Spring 2026: Math 831 Daily Homework

1. Let \(P\subseteq R\) be an ideal. Show that the following are equivalent:

1. (i) \(P\) is a prime ideal.
2. (ii) For all ideals \(I,J\subseteq R\), if \(IJ\subseteq P\), then \(I\subseteq P\) or \(J\subseteq P\).

Conclude that if \(P\) is a prime ideal and \(I\cap J \subseteq P\) for ideals \(I,J\subseteq R\), then \(I\subseteq P\) or \(J\subseteq P\).

2. For an ideal \(I\subseteq R\), show that:

1. (i) \(I\) is a prime ideal if and only if \(R/I\) is an integral domain.
2. (ii) \(I\) is a maximal ideal if and only if \(R/I\) is a field.

3. Let \(R[x]\) denote the polynomial ring with coefficients in \(R\). For an ideal \(I\subseteq R\), we let \(I[x]\) denote the polynomials \(f(x)\in R[x]\), all of whose coefficients are in \(I\).

1. (i) Prove that \(I[x]\) is an ideal of \(R[x]\) and equals the ideal in \(R[x]\) generated by \(I\).
2. (ii) Prove that \(R[x]/I[x]\) is isomorphic to \((R/I)[x]\).
3. (iii) Prove that if \(P\subseteq R\) is a prime ideal, then \(P[x]\) is a prime ideal in \(R[x]\).
4. (iv) Let \(M\subseteq R\) be a maximal ideal. Show that \(M[x]\) is never a maximal ideal.

1. Let \(S, T\subseteq R\) be multiplicatively closed subsets of \(R\) such that \(st \not = 0\), for all \(s\in S\) and \(t\in T\). Let \(T'\) denote the set of fractions \(\frac{t}{1}\) in \(R_S\) such that \(t\in T\). Prove that \(ST\) is a multiplicatively closed subset of \(R\), \(T'\) is a multiplicatively closed subset of \(R_S\) and \(R_{ST}\cong (R_S)_{T'}\).

2. Let \(I\subseteq R\) be an ideal. Show that there is a one-to-one correspondence between the ideals \(J\subseteq R\) containing \(I\) and the ideals \(C\subseteq R/I\). Conclude that every ideal \(C\subseteq R/I\) is of the form \(J/I\) for an ideal \(J\subseteq R\) containing \(I\). Show that this correspondence extends to prime ideals so that \(Q\subseteq R/I\) is a prime ideal if and only if \(Q = P/I\), for a prime ideal \(P\) of \(R\) containing \(I\).

3. Let \(S\subseteq R\) be a multiplicatively closed set, \(I\subseteq R\) an ideal and \(R[x]\) the polynomial ring in one variable over \(R\).

1. (i) Prove that \(R_S[x]\) is isomorphic to \(R[x]_S\).
2. (ii) Use (i), problem 2 above and problem 3 from January 20 to prove that there cannot exist a chain of prime ideals \(Q_1\subset Q_2\subset Q_3\) in \(R[x]\) contracting to same prime ideal in \(R\). Hint: Mod out \(Q_1\cap R\) to assume \(R\) is an integral domain, then localize at the complement of \(Q_1\cap R\).
3. (iii) Can you generalize the statements (i)-(ii) above to the polynomial ring in any finite set of indeterminates over \(R\)?

1. Suppose \(a\in R\) is a non-zerodivisor. Prove that \(R\) is an integral domain if and only if \(R_S\) is an integral domain, where \(S = \{1, a, a^2, \ldots \}\).

2. Let \(a \in R\) be a non-nilpotent element and set \(S := \{1, a, a^2, \ldots \}\).

1. (i) Prove that \(R_S\) is isomorphic to \(R[x]/\langle ax-1\rangle\). This is pretty tricky. Note that \(R_S = R[\frac{1}{a}]\), so there is a canonical map from the polynomial ring \(R[x]\) to \(R[\frac{1}{a}]\) such that \(ax-1\) is in the kernel. Then try to prove that if \(f(x)\) is in the kernel of the canonical map, some power of \(a\) times \(f(x)\) is a multiple of \(ax-1\) paying close attention to what that power is relative to the degree of \(f(x)\). It is helpful to think of \(f(\frac{1}{a})\) as \(\frac{g(a)}{a^n}\), for some polynomial \(g(x)\in R[x]\).
2. (ii) Use part (i) to show that if \(p \in \mathbb{Z}\) is a prime number and \(R\) denotes the set of rational numbers whose denominator is not divisible by \(p\), then \(\langle pX-1\rangle\) is a maximal ideal in \(R[X]\). Note that \(R\) is a local ring whose maximal ideal is generated by \(\frac{p}{1}\).

