ra.rings and algebras – Elements with equal annihilators

Let $R$ be a finite commutative ring with unity. Let $a in R$ and define $C_a = {b in R : ann(a) = ann(b)}$.

I want to know the cardinality of the set $C_a$.

For example,

If $a=0$ then $|C_a| = 1$.

If $a$ is a unit then $|C_a| = |U(R)|$ the set of units of $R$.

If $a$ is a zero divisor then I don’t now the answer.

If $R$ is the ring of integers modulo $n$ and $a$ is a non-zero zero divisor then $C_a = {x in mathbb Z_n : (x,n)=d}$ where $a = md$ for a proper divisor $d$ of $n$.

If $R$ is a reduced ring, then $R cong F_{q_1} times cdots times F_{q_k}$, product of finite fields. Define supp(a) = {1 le i le k : a_i is a unit} where $a = (a_1,dots,a_k) in R$. Then $C_a = {b in R: text{supp} (b) = text{supp} (a)}$. From this cardinality of $C_a$ can be obtained.

Thank you.

ra.rings and algebras – Representation theory terminology question

For a paper I’m writing, I need a term for a representation-theoretic concept that I’m sure someone has thought of before, so I thought I’d ask here rather than just make something up.

Let $G$ be a group and $R$ be a commutative ring. Consider an $R(G)$-module $V$. For any ideal $I$ of $R$, we have the submodule
$$I V = {text{$c cdot v$ $|$ $c in I$ and $v in V$}}.$$
What is the term for $R(G)$-modules $V$ such that all submodules are of this form? The ones I’m interested in have the additional property that if you ignore the $G$-action, then they are free $R$-modules (though not finitely generated!), but I doubt this matters for this question.

For an easy example, if $R = mathbb{Z}$ and $G = text{GL}(n,mathbb{mathbb{Z}})$, then $mathbb{Z}^n$ has this property.

If $R$ is a field, then this reduces the the usual notion of an irreducible representation, so I think of this as a version of irreducibility. But looking through my ring-theory books, I can’t find it anywhere.

ra.rings and algebras – Reference for a certain derivation on the ring of ordered series over a free monoid

Let $R$ be a (commutative or non-commutative) unital ring, $X$ be a non-empty set, and $R langle! langle X rangle! rangle$ be the ordered series ring (in fact, a ring of formal power series over $R$ in $|X|$ non-commuting variables) obtained by endowing the set of all functions $mathscr F(X) to R$ with the usual operations of pointwise addition and Cauchy product. Here, $mathscr F(X)$ is the free monoid on $X$, whose operation (that is, word concatenation) I’ll denote by $ast$.

While looking for a counterexample to a certain property in the class of local rings, I happened to note that, for each $z in X$, the mapping $partial_z$ that sends an ordered series $f in R langle! langle X rangle! rangle$ to the function
mathscr F(X) to R colon mathfrak z mapsto sum_{(mathfrak u, mathfrak v) in mathscr F(X) times mathscr F(X): mathfrak u ast z ast mathfrak v = mathfrak z} f(mathfrak u ast mathfrak v),

is a well-defined derivation of $R langle! langle X rangle! rangle$. In particular, the Leibniz identity follows from the fact that $mathscr F(X)$ is a cancellative monoid with trivial group of units and every $X$-word factors uniquely in $mathscr F(X)$ as a product of elements of $X$ (that is, $X$-words of length one).

My question is whether anyone here can offer a reference where $partial_z$ is being introduced: I thought I would have found $partial_z$ defined in Cohn’s book on FIRs (where ordered series rings are discussed in Sect. 1.5), but it’s not there (as far as I can see). I’ve also tried with Lam’s A First Course in Noncommutative Rings, but the conclusion is the same.

ra.rings and algebras – Union star symbol in set theory

In the slides Provenance for Database Transformations, page 24, they provide a semiring for lineage, which include a $cup^*$ symbol. However, I can not find any related materials about the meaning of this symbol. To be more specific, the elements in this semiring should be set of different symbols, e.g, ${a,b,c}$. Any help would be appreciated.

ra.rings and algebras – Show that $Z^2 + Y^3 + X^5$ is irreducible in $mathbb C[X,Y,Z].$ Conclude that $B$ is an integral domain

Here is the question I want to answer:

Let $mathbb C(X,Y,Z) cong mathbb C^{(3)}.$ Define rings $$ A = mathbb C(Y,Z)/(Z^2 + Y^3) text{ and } B = mathbb C(X,Y,Z)/(Z^2 + Y^3 + X^5) = mathbb C (x,y,z)$$
where $x,y,z$ are the images of $X,Y,Z$ under the standard projection $mathbb C (X,Y,Z) rightarrow B.$

$(b)$ Show that $Z^2 + Y^3 + X^5$ is irreducible in $mathbb C(X,Y,Z).$ Conclude that $B$ is an integral domain.

