Background: Consider the ring $R=\mathbb{Z}[\omega]=\{a+b\omega:\omega=\sqrt{-5},a\in\mathbb{Z},b\in\mathbb{Z}\}$, which is also a $\mathbb{Z}[\omega]$-module. Let $I=(1-\omega,2), J=(1+\omega,3)$ be two ideals in $R$ generated by the specified elements.
Question: I believe that in the tensor $R$-module, $I\underset{R}\otimes J$, the following statement is true:
$$(1-\omega)\underset{R}{\otimes}(1+\omega)\not=2\underset{R}{\otimes}3.$$
And I am looking for a bilinear mapping defined on $I\times J$ which maps $(1-\omega)\underset{R}{\otimes}(1+\omega)$ and $2\underset{R}{\otimes}3$ to distinct values. If such a map can be found, then by the universal property of tensor product, certainly $(1-\omega)\underset{R}{\otimes}(1+\omega)\not=2\underset{R}{\otimes}3$. Could anyone suggest such a bilinear map? Or is there a better way to determine whether the above two simple tensors are different?
Update: This question was inspired by Exercise $10.4.21$ in Abstract Algebra, 3rd edition, by Dummit and Foote. A counterexample in which the map $I\otimes J\simeq IJ$ sending $i\otimes j$ to $ij$ fails to be injective can be found in Exercise $10.4.17$ in the same book.
Best Answer
We have:
$$2 [(1 - \omega) \otimes (1 + \omega)] = 2(1 - \omega) \otimes (1 + \omega) = 2 \otimes (1 - \omega) (1 + \omega) = 2 \otimes 6 = 2 [2 \otimes 3].$$
Similarly,
$$3 [(1 - \omega) \otimes (1 + \omega)] = (1 - \omega) \otimes 3 (1 + \omega) = (1 - \omega) (1 + \omega) \otimes 3 = 6 \otimes 3 = 3 [2 \otimes 3].$$
Therefore, subtracting the two gives:
$$(1 - \omega) \otimes (1 + \omega) = 2 \otimes 3.$$
A possible brute force method to solve the problem would be:
First, $I$ is generated as an Abelian group by $1-\omega, \omega(1-\omega) = 5 + \omega, 2, 2\omega$. Now, you could find the relations between these elements by finding the kernel of the matrix $A = \begin{bmatrix} 1 & 5 & 2 & 0 \\ -1 & 1 & 0 & 2 \end{bmatrix}$ (as a subgroup of $\mathbb{Z}^4$). One way to do this would be to use a Smith normal form calculation to write $A = P \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix} Q$ for invertible matrices $P, Q$, and then the kernel of $A$ would be generated by $Q^{-1} e_3, Q^{-1} e_4$.
If I'm not mistaken (though I very well could be), this should give presentations $I \simeq \langle a, b \mid 2a = (1 - \omega) b, (1 + \omega) a = 3 b \rangle$ and $J \simeq \langle c, d \mid 3c = (1 + \omega) d, (1 - \omega) c = 2d \rangle$ for $I$ and $J$ as $R$-modules. Therefore, $I \otimes_R J$ would have a presentation in terms of 4 generators $a \otimes c, a\otimes d, b\otimes c, b\otimes d$, and eight relations $2a\otimes c = (1-\omega)b \otimes c, \ldots, (1-\omega)b\otimes c = 2b\otimes d$. You now want to know whether $a\otimes c - b\otimes d$ is zero in this tensor product, which is equivalent to asking whether $a\otimes c - b\otimes d$ is in the submodule of $R^4$ generated by these relations. That question should be straightforward if tedious to answer by a Smith normal form calculation on an $8 \times 16$ matrix (using that the tensor product is generated as an Abelian group by $a\otimes c, \omega a\otimes c, \ldots$ and similarly the relations are given by the original relations along with the relations times $\omega$).