1. Let \(x_1, \ldots, x_n\) be indeterminates over \(R\). For \(f(x_2, \ldots, x_n) \in R[x_2,\ldots, x_n]\) and \(u\) a unit in \(R\), set \(x_1' := ux_1+f(x_2,\ldots, x_n)\). Prove that \(R[x_1, \ldots, x_n] = R[x_1',x_2, \ldots, x_n]\).

2. Let \(R\) be an integral domain, \(n\geq 2\), and \(x_1,y_1, \ldots, x_n,y_n\) indeterminates over \(R\). Prove that the ring

is an integral domain. Conclude that if \(K\) is a field and \(x,y,z,w\) are indeterminates over \(K\), then the ring \(K[x,y,z,w]/\langle xy-zw\rangle\) is an integral domain. (Hint: Use the first problem from today's assignment and the first problem from the previous assignment.)

3. Give a rigorous proof that if \(K\) is a field and \(x,y,z,w\) are indeterminates over \(K\), then the ring \(K[x,y,z,w]/\langle xy-zw\rangle\) is not a unique factorization domain.

1. Prove that a UFD is a GCD domain.

2. Prove that if \(R\) is a GCD domain then given non-zero, non-units \(a, b\in R\), LCM\((a,b)\) exists and satisfies \(\textrm{GCD}(a,b)\cdot \textrm{LCM}(a,b) = ab\).

3. Set \(R := \mathbb{Z}+x\mathbb{Q}[x]\), i.e., polynomials in \(\mathbb{Q}\) whose constant term is an integer. Show that \(R\) is a GCD, but not a UFD. Hint: Does \(R\) satisfy ACC on principal ideals?

1. Consider the sequence of polynomial rings \(\mathbb{Q}[x]\subseteq \mathbb{Q}[x^{\frac{1}{2}}] \subseteq \mathbb{Q}[x^{\frac{1}{4}}]\subseteq \cdots\) and set \(R := \bigcup_{n\geq 1}\mathbb{Q}[x^{\frac{1}{2^n}}]\). Prove that if \(f \in R\) and the constant term of \(f\) is 0, then \(f\) is not an irreducible element in the integral domain \(R\). Conclude that no element in \(R\) with zero constant term can be written as a product of irreducible elements.

2. Let \(R\) be the ring in problem 1 and \(S\subseteq R\) be the multiplicatively closed set of polynomials with non-zero constant term. Prove that the ring \(R_S\) has no irreducible elements.

Aside. In the literature, irreducible elements are sometimes called atoms, for obvious reasons. An integral domain in which each non-zero, non-unit be factored (not necessarily uniquely) as a product of a finite number of atoms (irreducible elements) is called an atomic domain. An integral domain without any atoms is then called an anti-matter integral domain. Thus, the ring \(R_S\) in problem 2 is an anti-matter integral domain. This latter designation has absolutely nothing to do with physics, but was probably coined for the amusement of mathematicians who study such rings.

1. Let \(R\) be a UFD, \(n\geq 3\), and \(x_1,y_1, \ldots, x_n,y_n\) indeterminates over \(R\). Prove that the ring

is a UFD. Hint: Use the technique outlined in the homework from January 29.

2. For the coordinate ring of the unit circle in \(\mathbb{R}^2\), \(R = \mathbb{R}[x,y]/\langle x^2+y^2-1\rangle\), it is known that \(R_{\mathfrak{m}}\) is a PID for every maximal ideal \(\mathfrak{m}\), and thus a UFD for every maximal ideal. Therefore, \(R\) is locally a UFD, but not a UFD. Take a step in the direction of proving the local statement by showing the following:

1. (a) For \((\alpha,\beta) \in \mathbb{R}^2\) such that \(\alpha^2 + \beta^2 = 1\) and \(\mathfrak{m} := (x-\alpha,y-\beta)R\), show that \(\mathfrak{m}\) is a maximal ideal of \(R\).
2. (b) Prove that \(\mathfrak{m} R_{\mathfrak{m}}\) is a principal ideal.

