Let $R$ be a euclidean domain. Let $E_n(R)$ be the subgroup of $GL_n(R)$ generated by all matrices of the form $I + λ$ where $λ$ is a matrix with precisely one non zero entry and this entry does not occur on the diagonal and $I$ is the $n × n$ identity matrix. Show that $SL_n(R) = E_n(R)$.
Proving $SL_n(R) \geq E_n(R)$ is easy. For the other direction, I am thinking I wold have to use Smith normal forms or Jordan canonical forms. For Smith Normal forms, since $R$ is a Euclidean fomain, it is PID. we have that (without loss of generality) that any matrix $M \in SL_n(R)$ is $M = \mathrm{diag} (d_1,...,d_r,0,...,0)$ for some $r$. Because the determinant is $1$, we conclude that $r=0$ and $d_i = 1$ for all $i$. Hence the conclusion.
This seems way too simplistic. Am I missing something?