I am reading Gathmann's note on Algebraic Geometry and there is a remark that I really don't understand in the chapter 10. First I recall a result that is, apprently, needed to understand the remark.
Proposition 2.28 Let $X$ be a non-empty irreducible affine variety. If $f \in A(X)$ is non-zero every irreducible component of $V(f)$ has codimension $1$ in $X$ (and hence dimension $\text{dim}~ X - 1$ by a previous result).
And now the remark.
Remark 10.10 (b) It is a result of commutative algebra that a regular local ring as in (a) (here he means $\mathcal O_{X, a}$, the stalk of $\mathcal O_X$ at $a \in X$ an affine variety, which is a regular local ring iff $a$ is smooth by a previous discussion) is always an integral domain. Translating this into geometry as in the previous proposition (he means the above one), this yields the intuitively obvious statement that a variety is locally irreducible at every smooth point $a$, i.e. that $X$ has only one irreducible component meeting $a$. (...)
I really don't understand how we can translate the fact that $\mathcal O_{X, a}$ is an integral domain to the geometry and deduce that there is only one irreducible component of $X$ that meets $a$. I don't even have the intuition of this result (which is suppose to be obvious according to him)... Could one of you enlighten me on this?
Claim: the prime ideals of $\mathcal{O}_{X,a}$ are in bijection with the irreducible subvarieties of $X$ passing through $a$.
Proof: we can start by looking at an open affine neighborhood $Y$ of $a$ inside $X$. Then by remark 2.9 in the current 2021 version of Gathmann's notes, irreducible subvarieties of $Y$ are in bijection with prime ideals of $A(Y)$. Next, by lemma/definition 3.19, $\mathcal{O}_{X,a}$ is the localization of $A(Y)$ at the ideal corresponding to $a$ - as the prime ideals of a localization are the prime ideals not meeting the set of elements we invert, we have that the prime ideals of $\mathcal{O}_{X,a}$ are precisely the prime ideals of $A(Y)$ which correspond to irreducible subvarieties passing through $a$. $\blacksquare$
Now we're almost done: irreducible components are maximal irreducible closed subsets, so they correspond to minimal prime ideals. As a domain has a unique minimal prime, $(0)$, if $\mathcal{O}_{X,a}$ is a domain, $a$ is on a unique irreducible component of $X$. Since regular local rings are domains, this shows what you're after.