Integral Extensions

The Cayley–Hamilton Theorem

We begin by first stating the generalized Cayley–Hamilton theorem.

Theorem 1 Fix a ring \(A\) and an ideal \(\mathfrak{a}\subseteq A\), and let \(M\) be generated by \(n\) elements \(e_1,\ldots,e_n\) as an \(A\)-module. Then for any \(u\in\End_\rMod{A}(M)\), if \(u(M)\subseteq \mathfrak{a}M\) holds, there exists a monic polynomial

\[p(\x)=\x^n+p_1\x^{n-1}+\cdots+p_n,\qquad p_k\in \mathfrak{a}^k\]

satisfying \(p(u)=0\).

Proof

From the assumption \(u(M)\subseteq \mathfrak{a}M\) and \(\mathfrak{a}M=\sum_j \mathfrak{a}e_j\), for each \(i\), we can obtain \(u(e_i)=\sum_j a_{ij}e_j\) by choosing \(a_{ij}\in \mathfrak{a}\). Now let the standard basis of \(A^n\) be \(\epsilon_1,\ldots,\epsilon_n\), and consider the surjection given by \(\epsilon_i\mapsto e_i\), namely \(\pi: A^n \rightarrow M\), and the endomorphism given by \(\tilde{u}(\epsilon_i)=\sum_j a_{ij}\epsilon_j\), namely \(\tilde{u}\in \End_\rMod{A}(A^n)\); then \(\pi\circ \tilde{u}=u\circ \pi\) holds.

Meanwhile, since \(A^n\) is a free module, we may apply [Multilinear Algebra] §Determinants, ⁋Proposition 9 (Cayley–Hamilton) to \(\tilde{u}\), and therefore, the characteristic polynomial of \(\tilde{u}\), \(p=\chi_{\tilde{u}}\), satisfies \(p(\tilde{u})=0\) and is a monic polynomial of degree \(n\). Then from the commutation relation above, we have \(p(u)\circ \pi=\pi\circ p(\tilde{u})=0\), and since \(\pi\) is surjective, we obtain \(p(u)=0\).

What remains is the condition on the coefficients of \(p\), which follows from the fact that the coefficients of the characteristic polynomial are given by

\[p(\x)=\sum_{k=0}^n (-1)^k\tr\left({\bigwedge}^k(\tilde{u})\right)\x^{n-k}\]

Since the entries of the matrix representation of \(\tilde{u}\) are all \(a_{ij}\), the diagonal entries of the matrix representation of \({\bigwedge}^k(\tilde{u})\) are the \(k\times k\) minors of this matrix, each of which is a sum of products of elements of \(\mathfrak{a}\) consisting of \(k\) factors. That is, \(p_k=(-1)^k\tr({\bigwedge}^k(\tilde{u}))\in \mathfrak{a}^k\).

Then the following holds.

Proposition 2 For a ring \(A\), suppose that an ideal \(\mathfrak{a}\subseteq A[\x]\) is given, let \(B=A[\x]/\mathfrak{a}\), and write \(b\) for \(\x+\mathfrak{a}\in B\). Then the following hold.

1. \(B\) is generated as an \(A\)-module by at most \(n\) elements if and only if \(\mathfrak{a}\) contains a monic polynomial of degree at most \(n\). In this case, \(B\) is generated by \(1,b,\cdots,b^{n-1}\).
2. \(B\) is a free \(A\)-module of rank \(n\) if and only if \(\mathfrak{a}\) is generated by a monic polynomial of degree \(n\). In this case, \(1,b,\cdots,b^{n-1}\) forms a basis of \(B\).

