Tag Archives: maximal ideal

Factor a given ideal in an algebraic integer ring

Let K = \mathbb{Q}(\sqrt{-6}) and let \mathcal{O} = \mathbb{Z}[\sqrt{-6}] be the ring of integers in K. Factor the ideals (2) and (5) in \mathcal{O}.

We claim that (2) = (2,\sqrt{-6})^2. Indeed, 2 = -\sqrt{-6}^2 - 2^2, so that (2) \subseteq (2,\sqrt{-6})^2. The reverse inclusion is clear.

Now we claim that (2,\sqrt{-6}) is maximal. By Corollary 9.11, N((2)) = 4, and by Theorem 9.14, we have N((2,\sqrt{-6}))^2 = 4, so that N((2,\sqrt{-6})) = 2. By Corollary 9.15, (2,\sqrt{-6}) is a prime ideal.

Thus (2) = (2,\sqrt{-6})^2 is the prime factorization of (2) in \mathcal{O}.

Now we claim that (5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}). Indeed, we have (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) = (25,5+10\sqrt{-6}, 5-10\sqrt{-6}, 25) = (25, 5+10\sqrt{-6}, 10) = (5).

Next we claim that Q_1 = (5,1+2\sqrt{-6}) and Q_2 = (5,1-2\sqrt{-6}) are proper. Suppose to the contrary that 1 \in Q_1; then we have 1 = 5(a+b\sqrt{-6}) + (1+2\sqrt{-6})(h+k\sqrt{-6}) for some a,b,h,k \in \mathbb{Z}. Comparing coefficients, 5a+h-12k = 1 and 5b+k+2h = 0, which yields a contradiction mod 5. So Q_1 is proper. Likewise, Q_2 is proper. In particular, neither Q_1 nor Q_2 have norm 1 as ideals. Now 25 = N((5)) = N(Q_1)N(Q_2), and neither factor is 1, so that N(Q_1) = N(Q_2) = 5. By Corollary 9.15, Q_1 and Q_2 are prime ideals.

Thus (5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) is the prime factorization of (5) in \mathcal{O}.

Facts about the inverse of a maximal ideal in an algebraic integer ring

Let K be an algebraic number field with ring of integers \mathcal{O}. Fix an ideal A \subseteq \mathcal{O}. Recall that, for an ideal A \subseteq \mathcal{O}, A^{-1} = \{r \in K \ |\ rA \subseteq \mathcal{O}\}. Prove the following: (1) A^{-1} is an \mathcal{O}-submodule of K, (2) If B \subseteq A is an ideal, then A^{-1}B \subseteq \mathcal{O} is an ideal, and (3) If C \subseteq B \subseteq A are ideals, then A^{-1}C \subseteq A^{-1}B.

Let x,y \in A^{-1} and let r \in \mathcal{O}. Now for all k \in A, we have (x+ry)k = xk + ryk \in \mathcal{O}, since xk \in \mathcal{O} and y(rk) \in \mathcal{O} (because A is an ideal). Since 0 \in A^{-1}, by the submodule criterion, A^{-1} is an \mathcal{O}-submodule of K.

Suppose C \subseteq B \subseteq A. Note that if x \in A^{-1}C, then x = \sum r_ic_i where r_i \in A^{-1} and c_i \in C. Certainly then x \in A^{-1}B, as desired in (3).

Suppose B \subseteq A; by the previous argument, we have A^{-1}B \subseteq A^{-1}A \subseteq \mathcal{O}. Suppose x,y \in A^{-1}B and r \in \mathcal{O}, with x = \sum r_ib_i and y = \sum s_ic_i. Certainly then since B is an ideal we have x+ry = \sum r_ib_i + \sum s_irc_i \in A^{-1}B. Since 0 \in A^{-1}B, by the submodule criterion, A^{-1}B is an ideal in \mathcal{O}.

Over ZZ, if (a,b) is maximal then gcd(a,b) is prime

Let a,b \in \mathbb{Z} such that (a,b) is maximal as an ideal in \mathbb{Z}. What can be said about a and b?

Since \mathbb{Z} is a principal ideal domain, (a,b) = (d) for some d. If (d) is maximal, then it is prime, so that d is a prime element. Moreover, d is a greatest common divisor of a and b. So \mathsf{gcd}(a,b) is prime.

