abstract algebra – How to precisely define $b/a$ if $a mid b$ in a general commutative ring?

Let $R$ be a commutative ring and let $a,b in R$. Suppose that $a mid b$, which means that there exists some $k in R$ such that $kcdot a = b$. I now often see expressions of the form $b/a$ in this context. ( For example the important lcm-gcd-formula $gcd(a,b) = ab / lcm(a,b)$ ). My question:

Is it enough to simply define $b/a := k$ or are there some caveats? (I think one problem here might be that $b/a$ does not have to be unique with this naive definition.)

Could you please explain this to me?

rt.representation theory – Is it possible for the reduction modulo $p$ of an non-commutative semisimple algebra to be commutative?

Suppose that $I, X_1, ldots, X_{d-1}$ are $n times n$ matrices with integer entries whose $mathbb{Z}$-span is a subalgebra of $mathrm{Mat}_n(mathbb{Z})$. Suppose that, thought of as a subalgebra of $mathrm{Mat}_n(mathbb{C})$, this algebra is semisimple and non-commutative. Thus, by Wedderburn’s Theorem, it is isomorphic to a direct product of complete matrix algebras $mathrm{Mat}_r(mathbb{C})$, with $r ge 2$ for at least one factor.

It is possible that there exists a prime $p$ and a field $K$ of characteristic $p$ such that, regarding $I, X_1, ldots, X_{d-1}$ as elements of $mathrm{Mat}_n(K)$ by reduction modulo $p$, the subalgebra of $mathrm{Mat}_n(K)$ spanned by $I, X_1, ldots, X_{d-1}$ is commutative, and still of dimension $d$?

matrices – Can all finite dimensional non commutative algebras be embedded into matrix rings?

Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear “trace” function $t: R to k$. Can I find square matrices $A_1,dots,A_n$ (of some dimension $r$) so that I have an embedding $f: R to M_r(k)$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2dots a_n) = t(a_2dots a_na_1)$. Is this the only restriction?

commutative algebra – Properties of $k[x(x-1)]_{langle x(x-1) rangle} subseteq k[x]_{langle x rangle}$

Let $k$ be aa arbitrary field.

Let $R=k(x(x-1))_{langle x(x-1) rangle}$ and let $S=k(x)_{langle x rangle}$,
$m=x(x-1)R$, $n=xS$, $k(m)=R/m$, $k(n)=S/n$.

We have, $mS = n$ (since $x-1$ is invertible in $S$).

Now, $k(m) cong k$ and $k(n) cong k$, so ​$k(m) cong k(n)$.

I am trying to figure out if the residue field extension $k(m) to k(n)$ is finite-dimensional or not.

Question: Is $k(m) to k(n)$ finite-dimensional or not? and why?


(1) I think that it is infinite dimensional;
indeed, I know that $A=k(x(x-1)) subseteq k(x)=frac{k(x(x-1))(T)}{langle T^2-T-x(x-1) rangle}=B$ is not separable,
since $(T^2-T-x(x-1))’=2T-1$ evaluated at $x$, $2x-1$, is not a unit of $k(x)$.

$B$ is Noetherian+finitely generated $A$-algebra, hence $B otimes_A B$ is Noetherian, and so the kernel of $B otimes_A B to B$ is a finitely generated ideal. In the case separablity is equivalent to (formal) unramifiedness.
Therefore, being non-separable impies that $A subseteq B$ is not (formally) unramified,
so there exists a maximal ideal $N$ of $B$, such that $A_{N cap A} subseteq B_N$ is not unramified.

Now I am not sure for which maximal ideals $N$ of $B$, such localizations are not unramified.

(2) $A subseteq B$ is free with basis ${1,x}$, hence flat, hence every such localization $R subseteq S$ is flat.

Any hints and comments are welcome!

nt.number theory – What is this binary commutative operation on $mathbb{Q_+}$?

Write $0inmathbb{N}$. For $ninmathbb{Q}_+$, if $n=prod_{iinmathbb{N}}p_i^{alpha_i}$ is the prime factorization of $n$ and $D_n:mathbb{N}→mathbb{N}$ is $D_n(i)=alpha_i$, then for all $a,bgeq1$ we have

$$min(D_a,D_b) = D_{gcd(a,b)}$$
$$max(D_a,D_b) = D_{text{lcm}(a,b)}$$

What is $D_aD_b$? Since $D_a$ and $D_b$ have finite support, it must be $D_x$ for some $x$.

soft question – Latest “A Term of Commutative Algebra” by Altman and Kleiman?

(Moderator, please turn this question to a community-wiki. I’ll post my answer soon. TIA.)

Where can I find the latest revision of A term of Commutative Algebra by Allen B. ALTMAN and Steven L. KLEIMAN? Is my 2013 version ok?

It is hard to locate the latest one; many old revisions and pointers to them are randomly scattered across the web.

This free textbook is intended to be an update of, and an improvement to “A & M”, i.e. Introduction to Commutative Algebra by Atiyah and MacDonald.

commutative algebra – Reference book to study dimension theory.

I’m looking for a book to study Dimension theory, In particular I want a book(s) which covers the following topics:

1.Krull-dimension and Examples, Dimension of an integral extensions
2.Noether’s Normalization Lemma and its consequences.
3.Graded rings and modules, Hilbert functions and series, Hilbert’s Theorem.
4.Hilbert-Samuel functions and polynomials, System of parameters, Dimension Theorem.
5.Dimension of polynomial rings over noetherian rings.
6.Normal rings and their characterizations and properties. Finiteness of integral closure

Please suggest me a good book(s) which covers the above topics. (I’m planning to do self study, so if any of the book(s) is intended for self study it’s really good. Even if not, I hope I can do, so you don’t have to restrict to books intended for self study). Thanks.

gr.group theory – Cancellation property for commutative monoid

Consider the monoid $M=mathbb{N}times {0,1}$ where $$(n,a)*(m,b):=(n+m, acdot b).$$
The unit element is $e:=(0,1)$. Note that this monoid is torsion free. Now consider the maps
$$g:(M,*,e)rightarrow (mathbb{N}, +,0), g(n,a)=n$$
and $f: (mathbb{N}, +,0) rightarrow (M,*,e)$ such that $f(0)=e$ and $f(n)=(n,0)$. Then we have $fcirc g=id$, but $f(1)=(1,0)$ is not cancellative as
$$ (1,0)(0,0)=(1,0)=(1,0)(0,1).$