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 Corollary 4.23(b) below.
The proof of the following lemma uses 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, \[ \mathrm {det} M_j = \sum _{i=1}^n x_i \mathrm {det} [ m_1\ m_2\ \ldots \ m_{j-1}\ m_i\ m_{j+1}\ \ldots \ m_n] = x_j \mathrm {det} M. \]
Lemma 4.21. Let \(A \subset B\) be commutative rings and \(b\in B\). The following conditions are equivalent:
Proof. \(i.\) implies \(ii.\) : By the hypothesis, we have a monic polynomial \(g \in A[X]\) of degree \(m\) 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 \(\deg (r) < m\). Then \(f(b) = g(b) h(b) + r(b) = r(b)\). 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 \[ b c_i = \sum _{j=1}^n a_{ij} c_j \text { for some } a_{ij} \in A. \] Let \(N\) be the \(n\)-by-\(n\) matrix with entries \(a_{ij}\) and let \(x\) be the column vector with entries \(c_i\). Then \(Nx = b x\), or equivalently \((bI_n - N) x =0\). By Cramer’s rule applied to \(M = bI_n - N \in M_n(B)\) and \(y=0\), we get \(\mathrm {det} M_j = 0\) and hence \[ c_i \mathrm {det}(bI_n - N) =0 \text { for all } i. \] 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 \[ \mathrm {det}(bI_n - N) = \sum _{i=1}^n a_i c_i \mathrm {det}(bI_n - N) = 0. \] Let \(p(X) = \mathrm {det}(X I_n - N) \in A[X]\). Expanding the determinant shows that \(p(X)\) is monic of degree \(n\), and \(p(b)=0\). □
\(\blacktriangleright \) For example, let \(B=k[X,Y]/(X^2-Y^2)\), and let \(x, y \in B\) be the classes of \(X, Y\). The subring \(A = k[y] \subset B\) is isomorphic to \(k[Y]\), since \((X^2-Y^2) \cap k[Y] = \{0\}\), and \(B= A\cdot 1 + A \cdot x\) since \(x^2 = y^2\). Hence \(B\) is a finitely generated \(A\)-module, so \(B\) is integral over \(A\) by \(iii.\) of the lemma.
\(\blacktriangleright \) Geometrically, the inclusion \(k[y] \subset B\) in the previous example corresponds to the projection of \(\mathcal {V}(X^2-Y^2)\), the two lines \(x = \pm y\) in \(k^2\), onto the \(y\)-axis: every fibre is finite (two points, or one point over \(y=0\)). Compare this with \(k[x] \subset C = k[X,Y]/(XY-1)\), the projection of the hyperbola \(xy=1\) onto the \(x\)-axis: the fibre over \(x=0\) is empty, the hyperbola “escapes to infinity” as \(x \to 0\), and indeed \(y\) is not integral over \(k[x]\), since multiplying an equation \(y^m + a_{m-1}(x) y^{m-1} + \ldots + a_0(x) = 0\) by \(x^m\) gives \(1 + a_{m-1}(x) x + \ldots + a_0(x) x^m = 0\) in \(k[x] \subset C\), which is impossible. In general, for \(k\) algebraically closed, an integral extension of coordinate rings of affine varieties corresponds to a map of varieties which is onto with finite fibres; we will prove the surjectivity (the lying-over theorem) in Lecture 22.
Definition 4.22. Let \(k\) be a field. We say that \(A\) is a \(k\)-algebra if
The 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 it is generated by finitely many elements as an algebra: there exist \(a_1,\ldots , a_n \in A\) 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 \(k\)-algebra homomorphism \[ \pi : k[X_1,\ldots , X_n] \to A \] given by \(\pi (X_i) = a_i\), and \(A\) is isomorphic to \(k[X_1,\ldots , X_n] / \ker (\pi )\). Thus every finitely generated \(k\)-algebra is of the form \(k[X_1,\ldots , X_n]/I\).
More generally, let \(A \subset B\) be commutative rings. For \(b_1,\ldots , b_n \in B\) we write \(A[b_1,\ldots , b_n] = \{ f(b_1,\ldots , b_n) : f \in A[X_1,\ldots , X_n] \}\) for the subring of \(B\) generated by \(A\) and \(b_1,\ldots , b_n\). We say that \(B\) is finitely generated as an \(A\)-algebra if \(B=A[b_1,\ldots , b_n]\) for some \(b_1,\ldots ,b_n \in B\), or equivalently if \(B\) is isomorphic, as an \(A\)-algebra, to \(A[X_1,\ldots , X_n]/ I\) for some \(n\) and some ideal \(I\). It is important not to confuse this with the similar sounding statement that \(B\) is a finitely generated \(A\)-module: \(A[X]\) is finitely generated as an \(A\)-algebra but not as an \(A\)-module.
Corollary 4.23. Let \(A\) and \(B\) be commutative rings with \(A \subset B\).
