Definition 4.20. Suppose \(A \subset B\) are commutative rings. We say that \(b \in B\) is integral over \(A\) if there exist \(m \geq 1\) and \(a_0, a_1,\ldots , a_{m-1} \in A\) such that \[ b^m + a_{m-1} b^{m-1} + \ldots + a_1 b +a_0 = 0. \] We say that \(B\) is integral over \(A\) if all elements of \(B\) are integral over \(A\).
In other words, \(b \in B\) is integral over \(A\) if there exists a monic polynomial \(f \in A[X]\) such that \(f(b)=0\).
\(\blacktriangleright \) Let \(A=\mathbb {Z}\) and \(B=\mathbb {Q}\). The elements of \(\mathbb {Q}\) integral over \(\mathbb {Z}\) are exactly the integers: if \(b = p/q\) in lowest terms satisfies a monic equation of degree \(m\) over \(\mathbb {Z}\), then multiplying by \(q^m\) shows that \(q\) divides \(p^m\), so \(q = \pm 1\).
\(\blacktriangleright \) Let \(A = \mathbb {Z}\) and \(B = \mathbb {Q}(i)\). The elements of \(B\) integral over \(\mathbb {Z}\) are the Gaussian integers \(\mathbb {Z}[i]\).
\(\blacktriangleright \) Let \(A = \mathbb {Z}[\sqrt {-3}]\) and \(B= \mathbb {Q}(\sqrt {-3})\). Then \(\omega = (-1+\sqrt {-3})/2 \notin A\) is integral over \(A\), since \(\omega ^2 + \omega + 1 = 0\). The elements of \(B\) integral over \(A\) are the Eisenstein integers \(\mathbb {Z}[\omega ]\).
More generally, in algebraic number theory one studies the ring of integers \(\mathcal {O}_K\) of a number field \(K\) (a finite extension of \(\mathbb {Q}\)), defined as \[ \mathcal {O}_K = \{ b \in K : b \text {\ is integral over \ }\mathbb {Z} \}. \] That this is a ring is not obvious from the definition, but follows from the characterisation of integral elements given in Lemma 4.21 below.
Lemma 4.21. Let \(A \subset B\) be commutative rings and \(b\in B\). The following conditions are equivalent:
We will need Cramer’s rule from linear algebra, which we recall for convenience; it holds over any commutative ring \(R\). Let \(M \in M_{n}(R)\) be an \(n\)-by-\(n\) matrix with entries in \(R\) and let \(Mx = y\) be a system of \(n\) linear equations, where \(x\) and \(y\) are column vectors with entries in \(R\). Write \(M = [m_1\ m_2\ \ldots \ m_n]\) where \(m_i\) are the columns of \(M\). Then the \(n\) linear equations can be written as \[ x_1 m_1 + x_2 m_2 + \ldots + x_n m_n = y. \] Define new \(n\)-by-\(n\) matrices \(M_j = [ m_1\ m_2\ \ldots \ m_{j-1}\ y\ m_{j+1}\ \ldots \ m_n]\). Since the determinant is linear in each column and vanishes when two columns coincide, \[ \mathrm {det} M_j = \sum _{i=1}^n x_i \mathrm {det} [ m_1\ m_2\ \ldots \ m_{j-1}\ m_i\ m_{j+1}\ \ldots \ m_n] = x_j \mathrm {det} M. \]
Proof. (of Lemma 4.21) \(i.\) implies \(ii.\) : By the hypothesis, we have a monic polynomial \(g \in A[X]\) such that \(g(b)=0\). Given \(f \in A[X]\), we can divide by \(g\) (as \(g\) is monic!) and write \[ f = gh +r , \text { with } h, r \in A[X] \] where \(r=0\) or \(deg(r) < deg(g) =m\). Then, \(f(b) = g(b) h(b) + r(b)\). Hence, \(A[b] = \{ r(b) : r=0 \text { or } deg(r) < m \} =A\cdot 1 + A \cdot b + \ldots + A \cdot b^{m-1}\).
\(ii.\) implies \(iii.\) : Trivial. Take \(C= A[b]\).
\(iii.\) implies \(i.\) : We have \(C= Ac_1 + A c_2+ \ldots + A c_n\) with \(c_i \in C\). Since \(b \in C\) and \(C\) is a ring, we can write \[ b c_i = \sum _{j=1}^n a_{ij} c_j \text { for } a_{ij} \in A \] Consider the \(n\)-by-\(n\) matrix \(N\) with entries \(a_{ij}\) and \(x\) be the column vector with entries \(c_i\). Then, we have \(Nx = b x\) or equivalently \((N- b I_{n}) x =0\). By Cramer’s rule applied to \(M = N-bI_n\), we get the equations \[ c_i \mathrm {det}(N-bI_n) =0 , \text { for all } i. \] (\(\mathrm {det} M_j=0\) since \(y=0\).) On the other hand, \(1 \in C\), hence there exist \(a_i \in A\) such that \(1 =\sum _{i=1}^n a_i c_i\). It follows that \[ \mathrm {det}(N-b I_n) = \sum _{i=1}^n a_i c_i \mathrm {det}(N-bI_n) = 0. \] Let \(p(X) = \mathrm {det}(N- X I_n)\). We have \(p(X) \in A[X]\) is a monic polynomial satisfying \(p(b)=0\). □
\(\blacktriangleright \) For example, consider the ring \(B=k[X,Y]/(X^2-Y^2)\) and its subring isomorphic to \(A=k[Y]\) generated by \(Y + (X^2-Y^2)\). It is easy to see \(B\) is finite as an \(A\)-module. Indeed, \(B= A\cdot 1 + A \cdot X\). Hence, \(B\) is integral over \(A\).
