New answers tagged

2

First of all, the quotient $M/\ker f$ may not be free -- consider $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. The criterion relating injectivity of $f$ to linear independence of the columns of $A$ works in the case where both source and target are free, but not necessarily otherwise. If the quotient $M/\ker f$ is free, the matrix will not necessarily be the ...


0

I will quote some results and problems from the book Combinatorial Games: Tic-Tac-Toe Theory by József Beck, some of which were also quoted in this answer. The terms "win" and "draw" refer to the game as ordinarily played, i.e., the first player to complete a line wins. The term "Weak Win" refers to the corresponding Maker-Breaker game, where the first ...


2

Your question is how to compute the matrix exponential $\exp(a)$ where $a$ is an element of a finite-dimensional real associative algebra $A$. (Your phrasing is not more general: you're implicitly working in the algebra $M_n(A)$.) This reduces to the case of real matrices, because the left regular representation $$A \ni a \mapsto \left( L_a : b \mapsto ab \...


2

You ask: wouldn't it make the most sense to just use the monoidal product from the "mother category"? but this question does not make sense, really. To judge if something makes more sense or less sense than something else, you have to spell out what you are trying to achive. Sometimes, the direct sum is the correct operation to turn modules into a ...


2

Every object in an additive category is a group and a cogroup, canonically, with respect to the monoidal structure of direct sum. This reflects the fact that vector spaces, and more generally, modules, are actually groups. So there's nothing there. Hopf algebras weren't invented to cause pain, much less because of the desire to decorate the monoidal ...


1

If you have a basis $S$ for $P$, we know that $\operatorname{Hom}(P,C)\simeq C^S$, and similarly $\operatorname{Hom}(P,B)\simeq B^S$. Now if for each $s\in S$, you choose an element $b_s\in B$ such that $\;f(b_s)=g(s)$ (we're using the axiom of choice here if $S$ is not finite), the family $\;(b_s)_{s\in S}\in B^S$ defines a homomorphism $h$ from $P$ to $...


0

The direct sum of $k$-vector spaces makes $\mathbf{Vec}_k$ into a cocartesian monoidal category, not a cartesian one.


0

Say $L \overset\iota\hookrightarrow R$ is the natural inclusion. Injectivity of $A$ implies for any $g: L \to A$ there is a $G: R \to A$ such that $G\circ \iota = g$. Take $a = G(1_R)$; then for any $r \in L$, we have that $g(r) = (G\circ \iota)(r) = G(r) = rG(1_R) = ra$.


3

Let $\tilde g\colon R\longrightarrow A$ be an extension of $ g\colon L\longrightarrow A$ to $R$ and let $r\in L$. We have $$g(r)=\tilde g(r) = \tilde g(r\cdot 1)=r\tilde g(1),$$ so just set $a=\tilde g(1)$.


1

We show the implications (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a). Assume (a) and let $e$ be the unit of $R$. Let $S$ be any ring such that $R\subseteq S$ is an ideal. For any $s\in S$, we have $se, es\in R$ and in fact they are equal: $$ es = e(es) = (es)e = e(se) = (se)e = se, $$ where for the second and fourth equality we have used that $...


0

$\DeclareMathOperator{\Hom}{Hom}$ If $P$ is a finitely generated projective $R$-module, then it is also finitely presented: Take a surjection $R^n\twoheadrightarrow P$ with kernel $Q$. Then $R^n = Q\oplus P$ as $P$ is projective. As the projection $R^n\rightarrow Q$ is surjective, $Q$ is also finitely generated. Thus, we obtain an exact sequence $R^m\...


1

It is infact never true that $\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. An ideal $\mathfrak{p}\subseteq R$ is prime if and only if $R/\mathfrak{p}$ is an integral domain. Moreover $\mathfrak{p}$ is maximal if and only if $R/\mathfrak{p}$ is a field. The first statement is simple, the second ...


3

What's an element of $R[x]$? It's a finite sum that looks like this: $$a_0+a_1x + \cdots + a_n x^n,$$ where the $a_i \in R$. If we forget about the variable $x$ and note that all that really matters in this description is the coefficients $a_i$ and the order they appear in, we realize that this corresponds in a bijective fashion to an ordered tuple (with ...


2

A module is an abelian group, and so every subgroup (and therefore any submodule) is a normal subgroup. Further, it's easy to show that scalar multiplication is well-defined in the quotient module, and so we don't need any added conditions. For your second question, that condition (that only finitely many $x_i$ are nonzero) is basically saying it contains ...


