jmc

We start subtract the rightmost digits, keeping in mind that we are in base $5$.

Since $1$ is less than $4$, we must borrow $1$ from the $2$, which then becomes $1$. Since $11_5-4_5=2_5$, we have $2$ in the rightmost digit. Since the $1$ left over is less than $3$, we must borrow $1$ from the $3$, which becomes a $2$. Next, $11_5-3_5=3_5$, so we have $3$ in the second rightmost digit. Since $2-2=0$, the third digit is 0. We subtract $1$ from $4$ to get $3$ for the fourth digit. In column format, this process reads \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & &4 & \cancelto{2}{3}& \cancelto{1}{2} & 1_5\\ & -& 1 & 2 & 3 & 4_5\\ \cline{2-6} & & 3 & 0 & 3& 2_5\\ \end{array}The difference is $\boxed{3032_5}$.