Proof. Let \(m_1,\dots,m_n\) be generators of \(M\). We can then represent \(\varphi\) in this generating set, i.e., write \[\varphi(m_i) = \sum_{j=1}^n a_{i, j} m_i\] for some \(a_{i,j} \in I\), so we can write \[\begin{bmatrix} \varphi(m_1)\\ \vdots \\ \varphi(m_n) \end{bmatrix} = \begin{bmatrix} a_{1,1} & \cdots & a_{1,n}\\ \vdots & \ddots & \vdots\\ a_{n,1} & \cdots & a_{n,n} \end{bmatrix} \begin{bmatrix} m_1\\ \vdots \\ m_n \end{bmatrix}.\] We denote this matrix by \(A \in M_n(R)\), and we denote our column of \(m_i\) by \(\mathbf{m} \in M^n\). Upgrading our module \(M\) to an \(R[x]\)-module by defining \(x\cdot m = \varphi(m)\), we also get \[\begin{bmatrix} \varphi(m_1)\\ \vdots \\ \varphi(m_n) \end{bmatrix} = \begin{bmatrix} x & & 0 \\ & \ddots & \\ 0 & & x \end{bmatrix} \begin{bmatrix} m_1\\ \vdots \\ m_n \end{bmatrix} = xI_n \mathbf{m}\] Thus, we have \[(xI_n - A)\mathbf{m} = 0.\] Multiplying by the adjugate matrix in \(M_n(R[x])\), we get \[\det(xI_n - A)\mathbf{m} = 0.\] Expanding the determinant, we get that \(\det(xI_n - A)\) is a monic polynomial in \(R[x]\) with non-leading coefficients in \(\left\langle a_{i,j}\right\rangle \subseteq I\). Then, since it annihilates \(m_1,\dots,m_n\) which are generators for \(M\), so it annihilates all of \(M\). Thus, \(\chi := \det(xI_n - A)\) is the zero homomorphism in \(\mathrm{Hom}_R(M,M)\). ◻

Proof. The ideal \(S^{-1} I \trianglelefteq S^{-1} R\) is proper because \(I\cap S = \emptyset\). In more detail, otherwise we would have \(1\in S^{-1} I\), so there exists \(a\in I, s\in S\) so that \(\frac{a}{s}=1\). Then, this implies there exists \(v\in S\) such that \(va=vs\). However, this element would be in \(I\cap S\).

Then, since \(S^{-1} I\) is proper, it can be extended to a maximal ideal in \(S^{-1} R\). The contraction \(P\) will then be a prime (not necessarily maximal) ideal disjoint from \(S\). ◻

Proof. Assume \(R' = \left\langle a_1,\dots,a_n\right\rangle\) (is finitely generated) as an \(R\)-module. In particular, \(R'=R[a_1,\dots,a_n]\). We now just need to show that \(R'\) is integral over \(R\) to get the “moreover” and that \(a_1,\dots,a_n\) are integral over \(R\). Let \(a\in R'\). Applying Cayley-Hamltion theorem 2 to the \(R\)-module homomorphism \[\begin{aligned} m_a: R' &\longrightarrow R'\\ x &\longmapsto ax \end{aligned}\] to get a monic polynomial equation \[m_a^k + c_{k-1} m_a^{k-1} + \cdots + c_0 = 0,\] for some \(c_0,\dots,c_{k-1}\in R\), which applied to \(1\in R\) gives us \[a^k + c_{k-1} a^{k-1} + \cdots + c_0 = 0.\] Thus, \(a\) is integral over \(R\).

Now, assume \(R' = R[a_1,\dots,a_n]\) with \(a_1,\dots,a_n\) integral over \(R\). Thus, each \(a_i\) satisfies a monic relation of degree \(d_i\). Therefore, we can write every element of \(R'\) as a sum \[\sum_{0\leq k_i \leq d_i} c_{k_1,\dots,k_n} a_1^{k_1}\cdots a_n^{k_n}\] since the homomorphism \(R[x_1,\dots,x_n]\to R[a_1,\dots,a_n]\) is surjective and we can rewrite any higher degree terms in terms of lower degrees using our monic relation. Thus, \(R'=\left\langle a_1^{k_1}\cdots a_n^{k_n}\right\rangle\) for \(k_i=0,\dots,n\) as an \(R\)-module. Therefore, \(R'\) is a finitely generated module over \(R\). ◻

