A ring is called left (resp. right) Artinian if every descending chain of left (resp. right) ideals \(I_1\supset I_2\supset I_3\cdots\) eventually terminates (i.e. there is some \(N\) such that \(I_k=I_N\) for \(k\ge N\)). A ring is called Artinian if it is both left and right Artinian. a. Let \(A\) be a left Artinian ring and suppose that \(f: A\to R\) is a surjective homomorphism. Prove that \(R\) is left Artinian. b. Show that an Artinian integral domain \(A\) is a field. (Hint: for a nonzero element \(a\in A\), consider the ideals \((a),(a^2),(a^3),\dots\).) c. Show that every prime ideal in a commutative Artinian ring is maximal.
Solution
1. A surjective homomorphic image of a left Artinian ring is left Artinian.
Let \[ J_1\supseteq J_2\supseteq J_3\supseteq\cdots \] be a descending chain of left ideals of \(R\). Since \(f:A\to R\) is surjective, each preimage \(f^{-1}(J_i)\) is a left ideal of \(A\), and \[ f^{-1}(J_1)\supseteq f^{-1}(J_2)\supseteq f^{-1}(J_3)\supseteq\cdots. \] Because \(A\) is left Artinian, there is \(N\) such that \[ f^{-1}(J_i)=f^{-1}(J_N) \qquad(i\ge N). \] Applying \(f\) and using surjectivity gives \[ J_i=f(f^{-1}(J_i))=f(f^{-1}(J_N))=J_N \qquad(i\ge N). \] Thus every descending chain of left ideals of \(R\) stabilizes.
2. Every Artinian integral domain is a field.
Let \(A\) be an Artinian integral domain and let \(0\ne a\in A\). The descending chain \[ (a)\supseteq(a^2)\supseteq(a^3)\supseteq\cdots \] stabilizes, so for some \(n\ge1\), \[ (a^n)=(a^{n+1}). \] Hence \(a^n\in(a^{n+1})\), so there exists \(b\in A\) with \[ a^n=a^{n+1}b. \] Because \(A\) is a domain and \(a^n\ne0\), cancellation gives \[ 1=ab. \] Thus every nonzero element of \(A\) is a unit, so \(A\) is a field.
3. Every prime ideal in a commutative Artinian ring is maximal.
Let \(\mathfrak p\) be a prime ideal of the commutative Artinian ring \(A\). The quotient \(A/\mathfrak p\) is an integral domain. By 1, applied to the quotient map \[ A\twoheadrightarrow A/\mathfrak p, \] the quotient is Artinian. By 2 it is therefore a field. Hence \(\mathfrak p\) is maximal.