Carefully state the definition of Noetherian for a commutative ring \(R\).
Let \(R\) be a subset of \(\ZZ[x]\) consisting of all polynomials \[ f(x) = a_ 0 + a_1 x + a_2 x^2 + \cdots + a_nx^n \] such that \(a_k\) is even for \(1\leq k \leq n\). Show that \(R\) is a subring of \(\ZZ[x]\).
Show that \(R\) is not Noetherian.
Hint: consider the ideal generated by \(\ts{ 2x^k \st 1\leq k \in \ZZ }\).
Solution
A ring is Noetherian iff \(R\) satisfies the ascending chain condition: every chain of ideals \(A_1 \subseteq A_2 \subseteq \cdots\) eventually stabilizes, so \(A_m \subseteq A_{m+1} = A_{m+2} = \cdots\).
That \(R\) is a subring of \(\ZZ[x]\):
\((R, +)\) is an abelian subgroup: note that \(f(x) + g(x) = \sum a_k x^k + \sum b_k x^k = \sum (a_k + b_k) x^k\), so if \(a_k, b_k\) are even then \(a_k + b_k\) is even. It’s closed under inverses, since \(a_k\) is even iff \(-a_k\) is even, and contains zero.
\((R, \cdot)\) is a submonoid: \(f(x) g(x) = \sum_{n=1}^N \qty{ \sum_{k=1}^n a_k b_{n-k}} x^k\) where without loss of generality, \(\deg f = \deg g = n\) by setting coefficients to zero. Then sums and products of even integers are even, so \(fg \in R\).
That \(R\) is not Noetherian: it suffices to show that \(R\) contains an ideal that is not finitely generated.
The claim is that setting \(S \da \ts{2x^k}_{k\in \ZZ_{\geq 1}}\) and taking \[ I \da \gens{S} = \sum_{k\in \ZZ_{\geq 1}} R\cdot 2x^k \da \ts{ \sum_{i=1}^m r_k(x) 2x^k \st r_k(x) \in 2\ZZ[x], m\in \ZZ_{\geq 0}} \] yields an ideal that is not finitely generated.
Suppose toward a contradiction that \(\ts{g_1, g_2, \cdots, g_M}\) were a finite generating set, where each \(g_i \in I\).