Proof

1. One direction is trivial. Conversely, suppose that \(B\) is generated as an \(A\)-module by \(n\) elements. Now consider multiplying elements of \(B\) by \(b\) to obtain the \(A\)-module endomorphism \(b:B \rightarrow B\). Applying Theorem 1 to the ideal \(A\), we know that this endomorphism satisfies a degree-\(n\) monic polynomial \(p(x)\), and substituting \(b\) as an element must also yield \(0\). Therefore, by the definition of \(b\), we know that \(p(\x)\in \mathfrak{a}\).
2. First, suppose that \(\mathfrak{a}\) is generated by a monic polynomial of degree \(n\). Then by the previous result, we know that \(B\) is generated by \(1,b,\ldots, b^{n-1}\). It now suffices to show that these are linearly independent. In the \(A\)-module \(B\), if \(\sum_{i=0}^{n-1} a_i b^i=0\), then \(q(\x)=\sum_{i=0}^{n-1}a_i\x^i\) must belong to \(\mathfrak{a}\), and because of its degree, \(q=0\).
   Conversely, if \(B\) is a rank-\(n\) free \(A\)-module, then \(B\) can be generated by \(n\) elements, and again from the previous result, \(\mathfrak{a}\) contains a degree-\(n\) monic polynomial \(p\), from which we can also deduce that \(1,b,\ldots, b^{n-1}\) forms a basis of \(B\). It remains to show that \(p\) generates \(\mathfrak{a}\); for this, suppose that an arbitrary \(f\in \mathfrak{a}\) is given, and upon dividing by \(p\), consider the remainder \(r\). Since both polynomials \(f\) and \(p\) belong to \(\mathfrak{a}\), sending this remainder to \(B\) must also yield \(0\). However, this is equal to evaluating the polynomial \(r(\x)\) at \(\x=b\), and viewing this as a linear combination of the basis of \(B\), \(1,\ldots, b^{n-1}\), we know that all coefficients of \(r\) must be \(0\).

Integral Extensions

Earlier, we saw that an \(A\)-algebra \(E\) consists of the same data as a ring homomorphism \(\phi:A \rightarrow E\). More precisely, the ring homomorphism \(\phi:A \rightarrow E\) endows the \(E\)-module \(E\) via restriction of scalars with an \(A\)-module structure \(\phi^\ast E\), and this \(A\)-module structure preserves (by the commutativity of multiplication in \(E\)) the multiplication structure of \(E\), so that it can also be viewed as an \(A\)-algebra.

Meanwhile, since any composition of ring homomorphisms is a ring homomorphism, we know that for an \(A\)-algebra \(E\), the ring homomorphism defined by

\[A\overset{\phi}{\rightarrow}E\hookrightarrow E[\x]\]

gives \(E[\x]\) an \(A\)-algebra structure. Therefore, by [Algebraic Structures] §Algebras, ⁋Proposition 8, a unique \(A\)-algebra homomorphism \(\phi[\x]:A[\x]\rightarrow E[\x]\) is defined; concretely, this is the map that takes an arbitrary polynomial

\[p(\x)=a_n\x^n+\cdots+a_0\in A[\x]\]

and yields the polynomial

\[(\phi[\x](p))(\x)=\phi(a_n)\x^n+\cdots+\phi(a_0)\in E[\x]\]

given by this expression.

Definition 3 Let \(\phi:A\rightarrow E\) be a ring homomorphism.

1. If \(\phi\) is injective, we call \(E\) an extension of \(A\).
2. In \(E\), an element \(x\) is said to be integral over \(\phi\) if there exists a monic polynomial \(p\in A[\x]\) such that \((\phi[\x](p))(x)=0\). If every element of \(E\) is integral over \(\phi\), we call \(\phi\) an integral homomorphism. If an extension \(\phi\) is an integral homomorphism, we call \(E\) an integral extension of \(A\).
3. Among the elements of \(E\), the collection of all elements integral over \(\phi\) is called the integral closure in \(E\) of \(A\), or the normalization in \(E\) of \(A\). If \(A\) is an integral domain, the integral closure in \(\Frac(A)\) of \(A\) is called, with no further qualification, the normalization of \(A\). An integral domain \(A\) is a normal domain if the normalization of \(A\) is itself.
4. If the \(A\)-module \(\phi^\ast E\) is a finitely generated \(A\)-module, we call \(\phi: A \rightarrow E\) a finite homomorphism.
5. If the \(A\)-algebra \(E\) is a finitely generated \(A\)-algebra, we call \(\phi: A \rightarrow E\) a finite type homomorphism.

When the structure morphism \(\phi: A \rightarrow E\) is clear from context, instead of saying that \(x\) is integral over \(\phi\), we also use the expression that \(x\) is integral over \(A\). Similarly, in such cases, instead of saying that \(\phi\) is an integral homomorphism, we say that \(E\) is integral over \(A\).

Any finite homomorphism is a finite type homomorphism by definition. On the other hand, for a finite homomorphism \(\phi: A \rightarrow E\) and any \(x\in E\), since \(x\times-: E \rightarrow E\) is an \(A\)-module endomorphism, applying Theorem 1 shows that \(x\) is integral over \(\phi\). That is, a finite homomorphism is also integral. The following lemma shows the converse.