Over an integral domain, every maximal ideal is irreducible

Let R be an integral domain. Prove that every maximal ideal in R is irreducible.

We begin with a lemma. This lemma is actually true (with a minor modification) over a much larger class of rings. However, I am not entirely comfortable with the proof of the general version. I am, however, comfortable with this more specific version. I referred to a discussion at MathForum.org for a proof of the lemma.

Lemma: Let R be an integral domain and let I \subseteq R be an ideal which is finitely generated as a \mathbb{Z}-module. If I^2 = I, then either I = 0 or I = R. Proof: Suppose I \neq 0. Let A = \{\alpha_i\}_{i=1}^n be a generating set for A over \mathbb{Z}; in particular, some \alpha_i is nonzero. Since I^2 = I, there exist u_{i,j} \in A such that \alpha_i = \sum_j u_{i,j} \alpha_j for each i. In particular, 0 = \sum_j (u_{i,j} - \delta_{i,j})\alpha_j for each i, where \delta_{i,j} is the Kronecker delta. That is, [\alpha_1\ \cdots\ \alpha_n]^\mathsf{T} is a nontrivial solution to the matrix equation Mx = 0, where M = [u_{i,j} - \delta_{i,j}]. Thinking of R as embedded in its field F of fractions, we have \mathsf{det}(M) = 0. On the other hand, by the Leibniz expansion of \mathsf{det}(M), we have \mathsf{det}(M) = \beta - 1, where \beta \in I (using the fact that I is an ideal). In particular, -1 \in I, so that I = R. \square

Now let M \subseteq R be nonzero and maximal. If M = AB, then M \subseteq A and M \subseteq B, so that A,B \in \{M, R\}. If A = B = M, then M^2 = M. By the lemma, either M = 0 (a contradiction) or M = R (also a contradiction). So either A or B is R. Since R is a unit in the semigroup of ideals of R under ideal products, M is irreducible.

Prove that a given ideal is maximal

Prove that A = (1+\sqrt{-5}, 1-\sqrt{-5}) is maximal in \mathcal{O}_K, where K = \mathbb{Q}(\sqrt{-5}).

We will prove this in slightly more generality, beginning with a lemma.

Lemma: Let D be a squarefree integer with D \equiv 3 mod 4. Then \{1+\sqrt{D}, 1-\sqrt{D}\} is a basis for A = (1+\sqrt{D}, 1-\sqrt{D}) over \mathbb{Z}. Proof: Let \zeta \in A. Then \zeta = (a+b\sqrt{D})(1+\sqrt{D}) + (h+k\sqrt{D})(1-\sqrt{D}) for some a,b,h,k \in \mathbb{Z}. Evidently, \zeta = u(1+\sqrt{D}) + v(1-\sqrt{D}), where u = a+b\frac{D+1}{2} - k\frac{D-1}{2} and v = h + b\frac{D-1}{2} - k\frac{D+1}{2}. So A = (1+\sqrt{D},1-\sqrt{D})_\mathbb{Z}. Suppose now that h(1+\sqrt{D}) + k(1-\sqrt{D}) = 0; then h+k = 0 and h-k = 0, so that h = k = 0, as desired. \square

Let D be a squarefree integer with D \equiv 3 mod 4. Now the ring of integers in K = \mathbb{Q}(\sqrt{D}) is \mathcal{O} = \{a+b\sqrt{D} \ |\ a,b \in \mathbb{Z}\}. We claim that A = (1+\sqrt{D}, 1-\sqrt{D}) is maximal in \mathcal{O}.

To see this, note that 2 = (1+\sqrt{D}) + (1-\sqrt{D}) \in A. Now let a+b\sqrt{D} \in \mathcal{O}, and note that a+b\sqrt{D} = a-b+b+b\sqrt{D} = a-b + b(1+\sqrt{D}) \equiv a-b mod A. In particular, if a-b \equiv 0 mod 2, then a+b\sqrt{D} \equiv 0 mod A, and if a-b \equiv 1 mod 2, then a+b\sqrt{D} \equiv 1 mod A. We claim that these cosets are distinct. Indeed, if 1 \equiv 0 mod A, then 1 \in A. In this case, by the lemma we have h,k \in \mathbb{Z} such that h(1+\sqrt{D}) + k(1-\sqrt{D}) = (h+k) + (h-k)\sqrt{D} = 1. But then h+k = 1 and h-k = 0, so that 2h = 1, a contradiction. So \mathcal{O}/A = \{0+A, 1+A\}. Now \mathcal{O}/A is a unital ring with 1 \neq 0 having only two elements, so that \mathcal{O}/A \cong \mathbb{Z}/(2); that is, \mathcal{O}/A is a field, and thus A is a maximal ideal.