Proof of (a).. By proposition 7, we can write \(R'=R[a_1,\dots,a_n]\) and \(R''=R'[b_1,\dots,b_m]\). Then, we can write \(R''=R[a_ib_j \mid i=1,\dots,n, j=1,\dots,m]\), so \(R''\) is finite over \(R\). ◻

Proof of (b).. Let \(a\in R''\). Since \(R''\) is integral over \(R'\), there exists monic polynomial \[a^n + c_{n-1} a^{n-1} + \cdots + c_1 a^1 + c_0 = 0,\] for \(c_0,\dots,c_{n-1} \in R'\), so \[a\in R[c_0,\dots,c_{n-1}].\] Since \(R'\) is integral over \(R\), each of \(c_0,\dots,c_{n-1}\) are integral over \(R\). Therefore, \(R[c_0,\dots,c_{n-1}]\) is a finite extension of \(R\), so \(a\in R[c_0,\dots,c_{n-1}]\) is integral over \(R\). Thus, \(R''\) is integral over \(R\). ◻

Proof. Let \(\overline{a}\in R'\). Then, \(a\in R'\) satisfies some monic relation \[a^n + c_{n-1}a^{n-1} + \cdots + c_1 a + c_0 = 0\] for \(c_0,\dots,c_{n-1}\in R\), so \(\overline{a}\) satisfies the relation \[\overline{a}^n + \overline{c_{n-1}}\overline{a}^{n-1} + \cdots + \overline{c_1} \overline{a} + \overline{c_0} = 0\] for \(\overline{c_0},\dots,\overline{c_{n-1}} \in R/(I\cap R)\) (identified with subring of \(R'/I\)), so \(\overline{a}\) is integral over \(R/(I\cap R)\). ◻

Proof of (a).. Assume there exists \(P' \trianglelefteq R'\) such that \(P'\cap R = P\). Then, \[PR' \cap R = (P'\cap R)R' \cap R \subseteq P'R' \cap RR' \cap R = P' \cap R \cap R = P\] (watch out that product and intersection need not commute exactly).

Now, assume that \(PR'\cap R \subseteq P\). We are looking for a prime ideal \(P'\trianglelefteq R\) containing \(PR'\) and not much more. By proposition 4, since for \(S=R\setminus P\), we have \[PR' \cap S = PR'\cap R\setminus P = \emptyset,\] there exists a prime ideal \(P'\trianglelefteq R'\) containing \(PR'\supseteq P\) with \(P' \cap S = \emptyset\). This means \(P'\cap (R\setminus P) = \emptyset\), and since \(P\subseteq P'\), this means \(P'\cap R = P\). ◻

Proof of (b).. We show that \(PR' \cap R \subseteq P\). Let \(a\in PR' \cap R\). In particular, \(a\in PR'\), so we have \(R\)-module homomoprhism \[\begin{aligned} m_a: R' &\longrightarrow PR'\\ x &\longmapsto ax. \end{aligned}\] By Cayley Hamilton theorem 2, there exist \(c_0,\dots,c_{n-1} \in P\) such that \[\begin{aligned} m_a^n + c_{n-1}m_a^{n-1} + \cdots + c_1m_a + c_0 &= 0\\ a^n = -(c_{n-1}a^{n-1} + \cdots + c_1 a + c_0). \end{aligned}\] Since \(a\in R\), the left hand expression is in \(RP=P\), so \(a^n\in P\). Since \(P\) is prime, \(a^n\in P\) implies \(a\in P\). Thus, \(PR' \cap R\subseteq P\). ◻

Proof. Assume \(P' \cap R = Q' \cap R\) and \(P'\subseteq Q'\). We prove that \(Q'\subseteq P'\) so that \(P'=Q'\).

Suppose, to the contrary, that there exists \(a\in Q'\setminus P'\). Then, we consider \(\overline{a}\in R'/P'\), which is integral over \(R/(P'\cap R)\) by lemma 9. Our strategy is to show that this relation on \(\overline{a}\) has to be trivial, so \(\overline{a}=0\) and \(a\in P'\).

Take a relation of minimal degree, and since \(a\not\in P'\) the degree is at least 2. We have \[\begin{aligned} \overline{a}^n + \overline{c_{n-1}}\overline{a}^{n-1} + \cdots + \overline{c_1} \overline{a} + \overline{c_0} &= 0\\ \overline{a}^n + \overline{c_{n-1}}\overline{a}^{n-1} + \cdots + \overline{c_1} \overline{a} &= -\overline{c_0} \end{aligned}\] for some \(c_0,\dots,c_{n-1} \in R\). Since \(a\in Q'\), the left hand side is in \(Q'/P'\), so \(\overline{c_0}\in Q'/P'\). Since \(c_0\in R\), we get \[c_0 \in (R \cap Q')/(R\cap P') = 0\] (by our original assumption). Thus, we have no constant term in the relation, but since we are in an integral domain \(R'/P'\), we can cancel out a factor of \(\overline{a}\) to get a monic relation of strictly smaller degree.

