# Why RSA must make sure that $\gcd(e,\varphi(n)) = 1$ instead of $\gcd(e,n) = 1$?

By Euler's theorem, if $$\gcd(e,n) = 1$$, then $$e^{\varphi(n)} \equiv 1 \pmod n$$. But why does RSA need to make sure that $$\gcd(e,\varphi(n)) = 1$$?

$$m^{ed} \equiv m^1 \equiv m \pmod n$$
But why? Since $$ed \equiv 1 \pmod{\varphi(n)}$$
If $$\gcd(e,d) \neq 1$$ then $$e$$ and $$d$$ are not coprime then $$ed \not \equiv 1 \pmod{\varphi(n)}$$. So $$e$$ must be coprime with $$\varphi(n)$$ to have a modular multiplicative inverse.
Also $$\gcd(e,n)=1$$, since the public key is presented as $$(e,n)$$. Then you trivially can compute a factor of $$n$$ is $$e$$ is not coprime.