Let $k$ be an algebraically closed field and let $X$ be an irreducible projective variety over $k$. I am wondering what the module of differential 1-forms on $X$ is.
Since $X$ is a projective variety, its ring of regular functions $\mathcal{O}_X (X)$ is precisely the ring of locally constant functions on X. Now since $X$ is irreducible, it is connected, hence $\mathcal{O}_X (X)$ is the ring of constant functions on $X$. By identifying a function with its image we find $\mathcal{O}_X (X) \cong k$, thus $\mathcal{O}_X (X)$ is a $1$-dimensional $k$-vector space with as a basis the constant function $1$.
Now the $\mathcal{O}_X (X)$-module $\Omega^1(X)$ of differential 1-forms on $X$ and the Kahler differential $d\colon \mathcal{O}_X (X) \to \Omega^1(X)$ are defined as follows. $\Omega^1(X)$ and $d$ are such that for any $\mathcal{O}_X (X)$-module $M$ and derivation, i.e. a $k$-linear map satisfying the Leibniz rule, $D\colon \mathcal{O}_X (X) \to M$ there is a unique $k$-linear map $\varphi\colon \Omega^1(X) \to M$ such that $D=\varphi \circ d$.
Firstly, since $D$ and $d$ are derivations and $\mathcal{O}_X (X)$ is a $k$-vector space with basis $1$, we have $D(\mathcal{O}_X (X))=d(\mathcal{O}_X (X))=\{0\}$. This seems to imply that there is no restriction on what $\Omega^1(X)$ is other than a $k$-vector space, since I can always take $\varphi$ to be the zero map.
Now on the one hand, in the construction of $\Omega^1$ for a general $k$-algebra $A$, thus replacing $\mathcal{O}_X (X)$ with $A$ in the previous definition, it follows that all elements are of the form $adb$ with $a,b\in A$, implying that $\Omega^1(X)=0$.
On the other hand, I am trying to do an exercise in which I calculate $\Omega^1(X)$ for a smooth irreducible projective curve $X$ from a presentation of $X$ by glueing the local $\Omega^1(U_i )$ to the global set. However I found that the local $\Omega^1(U_i )$ are 1-dimensional and they glue to a 1-dimensional vector space.
In conlusion I have three different ways in which I can construct $\Omega^1(X)$ and they all seem to imply that $\Omega^1(X)$ is something else. Which method is correct?