$Ext_{\mathbb{Z}/2}^1(\mathbb{Z}/2,\mathbb{Z}/2)=0$?

Question

I am trying to find an example of a situation in which we obtain different $Ext$ for different base rings.

I am very new to this. I have found how to prove $Ext_{\mathbb{Z}}^1(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$.

But I do not know how to prove that $Ext_{\mathbb{Z}/2}^1(\mathbb{Z}/2,\mathbb{Z}/2)=0$.

It would be very useful if someone could explain how to do this and how the base ring is important in that calculation.

Thanks in advance.

Answer 1

If $A$ is a projective module, then $\mathrm{Ext}^n(A,-)$ vanishes in all positive degrees. This immediately follows from the definition of $\mathrm{Ext}$ which starts with a projective resolution of $A$. This applies to your example.