3. Assume \(R\) is a GCD domain. Prove the following statements:

1. (i) For \(a,b\in R\), if \(d = \gcd(a,b)\), then \(1 = \gcd (\frac{a}{d}, \frac{b}{d})\).
2. (ii) Suppose \(a,b,c\in R\), \(\gcd (a,b) = 1\) and \(a\mid bc\). Show that \(a\mid c\).

1. Let \(R\) be a commutative ring and \(X\) an infinite set of prime ideals. Suppose that \(J\subseteq R\) is an ideal maximal with respect to the property that \(J\) is contained in infinitely many elements of \(X\). Prove that \(J\) is a prime ideal. Conclude that if \(R\) is a Noetherian ring, then every ideal has only finitely many primes minimal over it.

2. Let \(M\) be a Noetherian \(R\)-module and \(\phi : M \rightarrow M\) a surjective \(R\)-module homomorphism. Prove that \(\phi\) is an isomorphism. Formulate and prove a version of this if \(M\) is Artinian.

1. A composition series of length n for \(M\) is a chain of submodules \[(0)= M_0 \subseteq M_1\subseteq \cdots \subseteq M_{n-1}\subseteq M_n = M\] such that each quotient \(M_i/M_{i-1}\) is a non-zero simple \(R\)-module, i.e., it has no proper submodules. In particular, this means that for each \(1\leq i\leq n\), we cannot insert an extra submodule in the chain above between \(M_{i-1}\) and \(M_i\). Prove that \(M\) has a composition series if and only if \(M\) is both Artinian and Noetherian.

2. Prove that if \(M\) has a composition series, then any two composition series have the same length. Comment: This is the module analogue of the Jordan-Hölder theorem from group theory. In this setting, the proof is easier than the one for groups. To prove this version, induct on the minimal length of a composition series.

3. If an \(R\)-module \(W\) has a composition series, we say that \(W\) has finite length. The length of \(W\), denoted \(\lambda(W)\), is the length of any composition series. Show that if \(0\to A\to B\to C\to 0\) is an exact sequence of \(R\)-modules, then \(B\) has finite length if and only if \(A\) and \(C\) have finite length. If these conditions hold, prove that \(\lambda(B) = \lambda(A)+\lambda(C)\).

1. Use the Krull principal ideal theorem and the prime avoidance lemma to prove the following statement. Let \(R\) be a Noetherian ring, and suppose there exist prime ideals \(P_0\subsetneq P_1\subsetneq P_2\). Then there exist infinitely many prime ideals \(P'\) satisfying \(P_0\subsetneq P'\subsetneq P_2\). Here is the version of prime avoidance to use: Let \(P_1, \ldots, P_n\) be prime ideals in \(R\). For \(I\subseteq R\) an ideal, assume \(I\subseteq P_1\cup \cdots\cup P_n\). Then \(I\subseteq P_i\), for some \(i\).

2. Let \(R\) be a commutative ring and \(f(x) = a_nx^n+\cdots + a_1x + a_0\) be a polynomial in \(R[x]\). Prove:

1. (i) \(f(x)\) is nilpotent in \(R[x]\) if and only if each \(a_i\) is nilpotent in \(R\).
2. (ii) \(f(x)\) is a unit in \(R[x]\) if and only if \(a_0\) is a unit and all other \(a_i\) are nilpotent.

Conclude that the nilradical of \(R[x]\) equals the Jacobson radical of \(R[x]\).

1. Let \(R\) be a Noetherian ring of Krull dimension \(d\). Prove that the polynomial ring \(R[x]\) has Krull dimension \(d+1\). Conclude that if \((R,\mathfrak{m})\) is a regular local ring and \(M := \langle \mathfrak{m},x\rangle\), then \(R[x]_{M}\) is a regular local ring. Hint: If \(M\subseteq R[x]\) is a maximal ideal, it is helpful to localize \(R\) at \(M\cap R\) and use systems of parameters.

2. Let \(R\) be a commutative ring and \(R[[x]]\) denote the formal power series ring over \(R\). Addition and multiplication of power series works just like for polynomials. In \(R[[x]]\), one often writes elements with the lower degree terms listed first, i.e., \(f(x) = a_0+a_1x+a_2x^2 +\cdots\). Show that \(f\in R[[x]]\) is a unit if and only if \(a_0\) is a unit in \(R\). Conclude that if \((R,\mathfrak{m})\) is quasi-local, then \(R[[x]]\) is also quasi-local.

