Throughout \(R\) will denote a commutative ring.
1. Let \(P\subseteq R\) be an ideal. Show that the following are equivalent:
- (i) \(P\) is a prime ideal.
- (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:
- (i) \(I\) is a prime ideal if and only if \(R/I\) is an integral domain.
- (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 \(I\).
- (i) Prove that \(I[x]\) is an ideal of \(R[x]\) and equals the ideal in \(R[x]\) generated by \(I\).
- (ii) Prove that \(R[x]/I[x]\) is isomorphic to \((R/I)[x]\).
- (iii) Prove that if \(P\subseteq R\) is a prime ideal, then \(P[x]\) is a prime ideal in \(R[x]\).
- (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 \(R\).
- (i) Prove that \(R_S[x]\) is isomorphic to \(R[x]_S\).
- (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 \(Q_1\cap R\).
- (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 \(S := \{1, a, a^2, \ldots \}\).
- (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 \(g(x)\in R[x]\).
- (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:
- (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 \(R\).
- (b) Prove that \(\mathfrak{m} R_{\mathfrak{m}}\) is a principal ideal.
3. Assume \(R\) is a GCD domain. Prove the following statements:
- (i) For \(a,b\in R\), if \(d = \gcd(a,b)\), then \(1 = \gcd (\frac{a}{d}, \frac{b}{d})\).
- (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:
- (i) \(f(x)\) is nilpotent in \(R[x]\) if and only if each \(a_i\) is nilpotent in \(R\).
- (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 \(M\).
- (i) Prove that \(\chi_R(M)\) is well defined, i.e., is independent of the FFR over \(M\). Hint: Use Schanuel's Lemma.
- (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:
- (i) If the annihilator of \(M\) contains a non-nilpotent element, then \(\chi(M) = 0\).
- (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.
- (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 \(m\leq n\).
- (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 \(\begin{pmatrix} x_1 & x_2 & \cdots & x_n\end{pmatrix}\).