Thus, we have a contradiction, so \(a\in P'\) and \(P'=Q'\). ◻

Proof. Assume that \(R\) is a field. Let \(P'\trianglelefteq R'\) be a maximal ideal. Then, \(P'\) must contract to the \(0\) ideal, since the contraction is a prime ideal. Therefore, \[0\cap R = P' \cap R,\] so by incomparability, \(P'=0\). Thus, \(R'\) has only maximal ideal \(0\), so it is a field.

Now, assume that \(R\) is not a field, so there exists a non-zero maximal ideal \(\mathfrak{m}\trianglelefteq R\). By Lying Over proposition 10, there exists a prime ideal \(P'\trianglelefteq R'\) with \(P'\cap R = \mathfrak{m}\). In particular, \(P' \neq 0\), so \(R'\) has a nontrivial proper ideal and is not a field. ◻

Proof. Let \(\lambda, a_1,\dots,a_n \in K\) be arbitrary. We compute \(f\) written in our new coordinates, and then we determine how to choose our parameters. Writing \[f = \sum_{k_1,\dots,k_n} c_{k_1,\dots,k_n} x_1^{k_1}\cdots x_n^{k_n},\] we get \[\begin{aligned} \lambda f(y_1+a_1 &y_n, y_2 + a_2 y_n, \dots, y_{n-1} + a_{n-1}y_n, y_n)\\ &= \lambda \sum_{k_1,\dots,k_n} c_{k_1,\dots,k_n} (y_1+a_1y_n)^{k_1}\cdots (y_{n-1}+a_{n-1}y_n)^{k_{n-1}} y_n^{k_n}. \end{aligned}\] Let \(d=\deg f\). Then, the maximum \(y_n\) degree is \(d\). Compute an expression for the \(y_n\) degree \(d\) part, we get \[\lambda \sum_{\substack{k_1,\dots,k_n\\ k_1+\cdots+k_n=d}} c_{k_1,\dots,k_n} a_1^{k_1}\cdots a_{n-1}^{k_{n-1}} y_n^{k_1+\dots+k_n} = \lambda f_d(a_1,\dots,a_{n-1},1) y^d,\] where \(f_d\) is the degree \(d\) part of \(f\). We can therefore guarantee \(f\) in the new coordinates to be monic, as long as we can choose \(f_d(a_1,\dots,a_{n-1},1) \in K\) nonzero and \(\lambda\) to be the inverse. The next lemma guarantees such a choice of \(a_1,\dots,a_{n-1}\) is possible. ◻

Proof. We prove the lemma by induction on \(n\). If \(n=1\), then \(f=cx\) for \(c\in K\) nonzero, so \(f(1)=c\). Now, assume \(n>1\). We can view \(f\) as a polynomial in \((K[x_2,\dots,x_n])[x_1]\) and write \[f=\sum_{i=0}^d f_i x_1^i,\] where \(d\) is the degree of \(f\), and the (potentially zero) \(f_i\) are homogeneous degree \(d-i\) polynomials in \(K[x_2,\dots,x_n]\). At least one \(f_i\) is nonzero because \(f\) is nonzero. Then, by induction we can choose \(a_2,\dots,a_{n-1}\in K\) such that for this \(i\), \[f_i(a_2,\dots,a_{n-1},1)\neq 0.\] Fixing these \(a_2,\dots,a_{n-1}\in K\), we get \[f(x_1,a_2,\dots,a_{n-1},1) \in K[x_1]\] a nonzero polynomial. It has at most finitely many roots (i.e., bounded by degree), and since \(K\) is infinite, we can choose \(x_1=a_1\) to not be a root. ◻

Proof. We prove the claim by induction on \(n\). For \(n=1\), \(f=c_n x^n + \cdots + c_1 x + c_0\). We can then choose \(\lambda = \frac{1}{c_n}\).