7

Please see the following paper by Bhargava and Satriano here, which contains the details of this computation and other related results. M. Bhargava and M. Satriano, On a notion of “Galois closure” for extensions of rings, Journal of the European Mathematical Society 16, 1881-1913 (2014). The aim of the authors is to define for commutative ring ...


0

Let $q$ be a fixed prime number and let $S$ be the multiplicative system $S = \mathbb Z \setminus q \mathbb Z$. We have $\prod_{p \in \mathbb P} S^{-1}(\mathbb Z/p\mathbb Z) = \mathbb Z/q\mathbb Z$ but in $S^{-1}\prod_{p \in \mathbb P}\mathbb Z/p\mathbb Z$ all elements of the form $(a,a,a, \dotsc)$ are non-zero, i.e. $S^{-1}\prod_{p \in \mathbb P}\mathbb Z/...


0

See this MO answer: http://mathoverflow.net/a/5906/6427 In particular the path algebras that Mariano describes are very concrete.


0

It's not true when $I$ is infinite, due exactly to the problem you encounter. Consider the case where $R = \mathbb{R}$, $N$ is an infinite dimensional vector space, and each $M_i$ is $\mathbb{R}$, with $I$ infinite. Convince yourself that $\operatorname{Hom}(N,\bigoplus_i M_i)$ corresponds to infinite matrices whose columns each have finitely many nonzero ...


1

Additionally, you will see in category theory (if you go down that road) that direct products and direct sums satisfy different universal mapping properties: https://en.wikipedia.org/wiki/Direct_sum_of_modules As you notice in the Wiki page, the direct sum of modules is a $coproduct$, meaning it satisfies the universal mapping property $opposite$ of that of ...


1

Once you find $\,x^2\equiv 1\iff x\equiv \pm1 \pmod 7\,$ and $\,x^3\equiv 1\iff x\equiv 1,4,7\pmod{9},\,$ you can lift each possible combination of values using the Chinese Remainder Theorem (CRT). It's usually quicker to do it generically first, i.e. solve $\,x\equiv a\pmod 7,\,$ $\,x\equiv b\pmod 9,\,$ then substitute all possible combinations of the ...


0

If you solve $x^2-1=0$ modulo $7$, you get $x=\pm 1$, just by factorization. Since the integers modulo $7$ form a field, you can use all of the algebra manipulations that you're accustomed to, as long as you interpret multiplicative inverses correctly. Modulo $9$, the situation is a little different, because the integers mod $9$ do not form a field. There ...


1

There are many differences between the two, for example you can prove yourself that the ring $\prod_{n\geq 1}\mathbb{Z}_{p^n}$ is not Noetherian. Since each $\mathbb{Z}_{p^n}$ is finite it is trivially Noetherian, we see that the property of being Noetherian is not preserved under direct products whereas it is preserved under direct sums. In some sense, ...


0

To remove this from the unanswered queue I shall enter user74230's comment from above. I wish he/she had taken rschwieb's advice to enter this as an answer so they could receive their just rewards. "No, this is false over every PID that is not a field: let $M=R\oplus(R/aR)$ for a nonzero nonunit $a\in R$ and let $L$ be the first summand (consisting of ...


1

Here there is a proof of equivalence of two definitions of Yetter-Drinfeld modules. Defining Yetter-Drinfeld modules on Hopf algebra using antipodal map $S$ is more natural, because there is a lot of properties of Yetter-Drinfeld modules related to properties of $S$ - for example in this article you can find some of them.


1

This is indeed not true. As you observe, $T(V\oplus W)$ is the coproduct of $T(V)$ and $T(W)$ in the category of $K$-algebras, which is typically larger than $T(V)\otimes_K T(W)$. For a very simple example, take $V=W=K$. Then $T(V)$ and $T(W)$ are both polynomial rings in one variable over $K$; write $T(V)=K[x]$ and $T(W)=K[y]$. Then $T(V)\otimes_K T(W)\...


3

Welcome! In general, please provide context to your question and let us know what you have already tried. Hint to the exercise: A module over $A$ is the same as a $k$-vector space $V$ together with a $k$-linear endomorphism $f: V\to A$ such that $f^n=0$. Now apply what you know about normal forms of nilpotent endomorphisms of finite-dimensional vector ...


2

A standard basis for $S^2(\mathbb{Z}^3)$ is given by $(e_1e_1,e_1e_2,e_1e_3,e_2e_2,e_2e_3,e_3e_3)$, where juxtaposition denotes the symmetric product and $(e_i)_{i=1}^3$ is the standard basis of $\mathbb{Z}^3$. A symmetric integral $3 \times 3$ matrix is uniquely determined by $6$ integer entries which lie on the coordinates $(1,1),(1,2)(1,3),(2,2),(2,3),(3,...


