Monthly Archives: September 2010

A sufficient condition for the ring property that every prime ideal is maximal

Let R be a commutative ring with 1 \neq 0 and suppose that for all a \in R, there exists an integer n > 1 such that a^n = a. Prove that every prime ideal of R is maximal.


Let P \subseteq R be a prime ideal. Now R/P is an integral domain. Let a + P \in R/P be nonzero; there exists n \geq 2 such that a^n = a. In particular, a^n + P = a + P, so that a(a^{n-1} - 1) + P = 0. Since a \notin P, we have a^{n-1} + P = 1+P, so that (a+P)(a^{n-2} + P) = 1. Thus R/P is a division ring, and since R is commutative, R/P is a field. Thus P \subseteq R is maximal.

In a Boolean ring, all finitely generated ideals are principal

Prove that in a Boolean ring, every finitely generated ideal is principal.


We begin with a lemma.

Lemma: Let R be a ring. Suppose that for all A \subseteq R with |A| = 2, (A) is principal. Then for all finite nonempty A \subseteq R, (A) is principal. Proof: We proceed by induction on |A|. For the base case, If |A| is 1 or 2, then (A) is principal. For the inductive step, suppose that for some n \geq 2, for all A \subseteq R with |A| \leq n, (A) is principal. Let A \subseteq R have cardinality n+1 and write A = \{x\} \cup A^\prime, where |A^\prime| = n. Now (A) = (x) + (A^\prime), and by the induction hypothesis, (A^\prime) = (y) for some y \in R. Now (A) = (x,y) = (z) for some z \in R by hypothesis. \square

Now let R be a Boolean ring, and let a,b \in R. We claim that (a,b) = (a+ab+b). It is clear that a+ab+b \in (a,b) since R has a 1 and is commutative. Moreover, note that (a+ab+b)(ab+a+1) = b and (a+ab+b)(ab+b+1) = a, so that (a,b) \subseteq (a+ab+b). Thus every subset of order 2 generates a principal ideal in R, and by the lemma, every finitely generated ideal in R is principal.

Every prime ideal in a Boolean ring is maximal

Prove that in a Boolean ring, every prime ideal is maximal.


Let R be a Boolean ring, and let P \subseteq R be a prime ideal. Now R/P is a Boolean ring and an integral domain, so that by a previous theorem, R/P \cong \mathbb{Z}/(2) is a field. Thus P \subseteq R is maximal.

The ZZ/(p) Quaternions are not an integral domain

Let p be a prime and let the \mathbb{F}_p Quaternions be the the set of all formal sums a + bi + cj + dk where a,b,c,d \in \mathbb{Z}/(p), with addition defined componentwise and with multiplication defined using the same relations as for the real Quaternions.

  1. Prove that the \mathbb{F}_p Quaternions are a homomorphic image of the \mathbb{Z} quaterions.
  2. Prove that the \mathbb{F}_p Quaternions contain zero divisors, and thus cannot be an integral domain.

We begin with a definition and some lemmas.

Let R be a ring. The R-Quaternions are the quotient \mathbb{H}(R) = R[Q_8]/(1 \cdot (-1) + 1 \cdot 1). Note that this coincides with the book’s definition of the \mathbb{R}, \mathbb{Z}, and \mathbb{Z}/(p) Quaternions when R is \mathbb{R}, \mathbb{Z}, and \mathbb{Z}/(p), respectively.

Lemma: Let \varphi : R \rightarrow S be a ring homomorphism, and let G be a group. Then the map \psi : R[G] \rightarrow S[G] given by \psi(\sum r_ig_i) = \sum \varphi(r_i)g_i is a ring homomorphism. Moreover, if \varphi is surjective, then \psi is surjective. Proof: This follows because \varphi is a ring homomorphism. \square

  1. This follows from the lemma by considering the natural projection \pi : \mathbb{Z} \rightarrow \mathbb{Z}/(p).
  2. [Embarassing confession: I had to consult this note by Aristidou and Demetre for this part. It completely escaped me that fields are commutative while R-quaternions in general are not.] Suppose the \mathbb{F}_p Quaternions have no zero divisors; by a previous theorem, they form a field. In particular, k = ij = ji = -k. If p \neq 2, this is a contradiction. If p = 2, then since (1+i)^2 = 0, \mathbb{H}(\mathbb{Z}/(2)) is not a field.

A finite unital ring with no zero divisors is a field

Prove that a finite ring with 1 \neq 0 which has no zero divisors is a field.


[We will assume Wedderburn’s Theorem.]

Let R be a finite ring (not necessarily commutative) with 1 \neq 0 having no zero divisors. Let a \in R be nonzero, and define \varphi_a : R \rightarrow R by \varphi_a(r) = ar. Note that if \varphi_a(r) = \varphi_a(s), we have a(r-s) = 0, and since a \neq 0, r = s. Thus \varphi_a is injective. Since R is finite, \varphi_a is surjective. Thus for some b \in R, we have \varphi_a(b) = ab = 1. Similarly, with \psi_a(r) = ra, there exists c \in R such that ca = 1. By a previous theorem, a is a unit in R. Thus R is a division ring.

By Wedderburn’s Theorem, R is commutative; thus R is a field.

A nonzero finite commutative ring with no zero divisors is a field

Prove that any nonzero finite commutative ring with no zero divisors is a field. (Do not assume that the ring has a 1.)


Let R be a finite commutative ring with no zero divisors.

Now let I \subseteq R be a nonzero ideal, with a \in I nonzero. Define \varphi_a : R \rightarrow R by \varphi_a(r) = ar. If \varphi_a(r) = \varphi_a(s), then ar = as, so that a(r-s) = 0. Since R has no zero divisors and a \neq 0, r-s=0, so that r=s. Thus \varphi_a is injective. Since R is finite, \varphi_a is also surjective, so that every element r \in R has the form as for some s \in R. In particular, R \subseteq I. Thus R has only the trivial ideals.