Lemma 4 A ring homomorphism \(\phi\) is finite if and only if \(\phi\) is an integral homomorphism of finite type.

Proof

One direction was shown above. Conversely, suppose \(E\) is generated by finitely many integral elements; we must show that \(E\) is also finitely generated as an \(A\)-module. For this, let us use induction on the number of generators of \(E\) (as an \(A\)-algebra). If \(E\) is generated as an \(A\)-algebra by \(n\) integral elements \(x_1,\ldots, x_n\), we can consider, generated by \(n-1\) of these elements \(x_1,\ldots, x_{n-1}\), the subalgebra of \(E\), namely the \(A\)-subalgebra \(E'\), which by the inductive hypothesis is finitely generated as an \(A\)-module. Let the generators of \(E'\) (as an \(A\)-module) be \(\{s_i\}\). Then, since \(x_n\) is integral over \(A\), it is also integral over \(E'\), and therefore as an \(E'\)-module, \(E\) must be finitely generated. If we denote these elements by \(\{t_j\}\), we see that \(\{s_i t_j\}\) finitely generates \(E\) as an \(A\)-module.

Meanwhile, it is natural to expect a relationship between \(E\) being integral over \(A\) and each element of \(E\) being integral. To this end, we first prove the following lemma.

Lemma 5 Let a ring homomorphism \(\phi: A \rightarrow E\) and an element of \(E\), \(x\), be given. Then \(x\) is integral over \(A\) if and only if there exist an \(E\)-module \(N\) and an \(A\)-submodule of \(N\), a finitely generated \(A\)-module \(M\), such that \(M\) is not annihilated by any nonzero element of \(E\) and the inclusion \(xM\subseteq M\) holds.

Proof

First, suppose \(x\) is integral over \(A\). Then taking \(N=E\), we know that \(M=A[x]\) is finitely generated by Proposition 2. For the converse, as in the proof of Proposition 2, regard multiplication by \(x\) as an endomorphism of \(M\) and apply Theorem 1.

The following theorem states a property that is natural to expect, but proving it directly from the definition without Lemma 5 is nearly impossible.

Theorem 6 For an \(A\)-algebra \(E\), the integral closure in \(E\) of \(A\) is again an \(A\)-algebra.

Proof

Suppose two elements \(x,y\in E\) are integral over \(A\). Then we must show that \(x+y\) and \(xy\) are integral over \(A\). Now let \(M=A[x]\) and \(M'=A[y]\) be two submodules of \(E\). If we take products of elements of \(M\) and elements of \(M'\) and consider the generated subalgebra of \(E\), \(MM'\), then since each of \(M,M'\) is finitely generated, \(MM'\) is also finitely generated. Now, since

\[(xy)MM'=(xM)(yM')\subseteq MM',\qquad (x+y)MM'\subseteq xMM'+M(yM')\subseteq MM'\]

by using Lemma 5, we obtain the desired result.

Nakayama’s Lemma

We now prove a very useful lemma. First, recall that the nilradical \(\mathfrak{N}(A)=\sqrt{(0)}\) of [Algebraic Structures] §Field of Fractions, ⁋Definition 14 is given by

\[\mathfrak{N}(A)=\bigcap_\text{\scriptsize$\mathfrak{p}$ prime} \mathfrak{p}\]

(§Properties of Localization, ⁋Corollary 8) In a similar fashion, we define the Jacobson radical of \(A\) by

\[J(A)=\bigcap_\text{\scriptsize$\mathfrak{m}$ maximal} \mathfrak{m}\]

To prove Nakayama’s lemma, we must first prove the following lemma.

Lemma 7 Suppose a finitely generated \(A\)-module \(M\) and an ideal of \(A\), \(\mathfrak{a}\), satisfy \(\mathfrak{a}M=M\). Then there exists some \(a\in \mathfrak{a}\) such that \((1-a)M=0\).

Proof

From the given condition, \(M\subseteq \mathfrak{a}M\), so by Theorem 1 there exists a monic polynomial

\[p(\x)=\x^n+p_1\x^{n-1}+\cdots+p_n,\qquad p_k\in \mathfrak{a}^k\]

such that \(p(\id_M)=0\). That is,

\[(1+p_1+\cdots+p_n)M=0\]

and setting \(a=-(p_1+\cdots+p_n)\) gives the desired result.

Now we can finally state Nakayama’s lemma.