Here is my trial:

enter image description here
enter image description here
1- Can anyone give me a feedback on my trial please? specifically,I feel like my reasoning that it is an integral domain is not correct.

2- I got a hint to solve it by finding an isomorphism and noting that the composition of some projections are the same as can be seen below:

(!(enter image description here)(3))(3)

Could anyone show me how to find this isomorphism and its kernel and image please?


3-I also found this question on MSE:

https://math.stackexchange.com/questions/3028756/show-that-fx-y-z-x2-y2z-is-irreducible-in-mathbbcx-y-z?rq=1 but still I am confused about the general procedure and the specific details I should calculate to solve those kinds of problems, could anyone clarify this to me please?

EDIT: I felt like the question is not easy his is why I posted it here.

ra.rings and algebras – Example of idempotent left quasigroups which are right-distributive but not left-distributive

I am looking for examples of the following algebraic structure: a set (X,.) which satisfy the axioms

(idempotent) x.x = x

(left quasigroup) the equation a.x = b has a unique solution denoted by x = a*b

(right distributive) (x ? y) ! z = (x ! z) ? (y ! z) , where ? and ! are any of the operations . or *

but not the axiom

(left distributive) x ! (y ? z) = (x ! y) ? (x ! z) for any choice of the operations ? and ! among . and *

With the help of the (idempotent) axiom, the (right distributive) and (left distributive) may be rewritten as (right medial) and (left medial) axioms, so the question may be rephrased as: give (as many as possible) examples of idempotent left quasigroups which are right medial but not left medial.

ra.rings and algebras – Is the standard proof that Euclidean Domains are PIDs false?

In the books “Modern Algebra and it’s Applications” and “A First Course in Abstract Algebra”, the same proof that Euclidean Domains are PIDs is given. I will state it below (ver Batum from “A First Course in Abstract Algebra”) for convenience:

Let $D$ be a Euclidean domain with a Euclidean valuation $v$, and let $N$ be an ideal of $D$. If $N={0}$, then $N=left<0right>$ and $N$ is principal. Suppose that $N neq 0$. Then there exists $bneq0$ in N. Let us chose $b$ such that $v(b)$ is minimal among all $v(n)$ for $nin N$. We claim that $N=left<bright>$. Let $ain N$. Then by Condition 1 for a Euclidean domain, there exist $q$ and $r$ in $D$ such that
where either $r=0$ or $v(r)<v(b)$. Now, $r=a-bq$ and $a,bin N$, so that $rin N$, since $N$ is an ideal. Thus $v(r)<v(b)$ is impossible by our choice of $b$. Hence, $r=0$, so $a=bq$. Since $a$ was any element of $N$, we see that $N=left<bright>$.

My issue is, at the beginning of this proof it is assumed that there is some $b$ such that $v(b)$ is minimal among all $v(n)$, which is not necessarily true. For example, we can take $R$ to be the complex numbers under standard multiplication with a classic Euclidean norm of $v(a+bi)=a^2+b^2$. $R$ its self is obviously an ideal of $R$, but it does NOT have a minimal element in terms of valuation since one can pick numbers arbitrarily close to $0$.

Am I wrong? This proof is at its core based on this assumption, and so if it is not true the entire proof breaks. I feel like I must be misunderstanding this since this is a very important theorem and this is the proof given in almost every standard textbook.

ra.rings and algebras – division of polynomials

Consider the set of polynomials $P[x^1,cdots,x^n]$ over any field $k$. Now, given $p,qin P[x^1,cdots,x^n]$, what are the necessary and suficient conditions in order to q divide p? That is: $exists r in P$ such that $p = rq$. I know that necessary condition is that every root of $q$ is also a root to $p$, but I don’t know if this is a sufficient condition since I only can prove it when the field is algebraic close $k=bar{k}$.

Is this even a good question to be done? If not please let me know why

ra.rings and algebras – Why is Hochschild homology interesting if its cohomology groups are infinite-dimensional?

I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) differential forms of a smooth commutative $k$-algebra and the Hochschild homology. This clearly implies that the homology of the complex is infinite-dimensional, which seems strange to me. How can Hochschild homology be a good invariant of an algebra if the dimensions of its homology groups are all infinite-dimensional?