The original problem follows since -5 \equiv 3 mod 4.

Irreducible modules over a commutative ring are precisely its maximal ideal quotients

Let R be a commutative ring with 1 and let M be a unital left R module. Prove that M is irreducible if and only if M \cong_R R/I for some maximal ideal I \subseteq R. (Where \cong_R means “isomorphic as an R-module.”)

Suppose M is irreducible, fix m \in M nonzero, and define \varphi_m : R \rightarrow M by \varphi_m(r) = r \cdot m. Note that for all x,y \in R and r \in R, \varphi_m(x + r \cdot y) = (x+r \cdot y) \cdot m = x \cdot m + r \cdot (y \cdot m) = \varphi_m(x) + r \cdot \varphi_m(y). So \varphi_m is an R-module homomorphism. Now recall that since m \neq 0 and M is irreducible, then by this previous exercise, M = Rm. In particular, if b \in M, then there exists a \in R such that b = a \cdot m = \varphi_m(a). Thus \varphi_m is surjective.

FInally, we claim that \mathsf{ker}\ \varphi_m is a maximal ideal. To that end, let x + \mathsf{ker}\ \varphi_m be nonzero. In particular, x \cdot m \neq 0. Since M is irreducible, we have M = R(x \cdot m). In particular, m = y \cdot (x \cdot m) = (yx) \cdot m for some y \in R. Then 1 \cdot m - (yx) \cdot m = 0, so that 1 - yx \in \mathsf{ker}\ \varphi_m. That is, (y + \mathsf{ker}\ \varphi_m)(x + \mathsf{ker}\ \varphi_m) = 1 + \mathsf{ker}\ \varphi_m. So every nonzero element of R/\mathsf{ker}\ \varphi_m has a left inverse. Since R/\mathsf{ker}\ \varphi_m is commutative, R/\mathsf{ker}\ \varphi_m is a field, so that \mathsf{ker}\ \varphi_m is a maximal ideal of R.

By the First Isomorphism Theorem, we have M \cong_R R/\mathsf{ker}\ \varphi_m, where \mathsf{ker}\ \varphi_m is a maximal ideal.

Conversely, suppose I \subseteq R is a maximal ideal. Now R/I is a field. Let x + I be nonzero. Then there exists y \in R such that (y+I)(x+I) = (1+I). Now let r+I \in R/I. Note that r+I = (ry+I)(x+I), so that R/I = R(x+I). That is, R/I is generated (as an R-module) by any nonzero element. By this previous exercise, R/I is an irreducible R-module.

Two prime ideals in ZZ[x,y]

Prove that (x,y) and (2,x,y) are prime ideals in \mathbb{Z}[x,y] but that only (2,x,y) is maximal.

In this previous exercise, we saw that \mathbb{Z}[x,y]/(x,y) \cong \mathbb{Z}. Since this quotient is a domain but not a field, (x,y) is a prime ideal but not maximal.

On the other hand, we have \mathbb{Z}[x,y]/(2,x,y) \cong \mathbb{Z}[x][y]/((2,x) + (y)) \cong \mathbb{Z}[x]/((2)+(x)) \cong \mathbb{Z}/(2), which is a field. Thus (2,x,y) is maximal and hence prime.

Characterization of the ideals in a discrete valuation ring

Let K be a field, let v be a discrete valuation on K, and let R be the valuation ring of v. For each integer k \geq 0, define A_k = \{ r \in R \ |\ v(r) \geq k \} \cup \{0\}.

  1. Prove that A_k is a principal ideal and that A_{k+1} \subseteq A_k for all k \in \mathbb{N}.
  2. Prove that if I is a nonzero ideal of R, then I = A_k for some k \geq 0. Deduce that R is a local ring with maximal ideal A_1.