Corollary 4.22. Let \(A\) and \(B\) be commutative rings with \(A \subset B\).
Proof. \((a)\) Let \(B= A[b,b']\) and \(b'\) is integral over \(A[b]\) and \(b\) is integral over \(A\). Then by Lemma 4.21, we see that \(A[b]\) is a finitely generated \(A\)-module, and \(A[b,b']\) is a finitely generated \(A[b]\)-module. It follows that \(A[b,b']\) is finitely generated as an \(A\)-module. Then, again by Lemma 4.21, it follows that \(B\) is integral over \(A\). The general case follows from this by recursion.
\((b)\) Given \(b,b' \in B\) integral over \(A\) from \((a)\) we get \(A[b,b'] \subset B\) is finitely generated over \(A\) and so integral over \(A\).
\((c)\) Let \(b \in B\) integral over \(C\), then we have an equation \[ b^m + c_{m-1} b^{m-1} + \ldots + c_1 b + c_0 = 0 \text {\ for some\ } c_0,\ldots , c_{m-1} \in C\] Let \(C' = A[b,c_0,\ldots , c_{m-1}] \subset B\). The ring \(A[c_0,\ldots , c_{m-1}]\) is integral over \(A\), and by part \((a)\), \(C'\) is integral over \(A\). Hence, \(b\) is integral over \(A\).
\((d)\) Let \(B\) be a field, and \(a \in A\) be a nonzero element. Then \(a^{-1} \in B\). We want to show that \(a^{-1}\in A\). Since \(B\) is integral over \(A\), there is a relation \[ a^{-n} + a_{n-1} a^{-n+1} + \ldots + a_0 =0, \text {\ with \ } a_i \in A \] Then, we have \(a^{-1} = - (a_{n-1} + a_{n-2} a + \ldots + a_0 a^{n-1}) \in A\).
Conversely, suppose \(A\) is a field and \(B\) is an integral domain. Let \(b \in B\) a non-zero element. We have an equation \[ b^m + a_{m-1}b^{m-1} + \ldots + a_1 b + a_0 =0 \text {\ with \ } a_i \in A \] Since \(B\) is an integral domain, we can assume that \(a_0 \neq 0\). Then, \[ b^{-1} = - a_0^{-1} (b^{m-1}+ a_{m-1} b^{m-2}+ \ldots + a_{1}). \] □
Definition 4.23. \(A\) is said to be integrally closed in \(B\) if \(\overline {A}_B = A\). Assuming that \(A\) is an integral domain, we can see \(A\) embedded in its field of fractions: \(A \subset Frac(A)\). We say that an integral domain is normal if it is integrally closed in its field of fractions. If \(A\) is an integral domain, the integral closure of \(A\) in \(Frac(A)\) is called the normalisation of \(A\).
\(\blacktriangleright \) The ring \(k[X_1,\ldots , X_n]\) is normal.
\(\sphericalangle \) More generally, a ring which is a UFD is normal.
\(\sphericalangle \) The ring \(A= k[x,y]/(y^2-x^3)\) is not normal. Indeed, the element \(t= y/x \in Frac(A)\) satisfies the monic equation \(t^2 - x=0\) (and \(t^3-y=0\)). Show that the normalisation of \(A\) is isomorphic to \(k[t]\).
Macaulay2 can tell you whether a ring is normal, and compute the normalisation if not.
R = QQ[x,y]/(y^2-x^3) isNormal R S = integralClosure R describe S
Definition 4.24. Let \(k\) be a field. We say that \(A\) is a \(k\)-algebra if
Standard example of a \(k\)-algebra is the polynomial ring \(k[X_1,\ldots , X_n]\) or more generally a quotient ring \(A = k[X_1,\ldots , X_n]/I\) by an ideal \(I\). We say that a \(k\)-algebra \(A\) is finitely generated if \(A\) is generated by a finitely many number of elements as an algebra. More precisely, there exist elements \(a_1,\ldots , a_n \in A\) such that \(A = \{ f(a_1,\ldots , a_n) : f\in k[X_1,\ldots , X_n] \}\). We write \(A=k[a_1,\ldots , a_n]\). In this case, we have a surjective homomorphism \[ \pi : k[X_1,\ldots , X_n] \to A \] given by \(\pi (X_i) = a_i\) and \(A\) is isomorphic to \(k[X_1,\ldots , X_n] / \mathrm {ker}(\pi )\).
More generally, if \(A \subset B\) are commutative rings. We say that \(B\) is finitely generated as an \(A\)-algebra if \(B\) is isomorphic to \(A[X_1,\ldots , X_n]/ I\) for some \(n\) and ideal \(I\) in which case we can write \(B=A[b_1,\ldots , b_n]\) for some \(b_i \in B\). It is important not to confuse this with the similar sounding statement that says that \(B\) is a finitely generated \(A\)-module.
\(\blacktriangleright \) Let \(A \subset B\) be commutative rings. It follows easily from Lemma 4.21 that \(B\) is finitely generated as an \(A\)-module if and only if \(B\) is finitely generated as an \(A\)-algebra and integral over \(A\).
We say that \(a_1,\ldots , a_n\) are algebraically independent if there is no non-zero polynomial \(F \in k[X_1,\ldots , X_n]\) such that \(F(a_1,\ldots , a_n)=0\). This is equivalent to the injectivity of the homomorphism \(\pi \).