 # Commutative algebra Questions and Answers

Recent questions in Commutative Algebra cricafh 2022-05-24 Answered

### Let $\mathcal{g}$ be a Lie algebra and let $a,b,c\in \mathcal{g}$ be such that $ab=ba$ and $\left[a,b\right]=c\ne 0$. Let . How to prove that $\mathcal{h}$ is isomorphic to the strictly upper triangular algebra $\mathcal{n}\left(3,F\right)$?Problem: If $\mathcal{h}\cong n\left(3,F\right)$ then $\mathrm{\exists }{a}^{\prime },{b}^{\prime },{c}^{\prime }\in \mathcal{n}\left(3,F\right)$ with ${a}^{\prime }{b}^{\prime }={b}^{\prime }{a}^{\prime }$ and $\left[{a}^{\prime },{b}^{\prime }\right]={c}^{\prime }$ as in $h$ But then ${c}^{\prime }$ must equal $0$ whereas $c\in h$ is not $0$? Nylah Burnett 2022-05-24 Answered

### Let $R$ be a commutative finite dimensional $K$-algebra over a field $K$ (for example the monoid ring of a a finite monoid over a field). Assume we have $R$ in GAP. Then we can check whether $R$ is semisimple using the command RadicalOfAlgebra(R). When the value is 0, $R$ is semisimple. Thus $R$ can be written as a finite product of finite field extensions of $K$.Question: Can we obtain those finite field extensions of $K$ or at least their number and $K$-dimensions using GAP? Hailey Newton 2022-05-23 Answered

### If $\mathcal{A}$ is a commutative ${C}^{\ast }$-subalgebra of $\mathcal{B}\left(\mathcal{H}\right)$, where $\mathcal{H}$ is a Hilbert space, then the weak operator closure of $\mathcal{A}$ is also commutative.I can not prove this. Waylon Ruiz 2022-05-23 Answered

### Is there any simple proof (one that does not use continuous functional calculus) for the statement that $\sigma \left({x}^{\ast }x\right)\subseteq \left[0,\mathrm{\infty }\right)$ for any $x\in \mathcal{A}$ where $\mathcal{A}$ is a commutative ${C}^{\ast }$-Algebra? wanaopatays 2022-05-22 Answered

### In general, we have functors ${\mathcal{S}\mathcal{C}\mathcal{R}}_{R/}\stackrel{\varphi }{\to }{\mathcal{D}\mathcal{G}\mathcal{A}}_{R}\stackrel{\psi }{\to }{\mathcal{E}\mathcal{I}}_{R/}$. If $R$ is a $\mathbf{Q}$-algebra, then $\psi$ is an equivalence of $\mathrm{\infty }$-categories, $\varphi$ is fully faithful, and the essential image of $\varphi$ consists of the connective objects of ${\mathcal{D}\mathcal{G}\mathcal{A}}_{R}\simeq {\mathcal{E}\mathcal{I}}_{R/}$ (that is, those algebras $A$ having ${\pi }_{i}A=0$ for $i<0$).What is the explicit functor $\varphi :{\mathcal{S}\mathcal{C}\mathcal{R}}_{R/}\to {\mathcal{D}\mathcal{G}\mathcal{A}}_{R}$? I suppose that the natural thing would be to take a simplicial $R$-algebra $A$ and assign it to$\begin{array}{r}\varphi \left(A\right)=\underset{i=0}{\overset{\mathrm{\infty }}{⨁}}{\pi }_{i}A,\end{array}$and take a map $f:A\to B$ and assign it to$\begin{array}{r}\varphi \left(f\right)=\underset{i=0}{\overset{\mathrm{\infty }}{⨁}}\left({f}_{i}:{\pi }_{i}A\to {\pi }_{i}B\right),\end{array}$but as far as I could find this isn't stated explicitly in DAG. Is this the case, and if so, do you have a source or proof? And how does one show that $⨁{\pi }_{i}A$ is a differential graded algebra? Mauricio Hayden 2022-05-21 Answered

