## Monthly Archives: July 2010

### Basic properties of nilpotent ring elements

Let $R$ be a commutative ring and let $x \in R$ be nilpotent – that is, $x^n = 0$ for some positive integer $n$. Prove the following.

1. $x$ is either zero or a zero divisor.
2. $rx$ is nilpotent for all $r \in R$.
3. $1+x$ is a unit in $R$.
4. The sum of a unit and a nilpotent element is a unit.

1. Say $m$ is minimal such that $x^m = 0$. If $m = 1$, then $x = 0$. If $m > 1$, then $x \neq 0$, $x^{m-1} \neq 0$ and $x \cdot x^{m-1} = 0$, so that $x$ is a zero divisor.
2. Since $R$ is commutative, we have $(rx)^m = r^mx^m = 0$.
3. Note that $(1 - (-x))(\sum_{i=0}^{m-1} (-x)^i)$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=0}^{m-1} (-x)^{i+1})$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=1}^{m} (-x)^i$ $= 1 + (\sum_{i=1}^{m-1} (-x)^i) - (\sum_{i=1}^{m-1} (-x)^i) - (-x)^m$ $= 1 - (-1)^mx^m$ $= 1$. Thus $1+x$ is a unit.
4. Let $u$ be a unit and $x$ nilpotent. Then $u^{-1}x$ is nilpotent, so $1+u^{-1}x$ is a unit, and thus $u(1+u^{-1}x) = u+x$ is a unit.

### Examples of nilpotent elements

An element $x \in R$, $R$ a ring, is called nilpotent if $x^m = 0$ for some positive integer $m$.

1. Show that if $n = a^kb$ for some integers $a,b$, then $\overline{ab}$ is nilpotent in $\mathbb{Z}/(n)$.
2. If $a$ is an integer, show that the element $\overline{a} \in \mathbb{Z}/(n)$ is nilpotent if and only if every prime divisor of $n$ also divides $a$. Find the nilpotent elements of $\mathbb{Z}/(72)$ explicitly.
3. Let $X$ be a nonempty set and $F$ a field, and let $R = {}^XF$ be the ring of functions $X \rightarrow F$. Prove that $R$ contains no nonzero nilpotent elements.

1. Suppose $n = a^kb$, where $k \geq 1$. Now $(ab)^k = a^kb^k$ $= (a^kb)b^{k-1}$ $= nb^{k-1}$ $\equiv 0$ mod $n$.
2. $(\Rightarrow)$ Suppose $\overline{a} \in \mathbb{Z}/(n)$ is nilpotent. Then $a^m = nk$ for some $m$ and $k$. Now if $p$ is a prime dividing $n$, then $p$ divides $a^m$, so that it divides $a$. Thus every prime dividing $n$ divides $a$. $(\Leftarrow)$ Let $n = p_1^{e_1} \cdots p_k^{e_k}$ and $a = p_1^{d_1} \cdots p_k^{d_k}m$, where $1 \leq e_i, d_i$ for all $i$ and $m$ is some integer. Let $t = \max \{ e_i \}$. Then $a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t$ $= p_1^{d_it} \cdots p_k^{d_it}m^t$, where $d_it \geq e_i$ for each $i$. Thus $a^t = nb$ for some integer $b$, and we have $a^t \equiv 0$ mod $n$.

The nilpotent elements of $\mathbb{Z}/(72)$, where $72 = 2^3 \cdot 3^2$, are as follows: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.

3. Suppose $\alpha \in R$ is nilpotent. If $\alpha \neq 0$, then there exists $x \in X$ such that $\alpha(x) \neq 0$. Let $m$ be minimal such that $\alpha(x)^m = 0$; note that $m \geq 1$. Then $\alpha(x) \alpha(x)^{m-1} = 0$, where $\alpha(x)$ $\alpha(x)^{m-1}$ are not zero. Thus $F$ contains zero divisors, a contradiction. Thus no nonzero element of $R$ is nilpotent.

### Every subring of a field which contains 1 is an integral domain

Prove that any subring of a field which contains the identity is an integral domain.

