Expressing a matrix ring as a direct product of matrix rings
Can we express an arbitrary matrix ring $M_n(R[x])$ over the polynomial ring $R[x]$, where $R$ is a commutative ring with unity, as a direct product of other matrix rings (nontrivial)? I am finding...
View ArticleHow to show that $\mathrm{Tor}_{2}^{\mathbb{Z}}=0$
Since $\mathrm{Tor}$ functor preserve direct limit and anyAbelian group is direct limit of its finitely generated subgroups, and by the fundamental theorem of f.g. abelian groups, it reduces to the...
View ArticleIs there exists a module which is not completely decomposable
I am confused as to whether there is a module that is not completely decomposable. If the module $M$ has finite length, this is easy: if $M$ isn't indecomposable, $M$ can be written as $M_1\oplus M_2$;...
View ArticleShow the formal power series ring is a faithfully flat algebra.
Suppose $S=P^{-1}F[x]$, the localization of $F[x]$ at $P$ where $F$ is a field and $P=(x)\backslash\{0\}$. Let $\hat{S}=F[[x]]$, the formal power series ring. Clearly there is a homomorphism $\Phi:...
View ArticleChanging scalars of a tensor product $N \otimes_B \ A$
So I was working on exercise 1.5.E. in Vakil (which is to show that restriction of scalars is right adjoint to the tensor product), but I got caught on understanding how to think of a tensor product...
View Articlerelation between $E[x]$-module and $F[x]$-module when $E$ is a subfield of $F$
Let $F$ be a field and $E$ be a subfield $F$. Let $X$ be a $n\times n$ matrix with entries in the field $E$. We can give the $n$-dimensional vector space $E^n$ over $E$, a $E[x]$-module structure by...
View ArticleWhat is the rank of the module $\mathbb F_p[[x,y]]$ over its subring $\mathbb...
Consider the ring $M:=\mathbb F_p[[x,y]]$ and subring $N:=\mathbb F_p[[x^{p^m},y^{p^n}]]$, where $\mathbb F_p$ is the field of $p$-elements and $\gcd(m,n)=1$. Then $M$ is a finitely generated module...
View ArticleAbout the notion of invertibility in non unitary rings (or semigroups).
Sorry if this one was already asked, I did not find it. Here is the story:Let us take a unitary ring $R$, the notion of unit element is very clear:$$r \in R^{\times} \iff \exists r' \in R \mid rr' = 1...
View ArticleExtension of a module $M$ with $\mathbb Z_{(p)}$ structure to a module with...
Let $R$ be a ring and $M$ an $R$-module. For each $r \in R$ we define$T_r^R: M \to M$ as $v \mapsto rv$.Let $A$ be a subring of the ring $B$, if $M$ is an $A$-module, we say that the structure of...
View ArticleKernel of a modules aplication
Let $A$ be a ring, $I$ an $A$-ideal. We define the aplication $A/I\otimes _AM\longrightarrow M/IM$, $[a]\otimes x\longmapsto [ax]$. I want to see it is injective.My try is the following: let $[a]\in...
View ArticleRank-preserving linear transformations over rings
Let $f:\operatorname{Mat}_{m \times n}(\mathbb{F}_q) \to \operatorname{Mat}_{m \times n}(\mathbb{F}_q)$ be an invertible, rank-preserving linear map. Then we have the characterisation that $f(A)=PAQ \...
View ArticleQuotients of modules over DVRs
I'm trying to prove the following: Let $A$ be a DVR with uniformizer $\pi\in A$ and finite residue field $A/(\pi)$.Consider $A$-modules $M_0 \subseteq M_1 \subseteq M_2$ which are generated by two...
View ArticleDecomposition of non-finitely generated modules over a finite dimensional...
Let $A$ be a finite dimensional algebra over some field $k$ and of finite representation type. A finite dimensional algebra $A$ is said to be of finite representation type, or of finite type for short,...
View ArticleIf $f:A\rightarrow B$ is a ring morphism, and $M$ is a flat $A$-module, then...
The statement is: If $f:A\rightarrow B$ is a ring morphism, and $M$ is a flat $A$-module, then $M_B := B \otimes_A M$ is a flat $B$-module.By a proposition in theory, proving that $M_B$ is a flat...
View ArticleOne-to-one correspondence between the prime ideals lying over $\mathfrak{p}$...
Let $A\to B$ be a ring homomorphism and let $\mathfrak{p}$ be a prime ideal of $A$. Then the prime ideals of $B$ lying over $\mathfrak{p}$ are in one-to-one correspondence with the prime ideals of...
