Let $k$ be an algebraically closed field and let $(X,\mathcal{O}_X)$ be a space with functions(*) such that every stalk $\mathcal{O}_{X,x}$ is local. Furthermore let $Y\subset X$ be an irreducible subset.
Now if we define $\mathcal{U}$ to be set of all open subsets $U\subset X$ with $U\cap Y \neq\emptyset$, then we have $U,V\in\mathcal{U}$ implies $U\cap V\in\mathcal{U}$ and we get the $k$-algebra $$\mathcal{O}_{X,Y}:=\varinjlim_{U\in\mathcal{U}}\mathcal{O}_X(U).$$ My question is, is this $k$-algebra local (as a ring)? I have tried to find candidates for a maximal ideal but did not found anything promising.
Any help/ hint/ advice is very much appreciated!
(*) By this I mean a topological space $X$ with a sheaf of $k$-algebras $\mathcal{O}_X$ such that for every open $U\subset X$ $\mathcal{O}_X(U)$ is a subalgebra of the $k$-algebra of all $k$-valued functions on $X$.
$\textbf{Edit:}$ Following the comment of KReiser, I first looked at the case $Y=\{y\}$. There $\mathcal{O}_{X,Y}$ reduces to the stalk at $y$ and the ring is local by assumption. I had difficulties defining a map $\mathcal{O}_{X,Y}\to k$ for general irreducible $Y$. But then I noticed, that $k=k(\{y\})$ is the function field of the space $\{y\}$. Now I am trying to find a surjective $k$-algebra map $$\mathcal{O}_{X,Y}\to k(Y),$$ but again, I am stuck...
The maximal ideal is the collection of functions $(U,f:U\to k)$ such that $f(y)=0$ for all $y\in Y\cap U$. These are precisely the non-units of the ring, as the way to construct an inverse for any $(U,f:U\to k)$ is to take $(U\setminus f^{-1}(0),1/f)$ where $1/f$ is defined point-wise. It's easy to check that the sum of any two functions of this form (after restricting to a common domain of definition) is again of this form, so by the characterization of a local ring as one where $1\neq 0$ and the sum of two non-units is a non-unit, we're done.