I have encountered an apparent contradiction: Let $Y$ be an affine variety of $\mathbb{A}^n$ and $P$ a point of $Y$. Then i have proved that $\mathcal{O}_P$ is an integral domain and it is also not an integral domain.
Prp: $\mathcal{O}_P$ is an integral domain. Prf: $\mathcal{O}_P$ is injected into $K(Y)$.
Prp: $\mathcal{O}_P$ is not an integral domain. Prf: We can define an injection $k \rightarrow \mathcal{O}_P$ by sending $c \in k$ to the germ $\langle Y, c \rangle$. But $k \cong \mathcal{O}_P/m_P$, so $m_P \in Ass(\mathcal{O}_P)$, which implies that all elements of $m_P$ are zero-divisors. Finally, it is clear that $m_P \neq 0$.
Question 1: Where is error in the above argument?
Question 2: If $\mathcal{O}_P$ is indeed an integral domain, then localizing at zero the injection $ \mathcal{O}_P \rightarrow K(Y)$ we obtain that $k(y_1,\dots,y_n) \subset K(Y)$, which does not seem right ($k[y_1,\dots,y_n]$ is the polynomial ring). What am i missing here?
$\mathcal{O}_P$ is an integral domain. The error in your argument is the assertion that $k = \mathcal{O}_P/m_P$ means that $m_P$ consists of zero-divisors.
It may help to look at an example: Let $Y = \mathbb{A}^1$ and $P$ be $0$. Then $K(Y)$ is $k(x)$, the field of rational polynomials in one variable. $\mathcal{O}_P$ is the ring consisting of rational polynomials whose denominator is not divisible by $x$, and $m_P$ is the ideal generated by $x \in \mathcal{O}_P$. You can see that $\mathcal{O}_P$ is an integral domain and the elements of $m_P$ are not zero-divisors (they are also not invertible).