View ArticleIs this torsion submodule sequence an exact sequence?
Let $0\rightarrow M'\xrightarrow{f} M\xrightarrow{g} M''\rightarrow 0$ be an exact sequence. Then, the sequence $0\rightarrow T(M'')\rightarrow T(M)\rightarrow T(M')$, where T is the torsion submodule...
View ArticleModules with no regular elements in his annihilator
Suppose $R$ is noetherian, suppose $M$ is a finitely generated $R$-module of grade 0. This means that there are no regular elements in his annihilator. Does this implies that...
View ArticleIs $R/I\otimes _RB\cong B/IB$ even if $B$ is not unitary where $B$ is a right...
In the textbook Algebra by Hungerford, there is an exercise:Let $I$ be a right ideal of a ring $R$ with identity and $B$ be an $R$-module. Then $R/I\otimes _RB\cong B/IB$ as abelian groups.To show...
View ArticleRevisiting proof that all bases of a free module $M$ over a commutative...
I am following closely the book of T.S. Blyth, Module Theory. There is a theorem which says that every free $R$-module $M$ where $R$ is a commutative unitary ring, has equipotant base.The process is a...
View Article$R^n$ and $R^{(I)}$ not isomorphic if $I$ is infinite
Let $R$ be a ring and $M$ a free and finitely generated $R$-module. I have to show that there cannot be an isomorphism $M \cong R^{(I)}$ with $I$ infinite.(To note: $R^{(I)} := \{(r_i)_{i \in I} |...
View ArticleUnimodular lattice
We are considering $\mathbb{R}^n$ with the inner product (the usual one), and $L$ an unimodular lattice (so $covolume(L)=1$ and $<u,v> \in \mathbb{Z}$ for all u,v in L).I have to show that, for...
View ArticleWhy natural map $M \rightarrow \prod M_{\mathfrak{p}}$ makes sense?
I am referring to this post where there is a natural map from $A-$module $M \rightarrow \prod M_p$ for $p$ are maximal ideals of $A$.I understand why the map is injective if it makes sense, but what I...
View ArticleSimple objects in the direct sum of abelian categories
I'm not familiar with the direct sum of abelian categories.I have a question:Let $A$ and $B$ be algebras. Let $\mathrm{Mod}_A$ be the category of left $A-$module and $\mathrm{Mod}_B$ be the category of...
View ArticleDiagonalizability results via the structure theorem for finitely generated...
I'm looking for a better understanding of results about when a linear transformation is diagonalizable by using the structure theorem for finitely generated modules over a Euclidean domain.Result 1:...
View Article$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ exact, then the...
I want to demonstrate that if $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ is an exact sequence, then the induced sequence $0\rightarrow M'\otimes_A N\rightarrow M\otimes_A N\rightarrow...
View ArticleIs a finitely generated module over a hereditary ring always finitely...
In the §6 Appendix II (2) of the article Gorenstein projective modules says that:Lemma 1 : Let $R$ be a ring. If $M$ is a finitely generated $R$-module, then$$Ext^{n}( M, \oplus_{i\in I}\ N_{i}) \cong...
View ArticleElementary Divisors and Invariant Factors of $(\mathbb{Z}/360\mathbb{Z})^*$
Let $M = (\mathbb{Z}/360\mathbb{Z})^*$ as a $\mathbb Z$-Module.The goal is to find the elementary divisors and invariant factors of $M_{tor}$.I concluded that $M = M_{tor}$ since $k^{ord(k)} = \bar 1$...
View ArticleA doubt about exact sequence
I read book ''A Term of Commutative Algebra'' p.35 last line, meet the statement as followinghttps://i.sstatic.net/wiwcUvaY.pngWhy $R^{\oplus\Sigma} \to R^{\oplus \Lambda} \to M \to 0$ is exact?How did...
View ArticleClarification needed on the multiplication of elements of a tensor algebra
I'm reading Rotman's Advanced Modern Algebra: Part 1 and which gives the definition of a tensor algebra as:If $M$ is a $k$-module, define$$T(M)=\bigoplus_{p \geq 0}\left(\bigotimes^p M\right)=k \oplus...
View ArticleExample where the last arrow in the sequence is not surjective.
Consider the exact sequence $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ and its induced sequence $0\rightarrow T(M')\rightarrow T(M)\rightarrow T(M'')$, where $T$ denotes the torsion...
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