# Tag Info

## Hot answers tagged tensor-products

3

Note that not all elements of $(U \otimes V) \otimes W$ are simple tensors. This is the problem you are encountering, as your map isn't defined for elements like $(u_1 \otimes v_1) \otimes w_1 + (u_2 \otimes v_2) \otimes w_2$. You can fix this by mapping each simple tensor as above and then extending the map linearly on all elements. This trivially gets ...

3


1

It probably cant get better than the author of the article answering himself but here a small addition that also answers the question and gives a direct interpretation of this Tor, namely it counts something. Since we do homological algebra, we can assume that the algebra is basic. As in Jeremy Rickards answer one has $Tor_d^A(M,A/radA)=DExt_A^d(M,D(A/radA))$...

1

One short way of getting the desired isomorphism is \begin{align*} (K \otimes A) \otimes_K (K \otimes B) &\cong (A \otimes K) \otimes_K (K \otimes B) \\ &\cong A \otimes (K \otimes_K (K \otimes B)) \tag{$\ast$} \\ &\cong A \otimes K \otimes B \\ &\cong K \otimes A \otimes B \\ &\cong K \otimes (A \otimes B) ...

1

The correct symbol $\otimes$ is produced with \otimes. A strategy is to first find the matrix of $T$ taken with respect to the basis $\mathcal{B}$, and then convert it to the standard basis. With respect to $\mathcal{B}$, it is clear that $$[T]_{\mathcal{B}} = \begin{pmatrix} 0 & 1 \\-1 & 2 \end{pmatrix}.$$Now, we use the tensor transformation law (...

1

You probably meant $\otimes$ instead of $\oplus$. Try using the fact that the matrix elements of a tensor $M=a\otimes b$ in the canonical basis are $M_{ij}=v_i\cdot Mv_j = (a\cdot v_i)(b\cdot v_j)$ where $\cdot$ is the standard dot product and $v_k$ are the basis vectors.

Only top voted, non community-wiki answers of a minimum length are eligible