Now, assume that \(\deg f > 1\). Let \(d=\deg_{x_1} f\). Then, we can write \[f = gx_1^d + h,\] where \(g\in K[x_2,\dots,x_n]\) and \(h\in K[x_1,\dots,x_n]\) with \(\deg_{x_1} h \leq d-1\). By our induction hypothesis, we can choose \(\lambda \in K\) and \(a_2,\dots,a_{n-1}\in \mathbb{N}\) such that \[\lambda g(y_2+y_n^{a_2},\dots,y_{n-1}+y_n^{a_n}, y_n)\] is monic in \(y_n\). Denote its leading \(y_n\) term by \(y_n^k\). Fix \(a_1\), we will see how to choose it later. We write \(\tilde{f}\) to denote the evaluation of \(f\) at our new coordinates, i.e., the image of \(f\) under the homomorphism \[\begin{aligned} \varphi: K[x_1,\dots,x_n] &\longrightarrow K[y_1,\dots,y_n]\\ x_i &\longmapsto y_i + y_n^{a_i} \qquad i\neq n\\ x_n &\longmapsto y_n. \end{aligned}\] Then, we have \[\tilde{f} = \tilde{g}(y_1+y_n^{a_1})^d + \tilde{h}.\] The leading \(y_n\) term of \(\tilde{g}(y_1+y_n^{a_1})\) is just \(\frac{1}{\lambda} y_n^{k+a_1 d}\). On the other hand, writing the \(y_n\) degree of \(\tilde{h}\) is bounded a function of \(a_2,\dots,a_{n-1}\) plus a term \(a_1 (d-1)\). Therefore, we can choose \(a_1\) sufficiently large so \(\frac{1}{\lambda} y_n^{k+a_1 d}\) dominates \(\tilde{h}\) in \(y_n\) degree (e.g., choose \(a_1> a_2\cdots a_{n-1}\deg f\)), giving us a choice of \(a_1,\dots,a_{n-1}\in \mathbb{N}\) and \(\lambda\in K\). ◻

Proof of Noether Normalization 14. By induction on the given number of generators, \(n\). For \(n=0\), we get \(R=K\) and we are done.

Otherwise, assume \(n>0\). Consider the map \[\begin{aligned} K[z_1,\dots,z_n] &\longrightarrow R\\ z_i &\longmapsto x_i \end{aligned}\] from the polynomial ring to \(R\). If the map is injective (i.e., no algebraic relations among the \(x_i\)), then \(R\cong K[z_1,\dots,z_r]\) for \(r=n\), and we are done.

Otherwise, the map is not injective and there is a nonzero relation \(f\in K[z_1,\dots,z_n]\), i.e., \(f(x_1,\dots,x_n)=0\). Using either lemma 15 or lemma 17 (depending on the size of \(K\)), pick new generators \(y_1,\dots,y_n\), i.e.,

• For lemma 15, choose \(y_i :=x_i - a_i x_n\) for \(i\neq n\), and \(y_n :=x_n\).

• For lemma 17, choose \(y_i :=x_i - x_n^{a_i}\) for \(i\neq n\), and \(y_n :=x_n\).

Now, these new generators satisfy the algebraic relation guaranteed by the lemmas, which is monic in \(y_n\). This shows that \(y_n\) is integral over the subalgebra \(K[y_1,\dots,y_{n-1}]\). Therefore, \(R=K[y_1,\dots,y_{n-1}][y_n]\) is finite over the subalgebra \(K[y_1,\dots,y_{n-1}]\). By induction, there exists \(r\in \mathbb{N}\) and an injective homomorphism \[K[z_1,\dots,z_r] \longrightarrow K[y_1,\dots,y_{n-1}]\] such that \(K[y_1,\dots,y_{n-1}]\) is finite over the image. Since finiteness is transitive by proposition 8, we get that \(R\) is finite over the image. This completes the induction.

If \(K\) is infinite, we will always do linear coordinate changes to get the images of \(z_1,\dots,z_r\) to be \(K\)-linear combinations of \(x_1,\dots,x_n\). ◻

Proof. By Noether Normalization in proposition 14, we know \(R\) is finite over a polynomial ring \(K[z_1,\dots,z_n]\), so it is in particular integral over \(K[z_1,\dots,z_n]\) by proposition 7. Since \(R\) is a field and the extension is integral, by corollary 12, \(K[z_1,\dots,z_n]\) must be field. Thus, \(r=0\) for \(K[z_1,\dots,z_r]\) to be a field (e.g., for \(z_1\) to be invertible).