3. If \((R,\mathfrak{m})\) is a \(d\)-dimensional local ring, show that \(R[[x]]\) has dimension \(d+1\). Conclude that \(R[[x]]\) is a regular local ring if \(R\) is a regular local ring.

1. Let \(R\) be a Noetherian ring. Show that \(R\) satisfies DCC on prime ideals. (Hint: Krull's height theorem.)

2. Let \((R, \mathfrak{m}, k)\) be a regular local ring of dimension three with \(\mathfrak{m} = \langle x, y, z\rangle\). Prove that the sequence

is exact, where \(\pi: R\to R/\mathfrak{m}\) is the canonical homomorphism.

1. Let \(F\) be a free \(R\)-module of finite rank over the local ring \((R, \mathfrak{m})\). Prove that a minimal generating set for \(F\) is a basis for \(F\). Is this true even if \(R\) is not local? What if \(R\) is not Noetherian?

2. Prove Schanuel's Lemma: Let \(M\) be an \(R\)-module and consider the two short exact sequences \[0\to K_1\to P_1 \overset{\alpha}\to M\to 0,\] and \[0\to K_2\to P_2 \overset{\beta}\to M\to 0,\] where \(P_1\) and \(P_2\) are projective \(R\)-modules. Then \(P_1\bigoplus K_2\) and \(P_2\bigoplus K_1\) are isomorphic. Hint: Let \(U\) be the submodule of \(P_1\bigoplus P_2\) consisting of the pairs \((a, b)\) for which \(\alpha(a) = \beta(b)\). Show that there exist short exact sequences \[0\to K_2 \to U \to P_1 \to 0 \quad \quad \text{and}\quad \quad 0\to K_1 \to U \to P_2 \to 0.\] Note that it follows from Schanuel's Lemma that if \(K_1\) is a projective \(R\)-module, then \(K_2\) is also a projective \(R\)-module. This enables one to give a well defined definition for the projective dimension of an \(R\)-module in case \(R\) is not a local ring. Note also that if \(K_1\) is a free \(R\)-module, it need not follow that \(K_2\) is also a free \(R\)-module.

3. Let \((R, \mathfrak{m}, k)\) be a local ring and \(M\) a finitely generated \(R\)-module. Prove that a minimal free resolution \(F\) of \(M\) is a direct summand of any other free resolution. In fact, any free resolution can be decomposed as \(F\bigoplus C\) where \(C\) is a direct sum of complexes of the form \(0\to R \overset{\mathrm{id}}\to R\to 0\). Conclude that a minimal free resolution of \(M\) is unique up to isomorphism.

1. Let \(0\to A\to B\to C\to 0\) be an exact sequence of \(R\)-modules. Prove that if \(A\) and \(C\) admit a FFR, then \(B\) admits a FFR.

2. Let \(M\) be an \(R\) module admitting a FFR \[0\to F_n\to F_{n-1}\to \cdots \to F_1\to F_0\to M\to 0.\] The integer \(\chi_R(M) := \sum_{i=0}^n (-1)^i\textrm{rank}(F_i)\) is called the Euler characteristic of \(M\).

1. (i) Prove that \(\chi_R(M)\) is well defined, i.e., is independent of the FFR over \(M\). Hint: Use Schanuel's Lemma.
2. (ii) Use problem 1 to show that if \(0\to A\overset{\alpha}\to B\overset{\beta}\to C\to 0\) is an exact sequence of \(R\) modules each admitting an FFR, then \(\chi(B) = \chi(A)+\chi(C)\).

1. Let \(M\) be an \(R\)-module admitting a FFR and let \(S\subseteq R\) a multiplicatively closed subset of \(R\). Prove \(\chi_R(M) = \chi_{R_S}(M_S)\). Conclude:

1. (i) If the annihilator of \(M\) contains a non-nilpotent element, then \(\chi(M) = 0\).
2. (ii) If \(R\) is an integral domain with quotient field \(K\) and \(S\) is the set of non-zero elements of \(R\) (so that \(R_S = K\)), then \(\chi_R(M) = \dim_K(V)\), where \(V := M_S\).

2. Let \(R\) be Noetherian and \(M\) an \(R\)-module admitting a FFR.

1. (i) Prove that \(\chi_R(M) \geq 0\). Conclude that if \(0\to R^m\to R^n\) is an injective homomorphism of free modules over \(R\), then \(m\leq n\).
2. (ii) Prove that if \(\chi_R(M) = 0\), then the annihilator of \(M\) contains a non-zerodivisor. Hint: It might be useful to contemplate the domain case first.

