# Euler characteristic of projective plane (rank$(H_1(RP^2))=$ rank$(Z_2)=0?$)

Let $$K$$ denote the $$2$$ dimentional projective plane $$RP^2$$ . Then we have $$H_0(K) \cong Z$$ $$H_1(K) \cong Z_2$$ $$H_2(K)=0$$
Let $$b_i$$ denote the Betti number of $$K$$ , $$b_i=rank(H_i(K))$$, then we have $$E=b_0-b_1+b_2$$
We know that the Euler number $$E$$ of $$K$$ is $$1$$ , $$b_0=1$$, $$b_2=0$$ . So we must have $$b_1=0$$ , which means that $$rank(Z_2)=0\neq1$$ . However , $$rank(Z_2)$$ should be $$1$$ since $$<1>=Z_2$$ .

There must be something wong in the argument above , but I can't see where it is . Any help would be very appreciate .

The rank of a module $$M$$ is the rank of its free part, or equivalently the maximal number of linearly independent elements in $$M$$. $$\mathbb{Z}_2$$ has no linearly independent elements, since $$2 \cdot 1 = 0$$ in $$\mathbb{Z}_2$$, and hence its rank is 0.
• Alternatively the rank of an abelian group ($\mathbb Z -$module) $A$ is the largest number $n$ such that there is an injective map $\mathbb Z^n \rightarrow A$ Feb 9 '20 at 13:27
You are working in homology with coefficients in $$\mathbb Z$$, where the chain groups and homology groups are modules over $$\mathbb Z$$. The rank of a $$\mathbb Z$$-module is equal to the maximum rank of a free submodule over $$\mathbb Z$$. So no, $$\text{rank}(\mathbb Z_2)=0$$, not $$1$$, because the only free submodule over $$\mathbb Z$$ of $$\mathbb Z_2$$ is trivial.
Perhaps your are also trying to think about homology with $$\mathbb Z_2$$ coefficients? If so, you get a different calculation with a consistent outcome: $$H_0(K;\mathbb Z_2) \cong \mathbb Z_2$$ $$H_1(K;\mathbb Z_2) \cong \mathbb Z_2$$ $$H_2(K;\mathbb Z_2) \cong \mathbb Z_2$$ all of which have rank $$1$$ as modules over $$\mathbb Z_2$$, and so the Euler characteristic over $$\mathbb Z_2$$ is $$1-1+1=1 \in \mathbb Z_2$$, which of course is equal to the reduction modulo $$2$$ of $$1 \in \mathbb Z$$.