Thus, \(K\subseteq R\) is a finite field extension, and if \(K\) is algebraically closed, this implies \(R=K\). ◻

Proof. Let \(\mathfrak{m}\trianglelefteq K[x_1,\dots,x_n]\) be a maximal ideal. Then, \(K[x_1,\dots,x_n]/\mathfrak{m}\) is a field as well as a finitely generated \(K\)-algebra (use cosets \(x_1,\dots,x_n\)). Thus, by Zariski’s lemma 18, \(K[x_1,\dots,x_n]/\mathfrak{m}= K\), i.e. identifying \(K\) with the image of \(K\subseteq K[x_1,\dots,x_n]\) under the quotient map. Then, taking \(a_i\) to be the image of \(x_i\) under quotienting, we get \[(x_1-a_1,\dots,x_n-a_n) \subseteq \mathfrak{m},\] and since the left hand ideal is already maximal, we must have \[\mathfrak{m}= I(a) = (x_1-a_1,\dots,x_n-a_n).\] ◻

Proof. The inclusion \(\subseteq\) is immediate because maximal ideals are prime, so they are radical, so \(I\subseteq \mathfrak{m}\) implies \(\sqrt{I}\subseteq \sqrt{\mathfrak{m}} =\mathfrak{m}\).

Now, we prove the \(\supseteq\) inclusion. Let \(f\not\in \sqrt{I}\). We want to find \(\mathfrak{m}\supseteq I\) (a closed point in \(V(I)\)) such that \(f\not\in \mathfrak{m}\) (\(f\) is nonzero at \(\mathfrak{m}\)). Thus, we want to find a closed point in \(D(f)\cap V(I)\). Taking \(S=\{1,f,f^2,\cdots\}\), equivalently we are looking for \(\mathfrak{m}\) such that \(\mathfrak{m}\cap S =\emptyset\) and \(f\not\in \mathfrak{m}\).

Since \(f\not\in \sqrt{I}\), we know \(S\cap I =\emptyset\), so by proposition 4, we can at least find a prime ideal \(P\) with \(P\cap S=\emptyset\), \(f\not\in P\), and \(P_f\) maximal. We now try to show \(P\) is in fact maximal anyways.

We consider the field extension which is the composition \(K\to R/P \to (R/P)_f = R_f/P_f\). This is indeed a field extension (i.e., not the zero map) because the second map is injective, since \(R/P\) is an integral domain.

Then, by construction \(R_f/P_f\) is a field, and it is a finitely generated \(K\)-algebra because it has generators consisting of those of \(R\hookrightarrow R_f\) together with \(\frac{1}{f}\). Therefore, by Zariski’s lemma 18, the extension is finite and in particular integral by proposition 7. Thus, \(R/P \subseteq R_f/P_f\) is integral as well, so \(R/P\) is a field by corollary 12. Therefore, \(P\) is maximal. ◻

Proof. Hard Part. We prove \(I(V(I)) \subseteq \sqrt{I}\). Assume that \(f\not\in \sqrt{I}\). We want to find a point \(a \in V(I)\) such that \(f(a)\neq 0\). By corollary 20, \[\sqrt{I} = \bigcap_{\substack{\mathfrak{m}\text{ maximal}\\ \mathfrak{m}\supseteq I}} \mathfrak{m},\] so there exists \(\mathfrak{m}\) maximal such that \(f \not\in \mathfrak{m}\). By corollary 19, this maximal ideal is \(I(a)\) for some \(a\in K^n\). Since \(I\subseteq I(a)\), we have \(a\in V(I)\). Since \(f\not\in I(a)\), we have \(f(a)\neq 0\). Therefore, \(f\not\in \sqrt{I}\).

Easy Part. Let \(f\in \sqrt{I}\), so \(f^n \in I\). Let \(a\in V(I)\). Since \(f^n\in I\), we know \(f^n(a)=0\). Thus, \(f(a)=0\), so \(f \in I(V(I))\).

This gives us \(I(V(I)) = \sqrt{I}\). We also have \(V(I(X))=X\) for any affine variety in \(\mathbb{A}_K^n\) because clearly \(X\subseteq V(I(X))\). Then conversely, we know that since \(X\) is an affine variety, \(X=V(J)\), so then we want to show \(V(J) \supseteq V(I(V(J)))\). By our previous discussion, \(J \subseteq \sqrt{J}\subseteq I(V(J))\), and applying \(V\) we get our desired subset relation.

Therefore, for radical ideals \(I\) and affine varieties \(X\), we have \(I(V(I)) = I\) and \(V(I(X)) = X\). Therefore, \(V(-),I(-)\) are inverse bijections. ◻