# Tag Info

7

Here's why I find it confusing: if we write $\mathbf{Q}_p/\mathbf{Z}_p = \lim_{\rightarrow n }\mathbf{Z}/p^n \mathbf{Z}$ and use the fact that tensor products commute with direct limits, we should get that $$T \otimes_{\mathbf{Z}} \mathbf{Q}_p/\mathbf{Z}_p \cong p\text{-Sylow subgroup of }T,$$ which is certainly not always zero. It's not true that the ...

3

This is true if $M$ is reflexive, which is the case if $M$ is maximal Cohen-Macaulay, as you assume $R$ to be Gorenstein. Indeed, without loss of generality, we may pass to the completion to suppose $R$ is complete. By Matlis duality, $M \otimes_R E(R/\mathfrak{m})$ is indecomposable if and only if $\operatorname{Hom}_R(M \otimes_R E(R/\mathfrak{m}),E(R/\... 3 In terms of extra structure, there is a bilinear map$V\times W \to V\otimes W$. However, if$F^n$has basis$e_1,\ldots,e_n$and$F^m$has basis$f_1,\ldots,f_m$then you can just take the standard basis of$F^{mn}$and label the basis elements$g_{ij}$for$i=1,\ldots,n$and$j=1,\dots,m$. Then you can define the bilinear map to send$(e_i,f_j)\mapsto g_{...

2

This definition is unfortunately rather poorly behaved for infinite index sets, and in particular is not associative. For instance, $R$ ought to be a "unit" for the tensor product, but a tensor product of infinitely many copies of $R$ is typically much larger than $R$. Indeed, $\bigotimes M_i$ is can be constructed explicitly by starting with the ...

2

This is checked directly. Define $\gamma:M_n(A)\to A\otimes M_n(\mathbb C)$ by $$\gamma:[A_{kj}]\longmapsto \sum_{kj}A_{kj}\otimes E_{kj}.$$ It is trivial to verify that this is a $*$-isomorphism.

2

No. Take two free modules of rank $2$.

2

By Chinese Remainder, $$\mathbb C[x]/(x^2+x)\cong \mathbb C[x]/(x)\oplus\mathbb C[x]/(x+1)\simeq \mathbb C\oplus \mathbb C$$ $$\mathbb C[x]/(x^2-1)\cong\mathbb C[x]/(x+1)\oplus \mathbb C[x]/(x-1)\simeq \mathbb C\oplus\mathbb C$$ These decompositions are $\mathbb C[x]$-module decompositions, therefore $\mathbb C[x^2]$-module decompositions too. Let $\mathbb ... 2 $$(A \otimes_k K) \otimes_K (B \otimes_k K) \cong (A \otimes_k K) \otimes_K (K \otimes_k B)\cong A \otimes_k (K \otimes_K K) \otimes_k B \cong A \otimes_k K \otimes_k B \cong (A\otimes_k B) \otimes_k K$$ 1 Given a complex vector space$V$, let me write$\overline{V}$for$V$with its scalar multiplication replaced by its complex conjugate (that is,$\alpha\cdot_\overline{V}v=\overline{\alpha}\cdot_V v$). Then the "direct multiplication" of$V_1,\dots,V_n$is just the space of multilinear maps$\overline{V_1}\times\dots\times \overline{V_n}\to\mathbb{...

1

I don't think your proof is really complete. It is true that it follows from the uniqueness of the expansion, but there are steps in between, such as in your link. In that case the map $b$ is well defined because of the uniqueness of the representation. The second part of $a$ follows immediately from the first since \sum_{i=1}^n m_i\otimes e_i = 0 = \...

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