4

This is somewhat similar to one of your previous questions. Note that $\mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}(X)[T]/\langle T^{2}- X\rangle$. This polynomial is irreducible by Eisenstein's criterion, and is separable if $p \neq 2$. In this case, we have $$\mathbb{F}_{p}(\sqrt{X}) \otimes_{\mathbb{F}_{p}(X)} \mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}...


1

The ring $\mathbb Z[\zeta]$ is called the ring of cyclotomic integers.


7

I don't know if you consider this a special name, but the ring $\mathbb{Z}[\zeta]$ is the ring of integers for the number field $\mathbb{Q}(\zeta)$. This is (IMO, anyway) a nontrivial fact, but you can find its proof in Neukirch (see below), or (for the case $n$ prime) in Samuel's ''Algebraic Theory of Numbers''—this link might also be helpful to you. ...


1

Even with the extra condition that $\mathcal{A}$ is a Grothendieck category, it may still have no simple objects. I think the following is the easiest example I know. Let $R$ be a (necessarily non-noetherian) commutative local ring with non-zero maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=\mathfrak{m}$. Let $\mathcal{C}$ be the category of $R$-...


0

How about checking that $X:=(R^{ \oplus A_1 })^{ \oplus A_2 }$ satisfies the universal property of the free $R$-module on $A_1 \times A_2$? Define $i:A_1\times A_2 \to X$ by $i(a_1,a_2) = \alpha_{a_1,a_2}$, where the mapping $(\alpha_{a_1,a_2}(b_2))(b_1) := 1$ if $a_i = b_i$ and vanishes otherwise. Note that this lets us decompose any $\alpha \in X$ as a ...


1

Let $f \in R^{\oplus(A_{1} \times A_{2})}$. Define a map $\varphi_{f} \colon A_{2} \to R^{\oplus A_{1}}$ by $\alpha \mapsto h_{f}$, where $h_{f} \in R^{\oplus A_{1}}$ is the map given by $h_{f}(\beta) = f(\beta, \alpha)$. Now, define a map $\Phi \colon R^{\oplus(A_{1} \times A_{2})} \to (R^{\oplus A_{1}})^{\oplus A_{2}}$ by $\Phi(f) = \varphi_{f}$. You ...


0

Have you ever heard of restriction of scalars? It works as follows: if $f:R\longrightarrow S$ is a homomorphism of rings then any left (right) $S$-module $N$ inherits a left (right) $R$-module structure if we define $$r\cdot x:=f(r)\cdot x\quad (x\in N, r\in R).$$ Now, given a left $R$-module $M$ and a left $S$-module $N$ then we can define a homomorphism ...


3

Other that simplifying into $\bigwedge R=R\oplus R$, you are completely right. You can also go for a different approach, by defining the degree$-n$ terms as follows. Namely by $\bigwedge^nM=M^{\otimes n}/N$, where $N$ is the submodule generated by the dublicate terms. Then we have $\bigwedge M=\bigoplus_n \bigwedge^n M$. From this definition, it is ...


1

Let's ask when, given a module $M$ with a proper nonzero submodule $L$, the set $(M\setminus L)\cup\{0\}$ is a submodule. Let $x\in M\setminus L$ and let $y\in L$, $y\ne0$. Then also $x+y\in M\setminus L$ and $-x\in M\setminus L$; on the other hand $$ (x+y)+(-x)=y\notin (M\setminus L)\cup\{0\} $$ Therefore the set is not even an additive subgroup. So your ...


4

Consider $X = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Then both $(1,1)$ and $(-1,0)$ are in $(X \setminus TX) \cup \{0\}$, but their sum isn't. In fact let $R$ be any ring that has a module $M$ containing a nonzero torsion element $m$. Then in $R \times M$, both $(1, m)$ and $(-1,0)$ are torsion-free, but their sum isn't. In other words, there is no ...


10

In general there is no reason for $f$ to be an $R$-module homomorphism just because it is an abelian group homomorphism. Consider complex conjugation $\bar{\cdot} : \mathbb{C} \to \mathbb{C}$. It is clearly an abelian group morphism since $\overline{z+z'} = \bar{z} + \bar{z}'$. But if you take $R = \mathbb{C}$, it's clearly not an $R$-module homomorphism, ...


1

Let $p$ be a prime. The Prufer $p$-group $\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ is an Artinian $\mathbb{Z}$-module, but is not a Noetherian $\mathbb{Z}$-module.