1. Let \(R\) be a Noetherian ring, \(M\) an \(R\)-module and \(S\subseteq R\) a multiplicatively closed set. Prove that \(\mathrm{Ass}_{R_S}(M_S)\) is the set of primes \(P_S\subseteq R_S\) such that \(P\in \mathrm{Ass}_R(M)\) is disjoint from \(S\).

2. Let \(R\) be a commutative ring and \(x_1, \ldots, x_n \in R\). The Koszul relations on \(x_1, \ldots, x_n\) are the vectors \(v_1, \ldots, v_N \in R^n\) listed (in order) below, where \(N = \binom{n}{2}\).

Let \(A\) denote the \(n\times N\) matrix whose columns are \(v_1, \ldots, v_N\) and write \(I\) for the ideal generated by \(x_1, \ldots, x_n\). First note that

is a complex, where \(\phi\) is the canonical map taking each \(e_i\) in \(R^n\) to \(x_i\), and then show the sequence is exact when \(x_1, \ldots, x_n\) form a regular sequence. Hint: Note that \(\phi\) is represented by the \(1\times n\) matrix \(\begin{pmatrix} x_1 & x_2 & \cdots & x_n\end{pmatrix}\).

Let \(R\subseteq S\) be an integral extension of commutative rings.

1. Show that \(R\) and \(S\) have the same Krull dimension.

2. Show that the extension of polynomial rings \(R[x]\subseteq S[x]\) is an integral extension.

3. Suppose \(R\subseteq S\subseteq T\) are commutative rings such that \(S\) is a finite \(R\)-module and \(T\) is a finite \(S\)-module. Prove that \(T\) is a finite \(R\)-module. Conclude that if \(S\) is integral over \(R\) and \(T\) is integral over \(S\), then \(T\) is integral over \(R\).

1. Prove that a one dimensional Noetherian domain is a Hilbert ring if and only if it has infinitely many maximal ideals.

2. Let \(R\) be a Hilbert ring in which every maximal ideal is generated by \(n\) or fewer elements. Prove that every maximal ideal in \(R[x]\) is generated by \(n+1\) or fewer elements. Conclude that if \(k\) is a field and \(R = k[x_1, \ldots, x_d]\) then every maximal ideal can be generated by \(d\) elements and no fewer than \(d\) elements.

3. Let \(R\) be a Hilbert ring with only finitely many maximal ideals. Then \(R\) is zero dimensional.

Let \(R = k[x_1, \ldots, x_n]\), for \(k\) a field. We use the same notation used in today's lecture (see the Daily Update).

1. Show that if \(k\) is algebraically closed and \(f_1, \ldots, f_t\in R\), then the equations \(f_1 = 0, \ldots, f_t = 0\) have a simultaneous solution if and only if there do not exist \(g_1, \ldots, g_t\in R\) such that \(g_1f_1+\cdots+g_tf_t = 1\).

2. An ideal \(I\subseteq R\) is a homogeneous ideal if it can be generated by homogeneous polynomials.

1. (i) Show that \(I\subseteq R\) is homogeneous if whenever \(f\in I\), all of its homogeneous constituents are in \(I\).
2. (ii) Suppose \(I\subseteq R\) is homogeneous. Show that \(V(I)\subseteq k^n\) is a cone, i.e., if \(\underline{\alpha} = (\alpha_1, \ldots, \alpha_n)\in V(I)\), then \(t\cdot\underline{\alpha}\in V(I)\), for all \(t\in k\).
3. (iii) Suppose \(I\subseteq R\) is an ideal such that \(V(I)\) is a cone. Prove that if \(k\) is infinite, then \(\mathcal{I}(V(I))\) is homogeneous. Show this is false if \(k\) is finite.

