Let \(k\) be a field. Show that \[ \mco(\AA^n_k) = k[x_1,\ldots,x_n] \] is a Noetherian ring.
Solution
1. The field \(k\) is Noetherian.
The only ideals of a field are \[ (0) \qquad\text{and}\qquad (1)=k. \] Both are finitely generated, so every ideal of \(k\) is finitely generated. Thus \(k\) is Noetherian.
2. For every integer \(r\) with \(0\leq r\leq n\), the polynomial ring\[ k[x_1,\ldots,x_r] \]is Noetherian.
We argue by induction on \(r\). The case \(r=0\) is step 1.
Assume \[ k[x_1,\ldots,x_r] \] is Noetherian. By the Hilbert basis theorem T-YYLPH, if a commutative ring \(R\) is Noetherian, then \[ R[x] \] is Noetherian. Applying it to \[ R=k[x_1,\ldots,x_r] \] gives that \[ k[x_1,\ldots,x_r][x_{r+1}] = k[x_1,\ldots,x_{r+1}] \] is Noetherian. This completes the induction.
3. The coordinate ring of affine \(n\)-space is Noetherian:\[ \boxed{ \mco(\AA^n_k)=k[x_1,\ldots,x_n] \text{ is Noetherian}. } \]
By definition, \[ \mco(\AA^n_k)=k[x_1,\ldots,x_n]. \] Step 2 with \(r=n\) shows that this polynomial ring is Noetherian.
4. Q.E.D.
Steps <1>1–<1>3 prove the stated Noetherianity of the affine-space coordinate ring.