| Popis: |
Fields with only finitely many maximal subrings are completely determined. We show that such fields are certain absolutely algebraic fields and give some characterization of them. In particular, we show that the following conditions are equivalent for a field $E$: 1. $E$ has only finitely many maximal subrings. 2. $E$ has a subfield $F$ which has no maximal subrings and $[E:F]$ is finite. 3. Every descending chain $\cdots\subset R_2\subset R_1\subset R_0=E$ where each $R_i$ is a maximal subring of $R_{i-1}$, $i\geq 1$, is finite. Moreover, if one of the above equivalent conditions holds, then $F$ is unique and contains all subfields of $E$ which have no maximal subrings. Furthermore, all chains in $(3)$ have the same length, $m$ say, and $R_m=F$, where $m$ is the sum of all powers of primes in the factorization of $[E:F]$ into prime numbers.\\ We also determine when certain affine rings have only finitely many maximal subrings. In particular, we prove that if $R=F[\alpha_1,\ldots,\alpha_n]$ is an affine integral domain over a field $F$, then $R$ has only finitely many maximal subrings if and only if $F$ has only finitely many maximal subrings and each $\alpha_i$ is algebraic over $F$, which is similar to the celebrated Zariski's Lemma. Finally, we show that if $R$ is an uncountable PID then $R$ has at least $|R|$-many maximal subrings. |
