About the tensor product identity: $A=B \otimes C = B \otimes I + I \otimes C$

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I am reading about Chern classes in Nakahara's Geometry, Topology and Physics, and am having trouble understanding the equation $$ A=B \otimes C = B \otimes I + I \otimes C \tag{1}$$ where $A,B,C$ are matrices. I understand that $$ (A \otimes B)(C \otimes D) = AC \otimes BD \tag{2}$$ and so would think that (1) should be $$ A=(B \otimes I)(I \otimes C)$$ The equation (1) appears in deriving

\begin{align} \tag{3} ch(B \otimes C) &= \sum_j \frac{1}{j!}\left(\frac{i}{2\pi}\right)^j tr(B \otimes I + I \otimes C)^j \\&= \sum_j \frac{1}{j!}\left(\frac{i}{2\pi}\right)^j \sum_{m=1}^j \binom{j}{m}tr(B^m)tr(C^{j-m}) \\&= \sum_m \frac{1}{m!} tr \left(\frac{iB}{2\pi} \right)^m \sum_n \frac{1}{n!} tr \left(\frac{iC}{2\pi} \right)^n = ch(B)ch(C) \end{align}

I thought perhaps $I$, and $+$ represented something other than the identity matrix and addition, but the second equality in (3) requires use of (2), the binomial expansion, and $tr(A \otimes B)=tr(A)tr(B)$, which makes me certain that $I$ is the identity matrix, and $+$ does indeed mean addition. I also tried writing out (1) explicitly for a $2\times 2$ matrix and it doesn't match up.

Any help would be appreciated, thanks.