Reductive and semisimple algebraic monoids (Q2716429)

Let \(M\) be an irreducible algebraic monoid over an algebraically closed field with \(G=G(M)\) the unit group. \(M\) is said to be reductive (solvable) if so is \(G\). One of the most striking results of the theory of algebraic monoids is the Putcha-Renner theorem stating that \(M\) is reductive implies \(M\) is regular (in the semigroup theory sense), and if \(M\) has zero then the converse is also true; if \(M\) is regular then \(M\) is reductive.NEWLINENEWLINENEWLINEThe main result of the paper (Theorem 2.1) gives a criterion for \(M\) to be reductive, no matter whether \(M\) has zero or not. It is worth to notice that in this criterion not the ``top'' of \(M\) (unit group) but the ``bottom'' of \(M\) (the kernel of \(M\) denoted by \(\text{ker}(M)\)) is involved (for any semigroup \(S\) the kernel \(\text{ker}(S)\) of \(S\) is the minimal two-sided ideal of \(S\) if it exists). Theorem 2.1 states that \(M\) is reductive if and only if \(M\) is regular and \(\text{ker}(M)\) is a reductive group. A similar criterion is proved for \(M\) to be semisimple. Recall that \(M\) is semisimple if and only if \(M\neq G\) and \(G\) is reductive with one dimensional center. Theorem 2.4 gives 4 conditions equivalent to the semisimplicity of \(M\), we formulate only one such criterion: \(M\) is semisimple if and only if the solvable radical of \(G\) is one dimensional and \(\text{ker}(M)\) is a semisimple algebraic group.NEWLINENEWLINENEWLINEThe above described occupies Chapters 1, 2. Chapter 3 is devoted to reductive monoids \(M\) with \(\dim M\leq 5\). It is shown that \(\dim M\leq 2\) implies \(M\) is solvable, and \(\dim M=3\) implies \(M\) is either solvable or semisimple. The case \(\dim M=4\) is considered in Theorem 3.1: in this case nonsolvable \(M\) with \(M\neq G\) must be semisimple. Suppose \(\dim M=5\), \(M\) is unsolvable, and \(G\neq M\). Then the following conditions are equivalent: (1) \(M\) is reductive, (2) \(\text{rank}(G)\neq 2\), (3) \(\text{rank}(G)=3\); moreover, if \(M\) is reductive then \(\dim C(G)=2\) (\(C(G)\) denotes the center of \(G\)) (Theorem 3.5).