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7

Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$. 2) If ...


5

Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to ...


4

Take $R$ to be a field $k$ and take $A = B = k^2$, with basis $e_1, e_2$. The tensor product $A \otimes_k B$ is $k^4$ with basis $e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, and most elements of it are indecomposable. For example, $e_1 \otimes e_1 + e_2 \otimes e_2$ is indecomposable. There's no more reason to expect all tensors to ...


3

when are ideals also rings with unity? Proposition: An ideal $I\lhd R$ will be a ring with identity iff there exists a central idempotent $e$ such that $eR=I$. Proof: ($\implies$) The identity of $I$, call it $e$, is an idempotent element of $R$ and satisfies $I=eI$. Then $I=eI\subseteq eR\subseteq I$, so $I=eR$. Since $e\in I$, we have ...


2

You're absolutely right that (ii) does not obviously follow from (i) as stated. What (i) should say is not just that $I_i$ is ring-isomorphic to $M(n_i,F)$ but that it is isomorphic to $M(n_i,F)$ as an $F$-algebra. Concretely, this means that if you take an element $a\in F$, consider it as an element of $FG$, project it to $I_i$, and then map it to ...


2

You are right: $P(A,t)=(1-t)^{-s}$, and this can be proved by induction on $s$. For the Hilbert polynomial just count the number of monomials of a fixed degree. For the characteristic polynomial note that the associated graded ring of $A$ with respect to $Q$ is nothing but $A$. Then the characteristic polynomial is $\binom{X+s}{s}$.


2

For $s\in U$ and $x\in M$ we have $sx=0\implies sx\in Q$, where $Q$ is a primary submodule of $M$ which appears in a primary decomposition of $(0)$ and $r_M(Q)\cap U=\emptyset$. Conclude that $x\in Q$. For the converse, $x\in\cap Q_j$ with $r_M(Q_j)\cap U=\emptyset$. On the other side, for some $Q_i$ such that $r_M(Q_i)\cap U\ne\emptyset$ we get an $s_i\in ...


2

Given a prime $p$, the Prüfer $p$-group is the direct limit of the cyclic groups of order a power of $p$. If $\Omega$ is an infinite set, the group of finitary permutations on $\Omega$ is the direct limit of all groups of permutations on the finite subsets of $\Omega$.


2

Direct limit is generalisation of the notion of union of a family of sets. Two other examples: Lazard's theorem in commutative algebra asserts that a flat $R$-module is a direct limit of free $R$-modules. Germs of continuous function at a point $a$ of a topological space is defined as the direct limit of the system of pairs $(U,f)$, where $U$ is an open ...


2

The "finitely generated" isn't really important here. You have to understand what are free modules. A $R$-module $M$ is free if it has a basis, i.e., there exists $\{x_i\}_I \subset M$ (where $I$ can be chosen the be finite in the finitely generated case) such that the $x_i$'s form a $R$-basis of $M$. This is a very special property, and every free module ...


1

Let $R=k[X,Y]$. We have $H_{R/I}(t)=1+t$, and $H_{R/J}(t)=1+2t+t^2$. On the other side, their Hilbert polynomial is $0$.


1

The notation here is a little awkward, so I'm going to adjust it a bit. You have that $V$ is a $\mathbb{C}G$-module, so for each $v\in V$ and $g\in G$, you should know what $g.v$ means (and similarly for $g.w$, $w\in W$). Now, starting from a linear transformation $L:V\to W$ and $x\in V$ we can compute $g^{-1}.L(g.x)$ (that is, act on $x$ with $g$, map it ...


1

The first step is to adapt the proof that $$End_R(V_1\oplus V_2)\cong End_R(V_1)\oplus End_R(V_2)$$ to the case of arbitrary $V_1$ and $V_2$ satisfying $Hom_R(V_1,V_2)=0$. Then, prove the following: For $\phi\in\mathrm{End}_R(S^n)$, define $$\phi_{ij}=\pi_j\circ\phi\circ \iota_i$$ where $\pi_j:S^n\to S$ is the projection onto the $j$th factor and ...


1

If $R$ is a ring then clearly is generated as a $R$ Module by $1\in R$. Considering $R^n$ the direct sum of n copies of $R$ we know that there are canonical injections of $i_i: R \hookrightarrow R^n$ sending $r \mapsto (0,0,..r,..)$. We denote $e_i=i_i(1)$ with this is clear that the set $\{e_i\}_{i=1}^n$ generates $R^{n}$ For your question, since you have ...


1

A direct sum of $FH$-modules is also a direct sum of vector spaces, so it is enough to consider a single term $M \otimes t_i$ in the decomposition. Now $M \otimes t_i = \{ m \otimes t_i : m \in M \}$. So, if $a_1,\ldots,a_n$ is a basis of $M$ as a vector space over $F$, then every element of $M$ can be written uniquely as $\sum_{i=1}^n f_i a_i$ with $f_i ...


1

Here's a counterexample. Let $k$ be a field and take $A=B=k[x]/(x^2)$ and $M=C=k$ (considered as an $A$-algebra via $x\mapsto 0$). Then $\mathrm{Hom}_B(M,B)\cong k$ (spanned by $1\mapsto x$), so $\mathrm{Hom}_B(M,B) \otimes_A C\cong k$ as well. But $1\mapsto x$ maps to $0$ in $\mathrm{Hom}_B(M,B \otimes_A C)$, since $x=0$ in $B\otimes_A C=C$. On the ...



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