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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, ...


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. ...


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 ...


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 ...


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, ...


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}...


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 ...


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 $$\...


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 ...


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 ...


3

Finite dimensional algebras have integrals, and using one you can show selfinjectivity at once —in fact, they are Frobenius. See for example the Lectures on Hopf algebras by Schneider, which Google will find for you, or Susan Montgomery's book in Hopf algebras.


3

Recall that if $M$ is a module over a ring $R$, then: $$\mathrm{ann}(M)=\{r\in R\mid \forall_{m\in M}rm=0\}$$ This is two sided ideal of $R$ called anihilator of $M$. Now if $M_1\cong M_2$ as $R$-modules, then: $$\mathrm{ann}(M_1)=\mathrm{ann}(M_2)$$ Annihilator of $\mathbb{C}[x,y]/(x,y)$ is $(x,y)$ and annihilator of $\mathbb{C}[x,y]/(x-1,y-1)$ is $(x-1,y-1)...


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 ...


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 $a=\tilde g(1)$.


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 ...


2

Since $\Bbb Q$ is a field, a module over it is also a vector space. Vector spaces always permit a basis (unless we are denied the Axiom of Choice). Hence every finitely generated module over $\Bbb Q$ is isomorphic to some $\Bbb Q^n$ with $n\in\Bbb N_0$. However, there are of course also infinte-dimensional vector spaces, i.e., for any set $A$, the set $\Bbb ...


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^*...


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,...


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 ...


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 ...


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 ...


1

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


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 ...


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 ...


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.


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

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 ...


1

All unitary modules over $\mathbb{Q}$ are $\mathbb{Q}$-vector spaces. There are non-unitary modules over $\mathbb{Q}$. In fact, every module over $\mathbb{Q}$ can be written as $V\oplus N$, where $V$ is a $\mathbb{Q}$-vector space and $N$ is a trivial module (i.e., an abelian group such that $q\cdot x=0$ for all $q\in\mathbb{Q}$ and $x\in N$). However, ...


1

The adjunction is the same in the non-commutative case. To construct it quickly, use the tensor-hom adjunction (which you'll certainly find stated and proven for non-commutative rings too). On one side, you can see that the restriction $f^*$ is obviously isomorphic to $f^*S ⊗_S -$, where $f^*S$ is $S$ equipped with the $(R, S)$ structure you mentioned, so ...


1

In fact, what happens if we only define a "subset" of $M$ as $S=\{x\in M|l_R(x)\neq 0\}$? You can certainly define this set, and it could be considered a set of "torsion elements" inside $M$, but the set does not have as many nice properties in general. As in the example of $\mathbb Z/6\mathbb Z$ given above, $2$ and $3$ are torsion but $3-2=1$ is not, so ...



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