What characterizes in general a basic solution is the presence of hydroxide ions, when adding a base to the water, the base B takes from the water molecule a proton H+, thus obtaining BH+ and OH-
B + H2O → BH+ +OH-
The same thing occurs with NH3: NH3 + H2O → NH4+ +OH-
but with NaOH it's different, we don't represent the water molecule in the equation, we just write:
NaOH → Na+ +OH-
What is the role of the water in this case, I know there's a phenomen of dilution, but there must be a proton exchange?
Thank you

You can safely say that the OH^- from NaOH reacts with the H^+ cations in water.:

\text{OH}^-\ +\ \text{H}^+\ \rightarrow\ \text{H}_2\text{O}

There is therefore an excess of HO^- and less H^+.



However, if you know about ionic equilibria and Le Chatelier's Principle this would be the better explanation:

\text{H}_2\text{O}\ \rightleftharpoons\ \text{H}^+\ +\ \text{OH}^-

The newly added OH^- react with the H^+, hence shifting the balance to the right and having more OH^- than H^+.

Same thing would happen with NH3, with NH3 reacting with the H^+ and shifting the balance to the right, thus making more OH^- ions and removing H^+ ions.