3. Here is a version of the Nullstellensatz for homogeneous ideals, or for projective varieties. Let \(\mathbb{P}^{n-1}\) denote the set of equivalence classes of non-zero elements \(\underline{\alpha}\in k^n\) under the equivalence relation \(\underline{\alpha}\sim\underline{\alpha}'\) if and only if \(\underline{\alpha}' = t\cdot\underline{\alpha}\), for some non-zero \(t\in k\). \(\mathbb{P}^{n-1}\) is called \((n-1)\)-dimensional projective space. If \(I\subseteq R\) is homogeneous, then we let \(V_{\mathbb{P}^{n-1}}(I)\) denote the set of equivalence classes \([\underline{\alpha}]\in \mathbb{P}^{n-1}\) such that \(f(\underline{\alpha}) = 0\), for all \(f\in I\). First verify that \(V_{\mathbb{P}^{n-1}}(I)\) is well defined, then prove that if \(k\) is algebraically closed, then \(V_{\mathbb{P}^{n-1}}(I) = \emptyset\) if and only if \(\sqrt{I} = \langle x_1, \ldots, x_n\rangle\), equivalently, \(I\) defines a non-empty projective variety in \(\mathbb{P}^{n-1}\) if and only if \(\sqrt{I}\not= \langle x_1, \ldots, x_n\rangle\). Hint: The Weak Nullstellensatz implies that as long as \(I\subseteq R\) is a proper ideal, then \(V(I)\subseteq k^n\) is non-empty.

Another ring obtained by localizing the polynomial ring \(R[x]\) at a multiplicatively closed set comes from the following. For \(f(x)\in R[x]\) define \(c(f(x))\), the content of \(f(x)\), to be the ideal of \(R\) generated by the coefficients of \(f(x)\). Let \(U\subseteq R[x]\) denote the set of polynomials \(f(x)\in R[x]\) with unit content, i.e., \(c(f(x)) = R\).

1. Show that \(U\) is a multiplicatively closed subset of \(R[x]\).

2. Set \(R(X) := R[x]_U\). Show that there is a 1-1 correspondence between the maximal ideals of \(R\) and the maximal ideals of \(R(X)\), given by \(M\subseteq R(X)\) is a maximal ideal if and only if \(M = NR(X)\), for some maximal ideal \(N\subseteq R\).

1. Let \(R\subseteq S\subseteq T\) be an extension of rings such that \(T\) is a finite \(S\)-module and \(S\) is a finite \(R\)-module. Prove that \(T\) is a finite \(R\)-module.

2. Let \(\mathbb{Z}_p\) denote the finite field with \(p\) elements, for \(p\in \mathbb{Z}\) prime, and take \(x, y\) indeterminates over \(\mathbb{Z}_p\). Set \(K = \mathbb{Z}_p(x^p, y^p)\) and \(L := \mathbb{Z}_p(x,y)\). Show that \(K\subseteq L\) is a finite field extension, but there is no primitive element for \(L\) over \(K\). Hint: In the field \(L\), we have \((a+b)^p = a^p+b^p\).

Let \(R\) be a Noetherian domain with quotient field \(K\) and let \(\mathcal{P}(R)\) denote the set of associated primes of non-zero principal ideals. The following problems lead to a proof of the so-called \(S_2\) plus \(R_1\) conditions for \(R\) to be integrally closed. We will use the following notation. If \(I\subseteq R\) is an ideal, then \(I^{-1}\) denotes the set of elements \(\alpha\in K\) such that \(I\alpha\subseteq R\). Note that \(II^{-1}\subseteq R\) is an ideal, possibly \(R\) itself.

1. Let \(\{V_i\}_{i\in I}\) be a family of rings between \(R\) and \(K\) such that each \(V_i\) is integrally closed and \(R = \bigcap_{i\in I} V_i\). Conclude that \(R\) is integrally closed.

2. Suppose that \((R, P)\) is a local domain and \(PP^{-1} = R\). Show that \(P\) is principal, and thus \(R\) is a one-dimensional regular local ring.

3. Give a direct proof that if \((R, P)\) is a one-dimensional regular local ring, then \(R\) is integrally closed.

4. Conclude that if \(R_P\) is a one-dimensional regular local ring for each \(P\in \mathcal{P}(R)\), then \(R\) is integrally closed.

5. Suppose \((R, P)\) is local and \(P = (aR:b) = (R:\frac{b}{a})\). Conclude that \(\frac{b}{a}\in P^{-1}\), and that if \(P\frac{b}{a}\subseteq P\), then \(\frac{b}{a}\) is integral over \(R\). Hint: Use the determinant trick.

6. Use the previous problems to show that if \(R\) is an integrally closed domain, and \(P\in \mathcal{P}(R)\), then \(R_P\) is a one-dimensional regular local ring.