### Let $A$ be a ${C}^{\ast }$-algebra. We say $A$ is “commutative“ if $a{b}^{\ast }c=c{b}^{\ast }a$ for all $a,b,c\in A$ and define “center” of $A$ as$Z\left(A\right)=\left\{v\in A:a{v}^{\ast }c=c{v}^{\ast }a\mathrm{\forall }a,c\in A\right\}$Are these notions of “commutativity“ and “center” same as usual notions of commutativity and center in ${C}^{\ast }$-algebras? Riley Yates 2022-05-21 Answered

### In rational homotopy theory, one uses various algebras and coalgebras to model (simply connected) spaces (topological spaces or simplicial sets, usually) up to rational equivalences.Two types of models one can use are dg cocommutative algebras (Quillen) and dg commutative algebras (Sullivan). I will leave the dg implicit from now on. The dual of a cocommutative coalgebra is always a commutative algebra (while the converse is only true in the finite dimensional case). I have the following question.Let $X$ be a simply connected space, and suppose $C$ is a cocommutative rational model for $X$. Under what assumptions is its linear dual ${C}^{\vee }$ a commutative rational model for $X$? aniawnua 2022-05-21 Answered

### Let $A$ be a commutative Banach algebra such that set of all characters is infinite. I want to prove that there exists an element in $A$ such that its spectrum is infinite.Why is there an element with infinite spectrum in a commutative Banach algebra with infinitely many characters? vestpn 2022-05-21 Answered

### Let $S$ be some base ring (a commutative ring or even just a field), and $R$ a commutative ring containing $S$ which is finitely generated (as an algebra) over $S$. What conditions guarantee that any two minimal systems of generators of $R$ over $S$ have the same size? shelohz0 2022-05-21 Answered

### Does Aluffi's book have enough commutative algebra for algebraic geometry? I understand that traditional graduate algebra course using Hungerford's book or Lang's book provides enough background for such a course. istupilo8k 2022-05-20 Answered

### Classifying all commutative $\mathbb{R}$-algebras of matrices over $\mathbb{R}$?I initially thought they were all isomorphic to some subring of the $n×n$ diagonal matrices $\mathcal{D}\cong \mathbb{R}×\cdots ×\mathbb{R}$, but this was wrong: Every commutative ring of matrices over $\mathbb{R}$ is isomorphic to the diagonals?. One counterexample is matrices of the form (using block matrix notation) $\left[\begin{array}{cc}\alpha {I}_{1}& A\\ 0& \alpha {I}_{n-1}\end{array}\right]$ for some $1×\left(n-1\right)$ real matrix block $A$ and some $\alpha \in \mathbb{R}$, which forms a commutative ring $\left(\mathcal{U},+,\ast \right)$.Are there other counterexamples? Can we classify all such rings up to isomorphism? Alani Conner 2022-05-19 Answered

### For a finitely generated algebra $A$, let $V$ be its finite-dimensional generating subspace. Then $A=\sum _{n=0}^{\mathrm{\infty }}{A}_{n}$ for ${A}_{n}=\mathbb{K}+V+\cdots +{V}^{n}$. Let the function ${d}_{V}\left(n\right)={\mathrm{dim}}_{\mathbb{K}}\left({A}_{n}\right)$. Then the growth of $A$, i.e. $\mathcal{G}\left(A\right)$ will be defind as $\mathcal{G}\left(A\right):=\mathcal{G}\left({d}_{V}\right)$.I cannot understand why the growth of the commutative polynomial algebra $\mathbb{K}\left[{x}_{1},\dots ,{x}_{d}\right]$ is polynomial of degree $d$; i.e. is ${P}_{d}$ and has a polynomial growth? starbright49ly 2022-05-19 Answered

