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Let $X$ be a Noetherian topological space. We know that $X$ is a union of finitley many irreducible components, i.e. $X=X_1\cup\ldots\cup X_n$, where $X_1,\ldots,X_n$ are the maximal closed irreducible subsets of $X$ (irreducible components of $X$).

  1. Prove that there exists a unique decompositon of $X$ into finitely many disjoint connected components i.e. $X=Y_1\sqcup\ldots\sqcup Y_m$, where $Y_1,\ldots,Y_m$ are maximal closed connected subsets of $X$. Moreover, every connected component is a union of some irreducible components.
  2. Prove that if $X$ is an affine algebraic group then every connected component of $X$ is irreducible.
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    $\begingroup$ I've done first question: We have this property: "irreducible implies connected", then $X$ is union of finitely many closed connected subsets $X_i$. If $X_i$, $X_j$ are not disjoint then combine this two connected (also irreducible) subsets $X_i,X_j$ to form a new connected subset. We continue do this until it stops then we have a decomposion into connected components of X. Clearly, by this process, every connected component is a union of some irreducible. I am not surely about the uniqueness. $\endgroup$ – math Nov 16 '17 at 14:24
  • $\begingroup$ In general, the property: "connected implies irreducible" is not true, but it is true with algebraic group. So I want to ask for the solution of this property in case of algebraic group. $\endgroup$ – math Nov 16 '17 at 14:33
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Let $X= X_1 \cup \cdots \cup X_n$ be a decomposition of $X$ into irreducible subsets. In particular each $X_i$ is connected.

We may choose an element $x\in X_1$ with the property $x \notin X_2 \cup \cdots \cup X_n$. Thus $x$ is contained in exactly one irreducible component. Due to homogeneity, all points of $X$ has this property, and therefore $X_i \cap X_j =0$ when $i \neq j$. Hence the irreducible components are exactly the connected components.

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