Proof. Let $m_1,\dots ,m_n$ be generators of $M$. We can then represent $\phi$ in this generating set, i.e., write $$\phi (m_i)=\sum _{j=1}^n{a}_{i,j}{m}_i$$ for some $a_{i,j}\in I$, so we can write $$\left[\begin{array}{c}\phi (m_1)\\ ⋮\\ \phi (m_n)\end{array}\right]=\left[\begin{array}{ccc}{a}_{1,1}& \cdots & a_{1,n}\\ ⋮& \ddots & ⋮\\ a_{n,1}& \cdots & a_{n,n}\end{array}\right]\left[\begin{array}{c}{m}_1\\ ⋮\\ m_n\end{array}\right].$$ 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 $x\cdot m=\phi (m)$, we also get $$\left[\begin{array}{c}\phi (m_1)\\ ⋮\\ \phi (m_n)\end{array}\right]=\left[\begin{array}{ccc}x& & 0\\ & \ddots & \\ 0& & x\end{array}\right]\left[\begin{array}{c}{m}_1\\ ⋮\\ m_n\end{array}\right]=x{I}_n\mathbf{m}$$ Thus, we have $$(x{I}_n-A)\mathbf{m}=0.$$ Multiplying by the adjugate matrix in $M_n(R[x])$, we get $$\det (x{I}_n-A)\mathbf{m}=0.$$ Expanding the determinant, we get that $\det (x{I}_n-A)$ is a monic polynomial in $R[x]$ with non-leading coefficients in $⟨a_{i,j}⟩\subseteq I$. Then, since it annihilates $m_1,\dots ,m_n$ which are generators for $M$, so it annihilates all of $M$. Thus, $\chi :=\det (x{I}_n-A)$ is the zero homomorphism in ${\mathrm{Hom}}_R(M,M)$. ◻

Proof. The ideal $S^{-1}I⊴S^{-1}R$ is proper because $I\cap S=\mathrm{\varnothing }$. 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 $R^{\prime }=⟨a_1,\dots ,a_n⟩$ (is finitely generated) as an $R$-module. In particular, $R^{\prime }=R[a_1,\dots ,a_n]$. We now just need to show that $R^{\prime }$ is integral over $R$ to get the “moreover” and that $a_1,\dots ,a_n$ are integral over $R$. Let $a\in R^{\prime }$. Applying Cayley-Hamltion theorem 2 to the $R$-module homomorphism $$\begin{array}{rl}{m}_a:R^{\prime }& ⟶R^{\prime }\\ x& ⟼ax\end{array}$$ 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 $R^{\prime }=R[a_1,\dots ,a_n]$ 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 $R^{\prime }$ as a sum $$\sum _{0\le k_i\le d_i}{c}_{k_1,\dots ,k_n}{a}_1^{k_1}\cdots a_n^{k_n}$$ 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, $R^{\prime }=⟨a_1^{k_1}\cdots a_n^{k_n}⟩$ for $k_i=0,\dots ,n$ as an $R$-module. Therefore, $R^{\prime }$ is a finitely generated module over $R$. ◻

Proof of (a).. By proposition 7, we can write $R^{\prime }=R[a_1,\dots ,a_n]$ and $R^{″}=R^{\prime }[b_1,\dots ,b_m]$. Then, we can write $R^{″}=R[a_i{b}_j\mid i=1,\dots ,n,j=1,\dots ,m]$, so $R^{″}$ is finite over $R$. ◻

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

Proof of (a).. Assume there exists $P^{\prime }⊴R^{\prime }$ such that $P^{\prime }\cap R=P$. Then, $$P{R}^{\prime }\cap R=(P^{\prime }\cap R)R^{\prime }\cap R\subseteq P^{\prime }{R}^{\prime }\cap R{R}^{\prime }\cap R=P^{\prime }\cap R\cap R=P$$ (watch out that product and intersection need not commute exactly).

Now, assume that $P{R}^{\prime }\cap R\subseteq P$. We are looking for a prime ideal $P^{\prime }⊴R$ containing $P{R}^{\prime }$ and not much more. By proposition 4, since for $S=R\setminus P$, we have $$P{R}^{\prime }\cap S=P{R}^{\prime }\cap R\setminus P=\mathrm{\varnothing },$$ there exists a prime ideal $P^{\prime }⊴R^{\prime }$ containing $P{R}^{\prime }\supseteq P$ with $P^{\prime }\cap S=\mathrm{\varnothing }$. This means $P^{\prime }\cap (R\setminus P)=\mathrm{\varnothing }$, and since $P\subseteq P^{\prime }$, this means $P^{\prime }\cap R=P$. ◻

Proof of (b).. We show that $P{R}^{\prime }\cap R\subseteq P$. Let $a\in P{R}^{\prime }\cap R$. In particular, $a\in P{R}^{\prime }$, so we have $R$-module homomoprhism $$\begin{array}{rl}{m}_a:R^{\prime }& ⟶P{R}^{\prime }\\ x& ⟼ax.\end{array}$$ By Cayley Hamilton theorem 2, there exist $c_0,\dots ,c_{n-1}\in P$ such that $$\begin{array}{rl}{m}_a^n+c_{n-1}{m}_a^{n-1}+\cdots +c_1{m}_a+c_0& =0\\ a^n=-(c_{n-1}{a}^{n-1}+\cdots +c_{1}a+c_0).\end{array}$$ 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, $P{R}^{\prime }\cap R\subseteq P$. ◻

Proof. Assume $P^{\prime }\cap R=Q^{\prime }\cap R$ and $P^{\prime }\subseteq Q^{\prime }$. We prove that $Q^{\prime }\subseteq P^{\prime }$ so that $P^{\prime }=Q^{\prime }$.

