I am trying to construct a proper non-projective surface following the indications in section III.5 in Hartshorne's 'Alebraic geometry'.

In $X=\mathbb{P}_k^2$ consider the sheaf of differential 2-forms $\omega_X$ and let $(X',\mathcal{I})$ be a nontrivial extension of $X$ by $\omega_X$,namely a scheme $X'$ together with a sheaf of ideals $\mathcal{I}$ with $\mathcal{I}^2=0$ such that $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$. This gives a short exact sequence of sheaves $0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence $\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$

Non-projectivity of $X'$ shall follow from the fact that $Pic(X')=0$ and in order to see this it suffices to prove that $\delta$ is injective and nonzero. Since $Pic X\cong \mathbb{Z}$, any invertible sheaf is of the form $\mathcal{L}=\mathcal{O}_X(d)\cong \mathcal{O}_X(1)^{\otimes d}$ and it suffices to see that $\delta(\mathcal{O}_X(1))\neq 0$. I am confused as to how to carry out this computation since I guess I still do not understand very well the correspondence between infinitessimal extensions and the cohomology group. What I intend to do is to compute $\delta$ explicitly in the standard way, namely via the diagram

$\begin{array}{ccccccccc} 0 & \rightarrow & \check{C}^1(U,\omega) & \rightarrow & \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^1(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \\ && \downarrow && \downarrow && \downarrow && \\ 0 & \rightarrow & \check{C}^2(U,\omega) & \rightarrow & \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^2(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \end{array}$

where $U$ is the standard cover of $\mathbb{P}^2$.

Let $X'$ be the non-trivial extension given by the cocyle $\xi\in H^1(X,\Omega_X^1)$ given by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$.

The 1-cocycle $\alpha$ corresponding to $\mathcal{O}_X(1)$ in $\check{C}^1(U,\mathcal{O}_X^{\ast})$ is $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ and we only have to prove that it maps to some nonzero element in $\check{C}^2(U,\omega_X)$.

In order to lift $\alpha$ to $\beta=(\beta_0,\beta_1,\beta_2)\in \check{C}^1(U,\mathcal{O}^{\ast}_{X'})$ we first need a description of $\check{C}^1(U,\mathcal{O}^{\ast}_{X'})=\bigoplus_{i<j} \Gamma(U_{ij},\mathcal{O}_{X'}^{\ast})$

I recall having read somewhere that

$\Gamma(U_{ij},\mathcal{O}_{X'})\cong \Gamma(U_{ij},\mathcal{O}_X)[\eta_{ij}]=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$

where $\eta_{ij}^2=0$. How does this isomorphism follow? What is the description of the lift $\beta$?

Thanks in advance for any insight.