Lemma 8 (Nakayama) Suppose an ideal of \(A\) contained in the Jacobson radical \(J(A)\), \(\mathfrak{a}\), is given, and let \(M\) be a finitely generated \(A\)-module. Then the following holds.

1. If \(\mathfrak{a}M=M\), then \(M=0\).
2. If the images of \(x_1,\ldots, x_n\) in \(M/\mathfrak{a}M\) generate \(M/\mathfrak{a}M\) as an \(A\)-module, then \(x_1,\ldots, x_n\) generate \(M\) as an \(A\)-module.

Proof

For the first statement, the element \(a\in \mathfrak{a}\) obtained from Lemma 7 belongs to every maximal ideal by assumption. In other words, \(1-a\) cannot belong to any maximal ideal, so \(1-a\) is a unit. Therefore, we obtain the desired result.

For the second statement, let \(N=M/\sum_i Ax_i\). Then one can show that \(N/\mathfrak{a}N=0\), and from the first statement we know that \(N=0\).

Localization

We now examine several results related to localization.

Proposition 9 A unique factorization domain is a normal domain.

Proof

For any \(a/b\in \Frac(A)\), suppose that \(a,b\) are coprime and \(a/b\) is contained in the normalization of \(A\). Then there exists a monic polynomial such that

\[\left(\frac{a}{b}\right)^n+a_{n-1}\left(\frac{a}{b}\right)^{n-1}+\cdots+a_1\left(\frac{a}{b}\right)+a_0=0\]

holds. Now from this, we know that

\[\x^n+a_{n-1}b \x^{n-1}+\cdots+a_1b^{n-1}\x+a_0b^n\in A[\x]\]

is a monic polynomial that, upon substituting \(\x=a\), becomes \(0\). That is, \(a^n\) is divisible by \(b\), and for this not to be a contradiction, \(b\) must be a unit. That is, \(A\) must be a normal domain, and this completes the proof.

More generally, the following holds.

Proposition 10 Let rings \(A\subseteq B\) be given and let a monic polynomial \(p\in A[\x]\) be given. If in \(B[\x]\) one can find, such that \(p=q_1q_2\), monic polynomials \(q_1,q_2\in B[\x]\), then the coefficients of \(q_1,q_2\) are integral over \(A\).

Proof

By adjoining roots of the equations, with \(B\) contained in a suitable ring \(C\), we can ensure that in \(C[\x]\), \(q_1,q_2\) both factor into the form \(\prod (\x-\alpha_i)\), \(\prod(\x-\beta_j)\). Then the \(\alpha_i,\beta_j\) are all roots of the monic polynomial in \(A[\x]\), \(p=q_1q_2\), so they are integral over \(A\), and therefore the subring of \(C\) generated by them, \(C'\), is an integral \(A\)-algebra. On the other hand, expanding \(q_1=\prod(\x-\alpha_i)\) and \(q_2=\prod(\x-\beta_j)\), we see that their coefficients are obtained by ring operations on the \(\alpha_i,\beta_j\), so they belong to \(C'\).

Therefore, the following holds.

Corollary 11 For a normal domain \(A\), every monic irreducible polynomial is prime.

Meanwhile, normalization commutes with localization, and its proof is also trivial.

Proposition 12 Fix rings \(A\subseteq B\) and a multiplicative subset of \(A\), \(S\). Then, for the integral closure of \(A\) in \(B\), \(A'\), \(S^{-1}A'\) is the integral closure of \(S^{-1}A\) in \(S^{-1}B\).

Another result related to localization is a slight strengthening of §Properties of Localization, ⁋Proposition 4. First, that a ring \(A\) is a semilocal ring means that \(A\) has only finitely many maximal ideals. Then the following holds.

Proposition 13 For a semilocal ring \(A\) and finitely presented \(A\)-modules \(M,N\), if \(M_\mathfrak{m}\cong N_\mathfrak{m}\) holds for every maximal ideal \(\mathfrak{m}\), then \(M\cong N\).

