First, let us show that the chain condition definition of a Noetherian ring is equivalent to every ideal being finitely generated.
Proposition 1.1
The two following definitions of a Noetherian ring are equivalent,
Proof. To show (1) implies (2), simply consider the following process of starting with an empty ideal \(I\) and then iteratively adding elements in \(A \setminus I\) as generators, then let \(I_x\) be the ideal obtained after the \(x\)th step. This process induces an ascending chain, \[ I_0 \subseteq I_1 \subseteq I_2 \subseteq \ldots \] which must stabilize, thus proving that \(I\) is finitely generated!
The (2) implies (1), let \(Z = \bigcup_{i = 1}^{\infty} Z_i\), which clearly is an ideal (verify this!) and thus finitely generated, since each generator will appear somewhere in the chain it follows that some \(Z_k\) will contain all the generators and thus imply that the chain stabilized!
Theorem 1.2
If \(A\) is Noetherian, then \(A[x]\) is as well.
Proof. Let us initiate a process, where we start with the empty ideal \(I\) and then iteratively add generators by finding lowest degree polynomials \(f\) that are not in \(I\). Let \(a_i\) be the coefficient in front of the largest exponent monomial of \(f_i\), then since \(A\) is Noetherian we may choose a large enough \(k\) such that \((a_1, \ldots, a_k) = A\).
Thus, we may write \(a_{k + 1} = \sum_{i = 1}^{k} b_i a_i\). Let \(f \in A[x] \setminus I\) be the lowest of such degree, then consider the following peculiar function, \[ g(x) = f_{k + 1}(x) - \sum_{i = 1}^{k} b_i f_i \cdot x^{\deg f_{k + 1} - \deg f_i} \] clearly all the exponents are non-negative since at each step the lowest degree function not present in \(I\) was added. Observe that \(\deg g < \deg f_{k + 1}\) since the largest exponent monomial in \(f_{k + 1}\) will be canceled by the sum of \(b_i a_i\) coefficients, however clearly \(g(x) \not\in I\) (since \(f_{k + 1} \not\in I\)), thus contradicting the minimality of \(f_{k + 1}\)!
Let us prove the following result,
Theorem 1.3 (Noether Normalization Lemma)
Let \(k\) be a field, and let \(A\) be a finitely generated \(k\)-algebra. Then there exist algebraically independent elements \(y_1, \ldots, y_d \in A\) such that the polynomial ring \(k[y_1, \ldots, y_d]\) injects into \(A\), \[ k[y_1, \ldots, y_d] \hookrightarrow A, \] and \(A\) is integral (and finitely generated as a module) over \(k[y_1, \ldots, y_d]\).
Proof. Let \(A = k[x_1, \ldots, x_k] / I\) (since \(A\) is finitely generated), let us now consider some polynomial \(P(X) = \sum_{i=0}^{n} \alpha_i X^i\), where \(\alpha_i \in k\), by Euclid we know that \(k[x_1, \ldots, x_k] = k[x_1 - x_k^m, x_2 - x_k^m, \ldots, x_{k - 1} - x_k^m, x_k^m]\) (let \(S_i = x_i - x_k^m\)), thus we may rewrite \(P\) as, \[ P(x) = \sum_{i =0}^{n} \alpha_i X^i = \sum_{i = 0}^{n} \alpha_i (S_1 + x_k^m)^{\alpha_{i1}}\ldots(S_{k - 1} + x_{k}^m)^{\alpha_{i(k-1)}} x_k^{\alpha_{ik}} \] by choosing \(m\) large enough we may force the polynomial to be monic in \(x_k\) (it will have a coefficient in \(k\) which is a field, thus all coefficients are units). Thus, we obtain that \(A\) is integral over \(k[S_1, \ldots, S_k]\) and thus it is integral over \(k[S_1, \ldots, S_{k - 1}]/(I \cap k[x_1, \ldots, x_k])\) and consequently we may repeat this until we quotient by an empty ideal by induction!
Let us first show a rather fundamental result about finite extensions.
Proposition 1.4
Let \(S \subseteq R\) be two commutative rings, then if \(R\) is a finitely generated \(S\)-module, it must be that \(R\) is integral over \(S\).
Proof. Let \(u \in R\), then consider the sequence \(u, u^2, u^3, \ldots\), since \(R\) is finitely generated there must be some linear relationship between \(\{u, \ldots, u^n\}\) (choosing the smallest \(n\) with a linear relationship) that gives, \[ u^n - \sum_{i = 1}^{n - 1} \alpha_i u^i = 0, \text{ where } \alpha_i \in S \] which is a monic polynomial with a root \(u\)!