Thus \(R\) is integrally closed if and only if every \(P\in \mathcal{P}(R)\) has the property that \(\mathrm{height}(P) = 1\) and \(R_P\) is a regular local ring. These are the Serre conditions \(S_2\) and \(R_1\).

1. Here is an example related to the theorem from the lecture of Thursday, April 9. Let \(B = k[x,y]\) be the polynomial ring in \(x, y\) over the field \(k\) of characteristic zero. Consider \(B[\omega]\), where \(\omega\) is a root of \(f(W) = W^3-xy\in B[W]\). Show directly that \(B[\omega]_P\) is a regular local ring for \(P\subseteq B[\omega]\) and \(f'(\omega)\not\in P\). Now find a prime \(Q\subseteq B[\omega]\) such that \(f'(\omega)\in Q\) and \(B[\omega]_Q\) is a regular local ring. Thus, the set of primes \(P\) such that \(B[\omega]_P\) is a regular local ring is larger than the set of primes not containing \(f'(\omega)\). Can you find an ideal \(I\subseteq B[\omega]\) such that \(B[\omega]_P\) is a regular local ring if and only if \(I\not\subseteq P\)? Does your analysis work over any field \(k\) whose characteristic is not 3?

2. Let \(R\) be a Noetherian ring and \(I, J\subseteq R\) ideals. Let \((I:J^{\infty})\) denote the stable value of the ascending sequence of ideals \((I:J)\subseteq (I:J^2)\subseteq \cdots\). Show that \((I:J^{\infty})\) equals the intersection of the primary components of \(I\) whose radicals do not contain \(J\).

1. Let \(a_1, \ldots, a_d\) be a regular sequence in \(R\) and set \(I := \langle a_1, \ldots, a_d\rangle\subseteq R\). Suppose \(f(x_1, \ldots, x_d)\) is a homogeneous polynomial of degree \(n\) such that \(f(a_1, \ldots, a_d)\in I^{n+1}\). Show that \(f(x_1, \ldots, x_d)\) has coefficients in \(I\). Hint: You'll probably have to induct on both \(n\) and \(d\).

2. Maintaining the notation from the previous problem, show that \(I^n/I^{n+1}\) is a free \(R/I\)-module for \(n\geq 1\). Conclude that \(\mathrm{Ass}(R/I^n) = \mathrm{Ass}(R/I)\) for all \(n\geq 1\). In particular, conclude that if \(P\) is a prime ideal generated by a regular sequence, then \(P^{(n)} = P^n\), for all \(n\geq 1\).

1. Let \(R\) be a Noetherian ring and \(I\subseteq R\) an ideal. Prove that the grade of \(I\) is less than or equal to the height of \(I\).

2. Let \(R = k[x,y,z]\) be the polynomial ring in three variables over the field \(k\). Show that \(x,\ y(1-x),\ z(1-x)\) is a regular sequence in \(R\), but \(y(1-x),\ z(1-x),\ x\) is not a regular sequence in \(R\).

1. Let \(R\) be a Noetherian ring and \(I\subseteq R\) an ideal whose grade equals \(t\). Let \(x\in I\) be a non-zerodivisor. Show that \(x\) can be extended to a regular sequence in \(I\) of length \(t\).

2. Let \(R\) be a Noetherian ring and \(I\subseteq R\) an ideal. Prove that the grade of \(I\) is less than or equal to the projective dimension of \(R/I\). When equality holds, \(I\) is said to be a perfect ideal. Can you extend this to the case when \(R/I\) is replaced by a finitely generated \(R\)-module \(M\) and \(I\) is replaced by the annihilator of \(M\)?

1. Let \(\phi: R^m\to R^n\) be an \(R\)-module homomorphism between finitely generated free \(R\)-modules. Let \(A\) and \(B\) be \(n\times m\) matrices representing \(\phi\). Take \(r\leq \mathrm{min}\{n,m\}\). Prove that \(I_r(A) = I_r(B)\), where \(I_r(C)\) for the \(n\times m\) matrix \(C\) is the ideal of \(R\) generated by the \(r\times r\) minors of \(C\). Thus, \(I_r(\phi)\) can be defined as \(I_r(A)\), for any matrix \(A\) representing \(\phi\).

2. Let \(\phi: R^m\to R^n\) be an \(R\)-module homomorphism between finitely generated free \(R\)-modules. Show that: (i) If \(\phi\) is surjective, \(m\geq n\) and (ii) If \(\phi\) is injective, then \(m\leq n\).