- What is the Jacobson radical \(J(R)\) of a ring \(R\)? State its equivalent characterizations.
- If \(R\) is a finitely generated commutative algebra over a field \(k\), what can you say about \(J(R)\) (Jacobson rings and Hilbert’s Nullstellensatz)?
Solution
For a commutative ring \(R\) with \(1\), the Jacobson radical is \[ J(R)=\bigcap_{\mathfrak m\in\operatorname{MaxSpec}R}\mathfrak m. \] Equivalently, \[ x\in J(R)\quad\Longleftrightarrow\quad 1-rx\in R^\times\text{ for every }r\in R. \] Also \(\sqrt{(0)}\subseteq J(R)\) because every maximal ideal is prime.
If \(R\) is a finitely generated commutative algebra over a field \(k\), then \(R\) is a Jacobson ring. Equivalently, for every prime ideal \(\mathfrak p\), \[ \mathfrak p=\bigcap_{\substack{\mathfrak m\supseteq\mathfrak p\\ \mathfrak m\text{ maximal}}}\mathfrak m. \] This is a standard consequence of Hilbert’s Nullstellensatz (the weak Nullstellensatz itself says that maximal ideals of a finite-type \(k\)-algebra have residue fields algebraic, hence finite, over \(k\)).
Taking \(\mathfrak p=(0)\) after passing to \(R/\sqrt{0}\), or intersecting over all primes, gives \[ J(R)=\bigcap_{\mathfrak m}\mathfrak m =\bigcap_{\mathfrak p}\mathfrak p =\sqrt{(0)}. \] Hence a reduced finite-type \(k\)-algebra has \(J(R)=0\).
If moreover \(R\) is finite-dimensional over \(k\), then \(R\) is Artinian, \(J(R)\) is nilpotent, and \(R/J(R)\) is a finite product of finite field extensions of \(k\).