First let us show the weak version of the theorem, (called The Weak Hilbert Theorem)
Theorem 1.5 (Weak Nullstellensatz)
If \(k\) is an algebraically closed field, then the maximal ideals \(\mathfrak{m}\) of \(k[x_1, \ldots, x_k]\) correspond to points in \(k^n\), so that, \[ k[x_1, \ldots, x_k] \supseteq \mathfrak{m} = (x_1 - a_1, x_2 - a_2, \ldots, x_k - a_k) \]
The proof will require to show that \(k[x_1, \ldots, x_k]/\mathfrak{m} \cong k\), then looking at the images of \(x_i\), let us assume that \(x_i \mapsto \alpha_i\), we will deduce that it must be that \(x_i - \alpha_i \mapsto 0\), which will imply that \((x_1 - \alpha_1, x_2 - \alpha_2, \ldots, x_k - \alpha_k) \subseteq \mathfrak{m}\), however since both are maximal ideals equality will follow!
To show the isomorphism we need to prove the following result,
Proposition 1.6
Let \(A\) be a finitely generated \(k\)-algebra (over algebraically closed field) and \(\mathfrak{m}\) a maximal ideal of \(A\), then \(A/\mathfrak{m}\) is integral over \(k\).
Proof. By Noethers Normalization Lemma \(A / \mathfrak{m}\) is integral over \(k[x_1, x_2, \ldots, x_k]\) for some \(k \in \mathbb{Z}\), where \(k = \dim A / \mathfrak{m} = 0\), thus we obtain the desired result!
Another way to proving this result without utilizing the formula for \(k\) is the following,
Proof. By Noethers Normalization Lemma, we have an injective homorphism \(k[x_1, \ldots, x_k] \hookrightarrow A/\mathfrak{m}\), then consider the image of \(x_1\) in this map, since \(A / \mathfrak{m}\) is a field it has an inverse \(1 / x_1\) (slight abuse of notation for the image of \(x_1\)), then since \(A / \mathfrak{m}\) is integral over \(k[x_1, \ldots, x_k]\) we have some monic polynomial with root \(1 / x_1\), \[ \left( \frac{1}{x_1} \right)^n + Q_{n - 1} \cdot \left( \frac{1}{x_1} \right)^{n - 1} + \ldots + Q_1\cdot \left( \frac{1}{x} \right) + Q_0 = 0 \] multiplying this equation by \(x_1^n\) one obtains, \[ 1 + x_1 Q_{n - 1} + \ldots + x_1^{n - 1}Q_1 + x_1^n = 1 + x_1 \cdot Q = 0 \] thus \(x_1\) is invertible in \(k[x_1, \ldots, x_k]\), which is of course nonsense, thus \(k = 0\)!
A very nice consequence of Weak Nullstellensatz is the following corrolary,
Proposition 1.7
If \(I \subseteq k[x_1, \ldots, x_n]\) has no common roots, then \(I\) is the entire ideal.
Proof. Since every ideal is part of a maximal ideal, which has has a common root given by the point it corresponds to, thus it must be the entire ideal!
Let \(V(I)\) be the set of common zeros of all the polynomials in \(I\) and \(Z(J)\) be the set of polynomials vanishing on \(J \subseteq k^n\). It turns out the following result holds,
Theorem 1.8 (Hilbert's Nullstellensatz)
For any ideal \(I \subseteq k[x_1, \ldots, x_n]\) we have that, \[ Z(V(I)) \cong \sqrt{I} \] where \(\sqrt{I}\) denotes the radical of \(I\).
Proof. It is clear that \(\sqrt{I} \subseteq Z(V(I))\), since the zeros of \(p^n\) are the same as \(p\). To show the other direction, one can utlize the so called Rabinowitsch trick.
Let \(p \in Z(V(I))\), the idea is that we can map \(I \subseteq k[x_1, \ldots, x_n]\) into \(k[x_1, \ldots, x_n, y]\) by considering, \[ J = I \cdot k[x_1, \ldots, x_n] + (1-yp) \] The zeros of \(J\) must be those that all polynomials in \(I\) vanish on the first coordinates and \(1 - yp = 0\), so \(yp = 1\), however since \(p\) vanishes on all common zeros it must be that \(p = 0\), thus there are no zeros! By the Weak Nullstellensatz this implies that \(1 \in J\), thus, \[ 1 = \sum_{i} f_i h_i(x_1, \ldots, x_n, y) + (1 - yp) h_{n+1}(x_1, \ldots, x_{n}, y) \] now we take \(y = \frac{1}{p}\), then multiplying by a sufficiently large power of \(p\), we will obtain that, \[ p^n = \sum_{i} f_i H_i(x_1, \ldots, x_{n}) \] which proves that \(p \in \sqrt{I}\)!