Norton Equivalent Circuit of Ideal Independent Voltage source
An ideal voltage source can provide unlimited current, so it cannot be replaced with an ideal current source of any specified value. No matter what load current is drawn, the terminal voltage is always equal to the open-circuit voltage in this ideal case. So there is no meaningful value of short-circuit current, and Norton can't be applied.
A practical voltage source has some non-zero source resistance, so in that case it is possible to calculate a Norton Equivalent current source. The resistor in the Thevenin model is actually required, it can't just be zeroed out.
Division by zero is not a meaningful operation, because it leads to ambiguous or inconsistent results. Whenever you have a formula that has division by zero, more analysis is required.
See https://math.stackexchange.com/questions/26445/division-by-0
Instead of dividing by R=0, use an infinitesimally  small R value and consider the Limit as R approaches 0. This models the practical voltage source approaching ideal. The product IR must equal V, so whichever non-zero R value is selected, it is always possible to calculate a specific I = V/R value that satisfies the Thevenin-Norton equivalence. However, there is a discontinuity at R=0. There is no meaningful way to produce an equivalent Norton current source for R=0.