Proof

First, let the maximal ideals of \(A\) be \(\mathfrak{m}_1,\ldots, \mathfrak{m}_n\), and suppose that for each \(k\) an isomorphism \(u_k: M_{\mathfrak{m}_k}\rightarrow N_{\mathfrak{m}_k}\) is given. Then \(A_{\mathfrak{m}_k}\) is a flat \(A\)-module, and since \(M\) is finitely presented by assumption, there exists an isomorphism

\[\Hom_{A_{\mathfrak{m}_k}}( M_{\mathfrak{m}_k}, N_{\mathfrak{m}_k}) \rightarrow \Hom_A(M,N)_{\mathfrak{m}_k}\]

(§Properties of Localization, ⁋Proposition 5). The image of \(u_k\) under this isomorphism is an element of \(\Hom_A(M,N)\) with an element of \(A\setminus \mathfrak{m}_k\) placed in the denominator, so by multiplying \(u_k\) by this denominator if necessary, we can treat \(v_k\) as an element of \(\Hom_A(M,N)\).

Meanwhile, since \(\mathfrak{m}_k\) is prime, if \(\bigcap_{l\neq k} \mathfrak{m}_l\subseteq \mathfrak{m}_k\), then for some \(l\) we must have \(\mathfrak{m}_l\subseteq \mathfrak{m}_k\), which is a contradiction. That is,

\[\bigcap_{l\neq k} \mathfrak{m}_l\not\subseteq \mathfrak{m}_k\]

holds, and from this, an element satisfying \(a_k\in \bigcap_{l\neq k}\mathfrak{m}_l\) but \(a_k\not\in \mathfrak{m}_k\), namely \(a_k\), exists. For these \(a_k\) so constructed, if we define \(v=\sum_{k=1}^n a_kv_k\), this becomes our desired isomorphism, and for its proof, by §Properties of Localization, ⁋Proposition 4, it suffices to examine the localization at each maximal ideal \(\mathfrak{m}_k\).

More generally, we show that for any local ring \((B, \mathfrak{n})\) and finitely generated \(B\)-modules with maps \(s,t:K \rightarrow L\) between them, if \(s\) is an isomorphism and \(t(K)\subseteq \mathfrak{n}L\), then \(s+t\) is also an isomorphism. Applying this result to the local ring \((A_{\mathfrak{m}_k}, \mathfrak{m}_kA_{\mathfrak{m}_k})\), and to the maps from \(M_{\mathfrak{m}_k}\) to \(N_{\mathfrak{m}_k}\) given by \(s=a_k v_k\) and \(t=\sum_{l\neq k} a_lv_l\), will then complete the proof.

Let us prove this claim. First, \(t\) can be viewed as the zero map from \(K\) to \(L/\mathfrak{n}L\), and \(s\) can be viewed as an epimorphism from \(K\) to \(L/\mathfrak{n}L\), so \(s+t\) can also be viewed as an epimorphism from \(K\) to \(L/\mathfrak{n}L\). Then by Lemma 8 (Nakayama), as a morphism from \(K\) to \(L\), \(s+t\) is also an epimorphism. Now, for the isomorphism \(s\), take its inverse \(s^{-1}\) and consider the surjective endomorphism \(w=s^{-1}(s+t): K \rightarrow K\). Then we can view \(K\), with \(\x\) acting as \(w\), as a \(B[\x]\)-module; since \(K\) is finitely generated as a \(B\)-module, it is also so as a \(B[\x]\)-module. Meanwhile, since \(w\) is surjective, for the ideal \(\mathfrak{a}=(\x)\subseteq B[\x]\) we have \(\mathfrak{a}K=w(K)=K\), and therefore by Lemma 7 there exists some \(q\in B[\x]\) such that \((1-\x q(\x))K=0\), that is, \(q(w)\circ w=\id_K\). From this, \(w\) is also a monomorphism and thus an isomorphism, and therefore \(s+t\) is a monomorphism, which gives the desired result.

Now let us examine the properties of the ring homomorphisms in Definition 3 a bit further. First, we begin with the following proposition.

Proposition 14 Let ring homomorphisms \(\phi:A \rightarrow E\) and \(\rho: A \rightarrow A'\) be given. Setting \(E'=A'\otimes_AE\), the following holds for \(\rho_!\phi: A' \rightarrow E'\).

1. If \(\phi\) is integral, then \(\rho_!\phi\) is also integral.
2. If \(\phi\) is finite, then \(\rho_!\phi\) is also finite.

Proof

Since the proofs of both claims are similar, we only prove the first. Assume that \(\phi\) is integral, and let the integral elements generating \(E\) be \(x_i\). Then each \(x_i\) satisfies a suitable monic polynomial \(p_i\in A[\x]\). Therefore, each \(1\otimes x_i\) generates \(E'\) and satisfies \((\rho[\x])(p_i)\). From this, we see that \(E'\) is integral over \(A'\).

The following also holds.