  1. It is clear that A_{k+1} \subseteq A_k for all k.

    Now we show that A_k is an ideal. We have 0 \in A_k, so A_k is nonempty. Now let a,b \in A_k; if one or both of a and b is 0, then a+b \in A_k. If a+b = 0, then a+b \in A_k. If a+b \neq 0, we have v(a+b) \geq \min(v(a),v(b)) \geq k, so that a+b \in A_k. Moreover, v(-a) = v(a), so that -a \in A_k. Finally, if r \in R, we have v(r) \geq 0. Then v(ra) = v(r) + v(a) \geq k, so that ra \in A_k. Since K is commutative, we have that A_k is an ideal of R.

    We now show that A_k is principal. Choose \alpha \in A_k such that v(\alpha) = k; such an element exists because v is surjective. Note that \alpha^{-1} exists in K, and that v(\alpha^{-1}) = -v(\alpha). Let r \in A_k; then v(r) \geq k. Note that v(\alpha^{-1}r) = v(\alpha^{-1}) + v(r) = v(r) - v(\alpha) \geq 0, so that \alpha^{-1}r \in R. Moreover, r = \alpha\alpha^{-1}r. Thus r \in (\alpha), and we have A_k \subseteq (\alpha), and thus A_k = (\alpha). In particular, A_k is generated by any element of valuation k.

  2. Let I \subseteq A_k be a nonzero ideal of R. Let k be minimal among v(r) for r \in I, and let a \in I such that v(a) = k. In particular, we have I \subseteq A_k. Moreover, since A_k = (a), we have A_k \subseteq I. Thus I = A_k.

    From part (1), every proper ideal of R is contained in A_1; thus A_1 is maximal, and moreover, is the unique maximal ideal of R.

If the set of all nonunits in a commutative ring is an ideal, then it forms the unique maximal ideal

A commutative ring R is called local if it has a unique maximal ideal. Prove that if R is a local ring with a maximal ideal M then every element of R \setminus M is a unit. Prove conversely that if R is a commutative ring with 1 \neq 0 in which the set of nonunits forms an ideal M, then R is a local ring with unique maximal ideal M.

Let u \in R \setminus M. Consider the ideal (u); if proper, this ideal must be contained in some maximal ideal, and M is the only maximal ideal. Thus (u) \subseteq M, a contradiction since u \notin M. Thus (u) = R, and for some v \in R we have uv = vu = 1. Thus u is a unit.

Suppose now that the set M of nonunits in R form an ideal. First we show that M is maximal. If M \subsetneq I for some ideal I, then I contains a unit, so that I = R. Now we show that M is the unique maximal ideal. Suppose that we have an ideal N \subseteq R such that N \not\subseteq M. Then N contains some element x not in M, which is a unit. Thus N = R. In particular, every proper ideal of R is contained in M. Thus M is the unique proper ideal of R, and R is local.

Use Zorn’s Lemma to construct an ideal which maximally does not contain a given finitely generated ideal

Let R be a ring with 1 \neq 0, and let A = (a_1,\ldots,a_n) be a nonzero finitely generated (proper) ideal of R. Prove that there is an ideal in R which is maximal with respect to the property “does not contain A“.

Let \mathcal{C} denote the set of all ideals in R which do not contain I; this set is partially ordered by inclusion. Moreover \mathcal{C} is nonempty since 0 does not contain A. Let \{C_i\}_\mathbb{N} be a chain in \mathcal{C}. Now \bigcup C_i is an ideal in R by this previous exercise. Suppose A \subseteq \bigcup C_i. Then for each generator a_k of A, we have a_k \in C_{i_k} for some i_k. Since \{C_i\}_\mathbb{N} is totally ordered and there are only finitely many a_k, there exists an inclusion-maximal element C_t such that a_k \in C_t for each k; thus A \subseteq C_t, a contradiction. Thus A \nsubseteq \bigcup C_i, and we have \bigcup C_i \in \mathcal{C}. So \bigcap C_i is an upper bound of the chain \{C_i\}_\mathbb{N} in \mathcal{C}. Thus every chain in \mathcal{C} has an upper bound, and by Zorn’s lemma there exists a maximal ideal B with respect to the property “does not contain A“.