Now suppose that a \in \mathfrak{N}(R) is nonzero, and let m be minimal such that a^m = 0. Now m \geq 2. But then aa^{m-1} = 0, and since a \neq 0, a^{m-1} = 0, a contradiction. Thus the nilradical of R is trivial.

Now let a \in R be nonzero; then (a) = R. In particular, note that since R is finite, there exists m \geq 2 such that a^m = a^n for some n < m. Let m be minimal with this property. Now a^m-a^n = 0. Suppose n \geq 2. Then a^{n-1}(a^{m-n+1} - a) = 0. Since a is not nilpotent, we have a^{m-n+1} = a, a contradiction. Thus n=1 and we have a^m = a.

Finally, let x = ra \in R. Now a^{m-1}x = a^{m-1}ra ra^m = ra = x. Then a^{m-1} satisfies the defining property of an identity element, and thus R is a finite integral domain. We proved in a lemma to a previous exercise that every finite integral domain is a field.

In a commutative ring, all prime ideals are maximal

Let R be a finite commutative ring with 1 \neq 0. Prove that every prime ideal of R is maximal.


We begin with a lemma.

Lemma: If R is a finite integral domain, then R is a field. Proof: Let u \in R be nonzero, and define \varphi_u : R \rightarrow R by \varphi_u(r) = ur. We claim that \varphi_u is injective. To see this, suppose \varphi_u(r) = \varphi_u(s); then ur = us, so that u(r-s) = 0. Since u \neq 0, r-s=0, so that r=s. Hence \varphi_u is injective. Since R is finite, \varphi_u is also surjective. In particular, there exists v \in R such that \varphi_u(v) = uv = 1, so that u is a unit. Thus every nonzero element of R is a unit, and R is a field. \square

Now if P \subseteq R is prime, then R/P is a finite integral domain. Then R/P is a field, and thus P \subseteq R is maximal.

In a ring of formal power series, the principal ideal generated by the indeterminate is maximal precisely when the coefficient ring is a field

Let R be an integral domain, and let R[[x]] be the ring of formal power series in x with coefficients in R. Prove that (x) is principal in R[[x]]. Prove that (x) is maximal in R[[x]] if and only if R is a field.


We first prove a lemma.

Lemma: The map \varphi : R[[x]] \rightarrow R given by \sum r_ix^i \mapsto r_0 is a surjective ring homomorphism and \mathsf{ker}\ \varphi = (x). Proof: \varphi is clearly a surjective homomorphism. Now suppose \sum r_ix^i \in \mathsf{ker}\ \varphi. Then r_0 = 0, and \sum r_ix^i = x \sum r_{i+1}x^i \in (x). If \alpha = x\sum a_ix^i, then \varphi(\alpha) = 0, so that \alpha \in \mathsf{ker}\ \varphi. \square

By the First Isomorphism Theorem for rings, R[[x]]/(x) \cong R. The problem at hand then follows from Propositions 12 and 13 in the text.

Prove that a given quotient ring is not an integral domain

Consider f(x) = x^3 - 2x+1 in the polynomial ring E = \mathbb{Z}[x] and use bars to denote the natural projection E \rightarrow E/(f(x)). Let p(x) = 2x^7 - 7x^5 + 4x^3 - 9x + 1 and q(x) = (x-1)^4.

  1. Express each of \overline{p(x)}, \overline{q(x)}, \overline{p(x) + q(x)}, and \overline{p(x)q(x)} as \overline{h(x)} for some h(x) \in E of degree at most 2.
  2. Prove that \overline{E} is not an integral domain.
  3. Prove that \overline{x} is a unit in \overline{E}.

  1. Evidently, \overline{p(x)} = \overline{-x^2-11x+3}, \overline{q(x)} = \overline{8x^2 - 13x + 5}, \overline{p(x)+q(x)} = \overline{7x^2 - 24x + 8}, and \overline{p(x)q(x)} = \overline{146x^2-236x+90}.
  2. Evidently, x^3 -2x+1 = (x-1)(x^2+x-1). Now neither \overline{x-1} nor \overline{x^2+x-1} is zero in \overline{E}, but their product is. Thus (for instance) \overline{x-1} is a zero divisor in \overline{E}. We found this factorization by noting that f(1)=0, so that x-1 divides f(x).
  3. Evidently \overline{x}\overline{-x^2+2} = 1, so that \overline{x} is a unit. To find this inverse, recall that if \overline{x} is a unit, its inverse is represented by a polynomial of degree at most 2- say \overline{ax^2+bx+c}. Suppose \overline{x}\overline{ax^2+bx+c} = \overline{bx^2 + (c+2a)x-a} = 1; comparing coefficients, we find that b=0, a=-1, and c=2.

Compute in a quotient of a polynomial ring

Consider f(x) = x^4 - 16 as an element of E = \mathbb{Z}[x] and use bars to denote the natural projection E \rightarrow E/(f(x)).

  1. Fine a polynomial of degree at most 3 which is congruent to g(x) = 7x^{13} - 11x^9 + 5x^5 - 2x^3 + 3 modulo x^4-16.
  2. Prove that \overline{x+2} and \overline{x-2} are zero divisors in \overline{E}.

  1. Evidently, \overline{g(x)} = \overline{-2x^3 + 25936x + 3}.
  2. Note that x^4-16 = (x+2)(x-2)(x^2+4). By §7.4 #14, \overline{x+2} and \overline{x-2} are zero divisors in \overline{E}.