## 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 $\overline{E}$.