I've took the opportunity to join the community, because I didn't find a satisfying explanation to the following fact.
Let $S^2$ the 2-sphere, let $H$ the tautological line bundle. Assume that $$ K(S^2) \approx \mathbb{Z}[x]/(x-1)^2$$ (thanks to the map $x \mapsto H$)
Then for every source I've found over the net, it's obvious that $\tilde{K}(X)$ is generated as an abelian group from $H-1$.
I can't prove this fact, according to me, if $K(X)$ has as additive basis {1,H} (because i'm killing every factor of degree $\geq 2$ right?), then thanks to the relation $$\tilde{K}(X) \approx K(X)/\mathbb{Z}$$ I'm tempted to conclude that I'm killing the $1$ of the additive basis and so I have only $H$ as a generator. But it seems that it is not the case. Thanks for every advice!
(I've tagged the questions as K-theory, if more tags are needed feel free to edit)
The equation you write $\tilde{K}(X)\cong K(X)/\mathbb{Z}$ is confusing, it should be
$$K(X)=\mathbb{Z}\oplus\tilde{K}(X)$$ its a small thing but I think its the source of your difficulty. $\tilde{K}(X)$ is the kernel of the degree map
$$\deg:K(X)\rightarrow \mathbb{Z}$$ now $H$ has dimension $1$ so its not in the kernel, but $H-1$ has dimension zero so it is in the kernel so in fact the decomposition reads,
$$K(S^2)=\mathbb{Z}\oplus \langle H-1\rangle$$