2

An obvious comment that nobody has mentioned: If you pass to a higher universe, then $\mathrm{Mod}(R)^\mathrm{op}$ is small, and so the embedding theorem tells us that it embeds exactly into some $\mathrm{Mod}(S)$. The catch is that (with respect to the original universe) $S$ will be a large (i.e. proper-class-sized) ring. You don't actually need the ...


0

Firstly, I want to thank Jyrki Lahtonen, Arctic Tern and mathreadler for elevating my understanding. The key reason of my confusion was usage of left matrix on vector multiplication in a right module $\mathbb{H}^n$. As it turns out in right modules it would safe only to use corresponding matrix ring acting on the module from the right. Hence, if we assume ...


0

Find the rational canonical form basis $\{v_1, ~v_2, ~T(v_2), ~T^2(v_2), ~T^3(v_2)\}$ for the $\Bbb{Q}$-vector space $\Bbb{Q}^5$. Then $\{v_1, v_2\}$ is a generating set for the $\Bbb{Q}[x]$-module $\Bbb{Q}^5$.


6

Sure, the keyword is "multivariable Chinese remainder theorem." A nice exposition can be found here. The general theorem states that if all mods $m_i$ are relatively prime and each row $i$ has an $a_{ij}$ that is relatively prime to $m_i$, then there's going to be at least one solution. The mechanical way of solving the system can be done by row reduction, ...


0

In the first part of your proof, $V$ being irreducible does not necessarily imply that $y$ has only one Jordan block. You have to consider $x$ also. Here I give an elementary proof and hope it makes sense. Let $V$ be an irreducible representation of $A$. Let $\lambda$ be an eigenvalue of $y$ and $\xi\in V$ an eigenvector. We contend that $\xi,x\xi,\cdots,x^{...


3

For Part (b), according to i707107's answer, the $(i,j)$-entry of $\textbf{H}_n:=\big(\textbf{A}_n(0)\big)^{-1}$ is given by $$h_{i,j}:=(-1)^{i+j}\,n\,\binom{n+i-1}{i-1}\,\binom{n-1}{j-1}\,\binom{n+j-1}{i+j-1}\,\binom{i+j-2}{i-1}\,.$$ Hence, $n$ is a divisor of the greatest common divisor $g_n$ over $\mathbb{Z}$ of the entries of $\textbf{H}_n$. Note that $$\...


2

First of all, we do not even have an isomorphism between $S\otimes_R f^*S = S\otimes_R S$ and $S$ as left $S$-modules (take for example $S$ to be a field and $R$ to be a subfield to see this). The counit is fortunately very easy to describe. We have the map from $S\otimes_R S$ to $S$ that sends $x\otimes y$ to $xy$, and as you noted, we can identify $f_!f^*...


6

This is a solution to (d), and partially for (b). We use Cauchy Matrix, and its evaluation of inverse given by Schechter: If $T$ is a $n\times n$ Cauchy matrix on the sequences $\{x_i\}$, $\{y_j\}$, then $S=T^{-1}=[s_{ij}]$ is given by: $$s_{ij} = (x_j - y_i) A_j(y_i) B_i(x_j) $$ where $$A_i(t) = \frac{A(t)}{A^\prime(x_i)(t-x_i)} \quad\text{and}\quad ...


1

Suppose $v=(x,y,w,z)$ is any vector. Observe that then $A$ sends $v$ to $(-y,x,-z,w)$, $B$ sends $v$ to $(-w,z,-x,y)$. Moreover $A^2=B^2=-1$ and $AB=-BA$ sends $(x,y,w,z)$ to $(z,w,y,x)$. Thus $\{1,A,B,AB=C\}$ is a basis of your algebra, which is in particular four dimensional. For the record, the relations we have are $$\begin{align*} A^2&=-1\\ B^2&...


1

You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated. Set $R=k[x]/(x^n)$. To solve your problem let us prove the following: Let $M$ be free a $R$-module. If $B\subseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$. As $...


1

It's just as easy to prove the more general result that for any $m \in M$, $0_R m = 0_M$. Proof: $$\begin{align} 0_R m &= (0_R + 0_R) m \\ &= 0_R m+ 0_R m \end{align}$$ Now add $-(0_R m)$ to both sides, and get $$0_R m + -(0_R m) = 0_R m + 0_R m + -(0_Rm)$$ $$0_M = 0_R m + 0_M$$ $$0_M = 0_R m$$ However, as noted in Aloizio Macedoo's comment on the ...



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