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

Take a relation of minimal degree, and since $a\notin P^{\prime }$ the degree is at least 2. We have $$\begin{array}{rl}{\bar{a}}^n+\overline{c_{n-1}}{\bar{a}}^{n-1}+\cdots +\overline{c_1}\bar{a}+\overline{c_0}& =0\\ {\bar{a}}^n+\overline{c_{n-1}}{\bar{a}}^{n-1}+\cdots +\overline{c_1}\bar{a}& =-\overline{c_0}\end{array}$$ for some $c_0,\dots ,c_{n-1}\in R$. Since $a\in Q^{\prime }$, the left hand side is in $Q^{\prime }/P^{\prime }$, so $\overline{c_0}\in Q^{\prime }/P^{\prime }$. Since $c_0\in R$, we get $$c_0\in (R\cap Q^{\prime })/(R\cap P^{\prime })=0$$ (by our original assumption). Thus, we have no constant term in the relation, but since we are in an integral domain $R^{\prime }/P^{\prime }$, we can cancel out a factor of $\bar{a}$ to get a monic relation of strictly smaller degree.

Thus, we have a contradiction, so $a\in P^{\prime }$ and $P^{\prime }=Q^{\prime }$. ◻

Proof. Assume that $R$ is a field. Let $P^{\prime }⊴R^{\prime }$ be a maximal ideal. Then, $P^{\prime }$ must contract to the $0$ ideal, since the contraction is a prime ideal. Therefore, $$0\cap R=P^{\prime }\cap R,$$ so by incomparability, $P^{\prime }=0$. Thus, $R^{\prime }$ 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 $\mathfrak{m}⊴R$. By Lying Over proposition 10, there exists a prime ideal $P^{\prime }⊴R^{\prime }$ with $P^{\prime }\cap R=\mathfrak{m}$. In particular, $P^{\prime }\ne 0$, so $R^{\prime }$ 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 $$\begin{array}{rl}\lambda f(y_1+a_1& y_n,y_2+a_2{y}_n,\dots ,y_{n-1}+a_{n-1}{y}_n,y_n)\\ & =\lambda \sum _{k_1,\dots ,k_n}{c}_{k_1,\dots ,k_n}(y_1+a_1{y}_n{)}^{k_1}\cdots (y_{n-1}+a_{n-1}{y}_n{)}^{k_{n-1}}{y}_n^{k_n}.\end{array}$$ Let $d=\mathrm{deg}f$. Then, the maximum $y_n$ degree is $d$. Compute an expression for the $y_n$ degree $d$ part, we get $$\lambda \sum _{\begin{array}{c}{k}_1,\dots ,k_n\\ k_1+\cdots +k_n=d\end{array}}{c}_{k_1,\dots ,k_n}{a}_1^{k_1}\cdots a_{n-1}^{k_{n-1}}{y}_n^{k_1+\cdots +k_n}=\lambda f_d(a_1,\dots ,a_{n-1},1)y^d,$$ 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 $i$, $$f_i(a_2,\dots ,a_{n-1},1)\ne 0.$$ 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 $\mathrm{deg}f>1$. Let $d={\mathrm{deg}}_{x_1}f$. Then, we can write $$f=g{x}_1^d+h,$$ where $g\in K[x_2,\dots ,x_n]$ and $h\in K[x_1,\dots ,x_n]$ with ${\mathrm{deg}}_{x_1}h\le d-1$. 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 $$\begin{array}{rl}\phi :K[x_1,\dots ,x_n]& ⟶K[y_1,\dots ,y_n]\\ x_i& ⟼y_i+y_n^{a_i}i\ne n\\ x_n& ⟼y_n.\end{array}$$ 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 $a_1>a_2\cdots a_{n-1}\mathrm{deg}f$), 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 $$\begin{array}{rl}K[z_1,\dots ,z_n]& ⟶R\\ z_i& ⟼x_i\end{array}$$ 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 $i\ne n$, and $y_n:=x_n$.

• For lemma 17, choose $y_i:=x_i-x_n^{a_i}$ for $i\ne n$, 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 $$K[z_1,\dots ,z_r]⟶K[y_1,\dots ,y_{n-1}]$$ 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 $\mathfrak{m}⊴K[x_1,\dots ,x_n]$ 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 $f\notin \sqrt{I}$. We want to find $\mathfrak{m}\supseteq I$ (a closed point in $V(I)$) such that $f\notin \mathfrak{m}$ ($f$ 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 $\mathfrak{m}\cap S=\mathrm{\varnothing }$ and $f\notin \mathfrak{m}$.

Since $f\notin \sqrt{I}$, we know $S\cap I=\mathrm{\varnothing }$, so by proposition 4, we can at least find a prime ideal $P$ with $P\cap S=\mathrm{\varnothing }$, $f\notin P$, 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 $f\notin \sqrt{I}$. We want to find a point $a\in V(I)$ such that $f(a)\ne 0$. By corollary 20, $$\sqrt{I}=\bigcap _{\begin{array}{c}\mathfrak{m}\text{ maximal}\\ \mathfrak{m}\supseteq I\end{array}}\mathfrak{m},$$ so there exists $\mathfrak{m}$ maximal such that $f\notin \mathfrak{m}$. 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 $f\notin I(a)$, we have $f(a)\ne 0$. Therefore, $f\notin \sqrt{I}$.

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. ◻