### Let $A$ be a domain. Assume that for any non-trivial finitely generated $A$-module $M$ we have ${\mathrm{Hom}}_{A}\left(M,A\right)\ne \left\{0\right\}$. Prove that $A$ is a field. babajijwerz 2022-05-18 Answered

### I have seen in a book Commutative Algebra by Reid mention of a "ring commutative with a 1". Does that mean that addition and multiplication are commutative and that the multiplicative identity is 1 or it means that it is in some way commutative with respect to 1? Can anyone explain? Noelle Wright 2022-05-14 Answered

### Let's fix the notation, $V=\underset{i\ge 0}{⨁}{V}^{i}$ is a graded vector space and $\mathrm{\Lambda }V$ is the free commutative graded algebra on $V$. I have been struggling to understand this example:Consider a graded vector space $V$ with basis $\left\{a,b\right\}$ such that $a\in {V}^{2}$ and $b\in {V}^{5}$. Now define a linear map $d$ (of degree 1) by $da=0$ and $db={a}^{3}$. It follows that d extends uniquely to a derivation .The point of the example is to show that the derivation on $\mathrm{\Lambda }V$ is completely determined by its values on $V$. So if i understand well, he considers a linear map $d:V⟶\mathrm{\Lambda }V$ of degree one defined by(here ${\mathrm{\Lambda }}^{k}V$ is the set of elements of word length $k$) and${d}_{5}:{V}^{5}⟶{\mathrm{\Lambda }}^{6}V;b↦{a}^{3}$The first question that i'm stuck on is for ${d}_{2}\left(b\right)={a}^{3}$, i mean ${a}^{3}$ is of length $3$, how it can be in ${\mathrm{\Lambda }}^{6}V$. Jazlyn Raymond 2022-05-14 Answered

### Determining Lie algebras from commutative diagrams of exact sequences.Suppose that we have the following commutative diagram of graded Lie algebras.$\begin{array}{}0& ⟶& {C}_{n}& ⟶& {A}_{n+1}& ⟶& {A}_{n}& ⟶& 0\\ & & ↓& & ↓& & ↓\\ 0& ⟶& {D}_{n}& \stackrel{}{⟶}& {B}_{n+1}& \stackrel{}{⟶}& {B}_{n}& ⟶& 0& \end{array}$for all $n\in {\mathbb{Z}}^{+}$, where the both rows are split exact sequences and every vertical map is onto.My question is, suppose that we know ${A}_{n}$, ${C}_{n}$, ${D}_{n}$ for all $n\in {\mathbb{Z}}^{+}$ and ${B}_{1}$, do we have enough information to uniquely determine up to isomorphism every ${B}_{n}$ (likely by induction)? Merati4tmjn 2022-05-12 Answered

### Question: What's the cleanest proof that ${H}^{\ast }\left(Sym\left(V\right)\right)\cong Sym\left({H}^{\ast }\left(V\right)\right)$? hovudverkocym6 2022-05-12 Answered

### Suppose that $A$ is a natural Banach function algebra on $K$, a compact Hausdorff space. So $A$ is realised as an algebra of continuous functions on $K$, is a Banach algebra for some norm (necessarily dominating the supremum norm) and each character on $A$ is given by evaluation at a point of $K$.If $F\subseteq K$ is closed, thenis a closed ideal in $A$. If e.g. $A=C\left(K\right)$ then every closed ideal is of this form.What's a simple example of an $A$ where not every closed ideal is of this form?Can we find an $A$ which is conjugate closed? bedblogi38am 2022-05-11 Answered

2022-05-03

### In basic terms, commutative algebra represents a complex study of the rings that take place in algebraic number theory. It also relates to solving problems and questions based on algebraic geometry. You will see that there are solutions related to Dedekind rings and various commutative patterns. Commutative algebra solutions will be quite short in most cases, yet one should start with examples that are represented and link them with various questions. Take your time to post your commutative algebra example problem as well and always compare your question to similar questions as it will help you provide the required data. When you are dealing with inequality and graph solution tasks as you are seeking commutative equ