Proof. \((a)\) We first show by induction on \(n\) that \(B\) is a finitely generated \(A\)-module. For \(n=1\) this is \(i. \Rightarrow ii.\) of Lemma 4.21. For \(n \geq 2\), put \(B' = A[b_1,\ldots , b_{n-1}]\), so that \(B = B'[b_n]\). By induction \(B' = \sum _i A u_i\) for finitely many \(u_i \in B'\), and since \(b_n\) is integral over \(B'\), Lemma 4.21 gives \(B = \sum _j B' v_j\) for finitely many \(v_j \in B\). Hence \(B = \sum _{i,j} A u_i v_j\) is a finitely generated \(A\)-module. Now for every \(b \in B\), condition \(iii.\) of Lemma 4.21 holds with \(C = B\), so \(b\) is integral over \(A\).
\((b)\) Every \(a \in A\) is a root of \(X - a\), so \(A \subset \overline {A}_B\); in particular \(1 \in \overline {A}_B\). If \(b, b' \in \overline {A}_B\), then \(b'\) is also integral over \(A[b]\), so \(A[b,b']\) is integral over \(A\) by \((a)\); in particular \(b \pm b'\) and \(bb'\) lie in \(\overline {A}_B\). Hence \(\overline {A}_B\) is a subring of \(B\).
\((c)\) Let \(b \in B\). Since \(b\) is integral over \(C\), we have an equation \[ b^m + c_{m-1} b^{m-1} + \ldots + c_1 b + c_0 = 0 \text {\ for some\ } c_0,\ldots , c_{m-1} \in C. \] Each \(c_i\) is integral over \(A\), hence over \(A[c_0,\ldots ,c_{i-1}]\), and the equation shows that \(b\) is integral over \(A[c_0,\ldots , c_{m-1}]\). By part \((a)\), applied to the generators \(c_0,\ldots ,c_{m-1},b\) in this order, the ring \(A[c_0,\ldots , c_{m-1}, b] \subset B\) is integral over \(A\). Hence \(b\) is integral over \(A\).
\((d)\) Let \(B\) be a field, and let \(a \in A\) be a non-zero element. Then \(a^{-1} \in B\), and we want to show that \(a^{-1}\in A\). Since \(B\) is integral over \(A\), there is a relation \[ a^{-n} + a_{n-1} a^{-n+1} + \ldots + a_0 =0 \text {\ with \ } a_i \in A. \] Multiplying by \(a^{n-1}\), we get \(a^{-1} = - (a_{n-1} + a_{n-2} a + \ldots + a_0 a^{n-1}) \in A\).
Conversely, suppose \(A\) is a field and \(B\) is an integral domain. Let \(b \in B\) be a non-zero element, and choose an equation \[ b^m + a_{m-1}b^{m-1} + \ldots + a_1 b + a_0 =0 \text {\ with \ } a_i \in A \] of smallest possible degree \(m\). If \(a_0 = 0\) then, since \(B\) is an integral domain and \(b \neq 0\), we may cancel \(b\) and obtain a monic equation of smaller degree, contradicting the choice of \(m\). Hence \(a_0 \neq 0\), and \[ b^{-1} = - a_0^{-1} (b^{m-1}+ a_{m-1} b^{m-2}+ \ldots + a_{1}) \in B. \] □
\(\blacktriangleright \) Let \(A \subset B\) be commutative rings. Then \(B\) is finitely generated as an \(A\)-module if and only if \(B\) is finitely generated as an \(A\)-algebra and integral over \(A\). Indeed, module generators are algebra generators, and integrality follows from \(iii.\) of Lemma 4.21; the converse is Corollary 4.23(a).
Definition 4.24. \(A\) is said to be integrally closed in \(B\) if \(\overline {A}_B = A\). If \(A\) is an integral domain, we regard \(A\) as a subring of its field of fractions \(\mathrm {Frac}(A)\). We say that an integral domain \(A\) is normal if it is integrally closed in \(\mathrm {Frac}(A)\), and the integral closure of \(A\) in \(\mathrm {Frac}(A)\) is called the normalisation of \(A\).
\(\sphericalangle \) Show that a UFD is normal (the argument given for \(\mathbb {Z}\) at the start of this lecture works verbatim). In particular, \(k[X_1,\ldots , X_n]\) is normal.
\(\sphericalangle \) Let \(A= k[X,Y]/(Y^2-X^3)\), and let \(x, y \in A\) be the classes of \(X, Y\). Show that \(x \mapsto t^2\), \(y \mapsto t^3\) identifies \(A\) with the subring \(k[t^2,t^3]\) of \(k[t]\), so that \(A\) is an integral domain with \(\mathrm {Frac}(A) = k(t)\). The element \(t= y/x \in \mathrm {Frac}(A)\) is not in \(A\) but satisfies the monic equation \(t^2 - x=0\) (and \(t^3-y=0\)), so \(A\) is not normal. Show that the normalisation of \(A\) is \(k[t]\).
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 icFractions R
The last command shows that the new variable \(w\) of \(S\) is the fraction \(y/x\), and \[ S = \mathbb {Q}[w,x,y]/(wy-x^2,\, wx-y,\, w^2-x) \simeq \mathbb {Q}[w], \] as the relations allow one to eliminate \(x = w^2\) and \(y = w^3\).