following first invariant of the additive group of $A=\mathbf{Z}[a]$: the degree $u(a)$, which is the $\mathbf{Q}$-rank of $A$, and for each prime $p$ the number $u_p(a)\in\{0,\dots,u(a)\}$ of roots of the minimal polynomial $P_a$ of $a$ that have $p$-adic norm $>1$ (it is 0 for $p$ large enough). Then $\mathbf{Z}[a]$ can be embedded as a discrete cocompact subgroup in $\mathbf{R}^{u(a)}\times\prod_p\mathbf{Q}_p^{u_p(a)}$. In my above example where $P_a=2X^2+X+5$, $u(a)=2$, $u_2(a)=1$ and $u_p(a)=0$ for odd $p$. | |
| Apr 24, 2017 at 23:34 | history | edited | YCor | CC BY-SA 3.0 |
added initial version of the question
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| Apr 24, 2017 at 22:54 | comment | added | David Handelman | Even the special case that $a$ is the reciprocal of an algebraic integer is interesting. Suppose $a = b^{-1}$ where $b$ has irreducible polynomial $f$ (these types of things arise in stationary dimension groups). Then a prime $p$ is invertible in $G:= {\mathbf Z}[a]$ iff all the non-leading coefficients are divisible by $p$. And $G$ is strongly indecomposable (as an abelian group) and its endomorphism ring (as an abelian group) is commutative under some circumstances, e.g., if all the non-leading coefficients are relatively prime to the determinant. | |
| Apr 24, 2017 at 20:43 | comment | added | Will Sawin | @Ycor: I would guess many of them, like me, computed fairly quickly the structure of the tensor product with $\mathbb Z_p$ for each $p$ and assumed this more-or-less fixed the structure, forgetting the huge gulf you note between the purely $p$-local information contained in a $\mathbb Z$-module and the information that can be recovered after tensoring with $\mathbb Z_p$. | |
| Apr 24, 2017 at 20:21 | answer | added | Will Sawin | timeline score: 10 | |
| Apr 24, 2017 at 19:48 | comment | added | YCor | One useful invariant of the underlying abelian group of $\mathbf{Z}[a]$ will be the ring of endomorphisms of the underlying abelian group, which contains $\mathbf{Z}[a]$ as a subring. In some cases it will be the whole matrix group $M_n(\mathbf{Z}[1/k])$, but in some others it might remember $\mathbf{Z}[a]$. | |
| Apr 24, 2017 at 19:43 | history | edited | YCor | CC BY-SA 3.0 |
fixed unclear title
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| Apr 24, 2017 at 19:43 | history | edited | Joe Silverman | CC BY-SA 3.0 |
Improved formatting to use TeX, removed extraneous comment
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| Apr 24, 2017 at 19:35 | history | reopened |
YCor Derek Holt Benjamin Steinberg R.P. Stefan Kohl♦ |
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| Apr 24, 2017 at 18:38 | review | Reopen votes | |||
| Apr 24, 2017 at 19:35 | |||||
| Apr 24, 2017 at 17:58 | history | edited | YCor |
edited tags; edited tags
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| Apr 24, 2017 at 17:41 | comment | added | YCor | So people who downvoted/closed should clarify why. | |
| Apr 24, 2017 at 17:40 | comment | added | YCor | So, let's say that $P=2X^2+bX+c$. If $P$ is irreducible over $\mathbf{Q}_2$, then it's easy to check we obtain a group isomorphic to $\mathbf{Z}[1/2]^2$. But let's take $P(X)=2X^2+X+5$. It's split in $\mathbf{Q}_2$, with exactly one root in $\mathbf{Z}_2$. So in this case $A=\mathbf{Z}[t]$ is additively isomorphic to the kernel of some additive homomorphism $\mathbf{Z}[1/p]^2\to C_{p^\infty}:=\mathbf{Z}[1/p]/\mathbf{Z}$, and also lies in a short exact sequence $0\to \mathbf{Z}[1/p]^2\to A\to C_{p^\infty}\to 0$. | |
| Apr 24, 2017 at 17:40 | comment | added | YCor | It's not clear for me why the question was closed: the additive group structure of the ring generated by an algebraic number sounds a reasonable question. If the integral minimal polynomial is $P=a_nX^n+\dots+a_0$ with $\gcd(a_0,\dots,a_n)=1$, then clearly we obtain a group isomorphic to a subgroup of $\mathbf{Z}[1/a_n]^n$ of $\mathbf{Q}$-rank $n$. If $n=1$, say $P=bX-a$ this is exactly $\mathbf{Z}[1/b]$. But even $n=2$ is worth looking. For $p$ prime, $\mathbf{Z}[1/p]^2$ admits uncountably many non-isomorphic subgroups. | |
| Apr 24, 2017 at 16:08 | review | Reopen votes | |||
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| Apr 24, 2017 at 15:52 | history | edited | user108921 | CC BY-SA 3.0 |
added 151 characters in body
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| Apr 24, 2017 at 15:37 | history | closed |
Stanley Yao Xiao js21 Qfwfq Will Sawin Chris Wuthrich |
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| Apr 24, 2017 at 15:32 | review | Close votes | |||
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| Apr 24, 2017 at 15:17 | review | First posts | |||
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| Apr 24, 2017 at 15:14 | history | asked | user108921 | CC BY-SA 3.0 |