Proposition 15 Let a ring homomorphism \(\phi:A \rightarrow E\) be given. If \((a_1, \ldots, a_n)=A\), then the following holds.

1. If each \(A_{a_i} \rightarrow E_{a_i}\) is integral, then so is \(A \rightarrow E\).
2. If each \(A_{a_i} \rightarrow E_{a_i}\) is finite, then so is \(A \rightarrow E\).

Proof

As above, since the proofs of the two claims are similar, we only prove the first claim. For any \(x\in E\), let the ideal of polynomials in \(A[\x]\) satisfying \(p(x)=0\) be \(\mathfrak{A}\). Then one can verify that the subset of \(A\) consisting of the leading coefficients of these polynomials, \(\mathfrak{a}\), is an ideal of \(A\).

Meanwhile, from the assumption that each \(A_{a_i} \rightarrow E_{a_i}\) is integral, the image of \(x\) in \(E_{a_i}\) is integral over \(A_{a_i}\); thus, after writing down an integral equation for this element and multiplying by all the powers of \(a_i\) in the denominators, we know that for each given \(i\), there exists an appropriate \(n_i\) such that \(a_i^{n_i}\in \mathfrak{a}\) holds. On the other hand, raising both sides of the equation

\[1=\sum_{i=1}^n \alpha_i a_i\]

to a sufficiently high power, we know that these \(a_i^{n_i}\) also generate the unit ideal, and therefore \(1\in \mathfrak{a}\). From this, we know that \(x\) is integral by definition.

Proposition 16 For a ring homomorphism \(\phi: A \rightarrow E\) and \(x\in E\), \(x\) is integral over \(\phi\) if and only if, for every prime ideal of \(A\), \(\mathfrak{p}\), the image of \(x\) in \(E_\mathfrak{p}\) is integral over \(\phi_\mathfrak{p}: A_\mathfrak{p} \rightarrow E_\mathfrak{p}\).

Proof

One direction follows from Proposition 14. Therefore, for any prime ideal \(\mathfrak{p}\subseteq A\), suppose that the image of \(x\) in \(E_\mathfrak{p}\) is integral over \(A_\mathfrak{p}\), and let us show that \(x\) is integral. For this, after taking the subalgebra generated by \(\phi(A)\) and \(x\) in \(E\), namely the \(A\)-subalgebra \(E'\), it suffices to show that \(A \rightarrow E'\) is integral.

First, from the assumption, in \(E_\mathfrak{p}\) the equation

\[x^d+\phi_\mathfrak{p}(a_{d-1})x^{d-1}+\cdots+\phi_\mathfrak{p}(a_0)=0\]

holds for some \(a_i\in A_\mathfrak{p}\). Now, finding a common denominator for the \(\phi_\mathfrak{p}(a_k)\), for a suitable \(f\in A\setminus \mathfrak{p}\) we can view each \(a_k\) as belonging to \(A_f\). Then the left-hand side of the above equation can be viewed as an element of \(E_f\), and when sent to \(E_\mathfrak{p}\), this element becomes \(0\). However, since \(E_\mathfrak{p}\) is the localization of \(E_f\) with respect to the image of \(A\setminus \mathfrak{p}\), there exists some \(h\in A\setminus \mathfrak{p}\) such that multiplying this left-hand side by \(h\) is, in \(E_f\), equal to \(0\), and therefore, replacing \(f\) by \(fh\), the above equation becomes an equation that holds in \(E_f\). Then the image of \(x\) in \(E_f\) is integral over \(A_f\), and since \(E_f'\) is generated by this element as an \(A_f\)-algebra, by Lemma 4 we know that \(A_f \rightarrow E_f'\) is a finite homomorphism.

In this way, for each prime ideal \(\mathfrak{p}\), we obtain an element \(f_\mathfrak{p}\in A\setminus \mathfrak{p}\); the ideal of \(A\) generated by these elements is contained in no prime ideal, so it is \(A\) itself. We can therefore choose finitely many prime ideals \(\mathfrak{p}_1,\ldots, \mathfrak{p}_m\) such that \((f_{\mathfrak{p}_1},\ldots, f_{\mathfrak{p}_m})=A\), and since each \(A_{f_{\mathfrak{p}_i}} \rightarrow E_{f_{\mathfrak{p}_i}}'\) is finite, it is integral. Then by Proposition 15, \(A \rightarrow E'\) is integral, and in particular \(x\) is integral over \(\phi\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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