Questions about the proofs of some basic facts about singular and nonsingular points of a variety $ X. $

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The following material comes from "Basic Algebraic Geometry I" by Shafarevich.

Theorem: If $ x $ is a nonsingular point of $ X $ then there is a unique component of $ X $ passing through $ X. $

I have a small difficulty with the proof.

Proof:

We replace $ X $ by an affine neighbourhood $ U $ of $ x $ contained in $ X' = X \backslash \bigcup_{i} Z_{i}, $ where $ Z_{i} $ are the components of $ X $ not passing through $ x. $ Then $ k[U] \subset \mathcal{O}_{x}. $ Since a function $ f \in \mathcal{O}_{x} $ is uniquely determined by any of its Taylor series, $ \mathcal{O}_{x} $ is isomorphic to a subring of the formal power series ring $ k[[T]]. $ Since $ k[[T]] $ has no zerodivisors, the same holds for $ k[U], $ which is isomorphic to a subring of $ k[[T]]. $ Hence $ U $ is irreducible, as asserted in the theorem.

I don't understand why we can replace $ X $ by such an affine neighbourhood $ U. $

Additionally, there is a corollary which states:

The set of singular points of an algebraic variety $ X $ is closed.

It states in the proof that if $ X = \bigcup X_{i} $ is a decomposition into irreducible components, then by the previously stated theorem, the set of singular points of $ X $ is the union of the sets $ X_{i} \cap X_{j} $ for $ i \neq j $ and the sets of singular points of $ X_{i}. $

I don't understand how this follows from the theorem.

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It is a basic fact that any variety is covered by affine varieties and any scheme is covered by affine schemes. This gets you the appropriate $U$ in part 1.

In part 2, use the contrapositive: if there's not a unique irreducible component passing through a point, it is singular. So a singular point of a variety $X$ is either only on one irreducible component or it's on multiple irreducible components (and in fact every point on multiple irreducible components is singular). In the first case, the point is a singular point of that irreducible component; in the second, it's a member of $X_i\cap X_j$ for some $i\neq j$. So the singular set is the union of the singular sets of each irreducible component with the points on multiple components.