Let $R \subseteq F$ be a subring of a field. (We need not yet assume that $1 \in R$). Suppose $x,y \in R$ with $xy = 0$. since $x,y \in F$ and the zero element in $R$ is the same as that in $F$, either $x = 0$ or $y = 0$. Thus $R$ has no zero divisors. If $R$ also contains 1, then $R$ is an integral domain.

### In an integral domain, there are at most two square roots of 1

Prove that if $R$ is an integral domain and $x^2 = 1$ for some $x \in R$, then $x = 1$ or $x = -1$.

If $x^2 = 1$, then $x^2 - 1 = 0$. Evidently, then, $(x-1)(x+1) = 0$. Since $R$ is an integral domain, we must have $x-1 = 0$ or $x+1 = 0$; thus $x = 1$ or $x = -1$.

### In a division ring, every centralizer is a division ring

Prove that if $D$ is a division ring, then $C_D(a)$ is a division ring for all $a \in D$.

We saw in the previous exercise that $C_D(a)$ is a subring which contains 1. It remains to be shown that every nonzero element of $C_D(a)$ has an inverse in $C_D(a)$. To that end, let $x \in C_D(a)$ be nonzero. Then $xa = ax$. Now $x^{-1}$ exists in $D$, and we have $x^{-1}xa = x^{-1}ax$, so that $a = x^{-1}ax$. Similarly, right multiplying by $x^{-1}$ yields $ax^{-1} = x^{-1}a$. Thus $x^{-1} \in C_D(a)$. Thus $C_D(a)$ is a division ring.

### Basic properties of centralizers in a ring

Let $R$ be a ring. For a fixed element $a \in R$, define $C_R(a) = \{r \in R \ |\ ra = ar \}$. Prove that $C_R(a)$ is a subring of $R$ containing $a$. Prove that $Z(R) = \bigcap_{a \in R} C_R(a)$.

(We need not assume that $R$ has a 1.)

Note that $aa = aa$, so that $a \in C_R(a)$. Similarly, $0a = a0$, so $0 \in C_R(a)$. If $R$ has a 1, then since $1a = a1$, $1 \in C_R(a)$.

Now if $x,y \in C_R(a)$, then $(x-y)a = xa-ya = ax-ay = a(x-y)$, so that $x-y \in C_R(a)$, and $xya = xay = axy$, so that $xy \in C_R(a)$. Thus $C_R(a)$ is a subring.

If $x \in Z(R)$, then for all $r \in R$, $xr = rx$. Thus for all $r \in R$, $x \in C_R(r)$, and we have $x \in \bigcap_{r \in R} C_R(r)$.

Now if $x \in \bigcap_{r \in R} C_R(r)$, then for all $r \in R$, $x \in C_R(r)$. So $xr = rx$, and we have $x \in Z(R)$.

### Compute the center of the Hamiltonian quaternions

Describe $Z(\mathbb{H})$, where $\mathbb{H}$ denotes the Hamiltonian Quaternions. Prove that $\{a+bi \ |\ a,b \in \mathbb{R} \}$ is a subring of $\mathbb{H}$ which is a field but is not contained in $Z(\mathbb{H})$.

We claim that $Z(\mathbb{H}) = \{ x + 0i + 0j + 0k \ |\ x \in \mathbb{R} \}$. The $(\supseteq)$ direction is clear. To see the $(\subseteq)$ direction, let $\alpha = a + bi + cj + dk \in \mathbb{H}$ and $\beta = x + yi + zj + wk \in Z(\mathbb{H})$.

Using the distributive property and the fact that multiplication of real numbers is commutative, we see that $\alpha \beta = (ax - by - cz - dw)$ $+ (ay + bx + cw - dz)i$ $+ (az - bw + cx + dy)j$ $+ (aw + bz - cy + dx)k$, while $\beta \alpha = (xa - yb - zc - wd)$ $+ (xb + ya + zd - wc)i$ $+ (xc - yd + za + wb)j$ $+ (xd + yc - zb + wa)k$.

Since $\beta$ is in the center, these elements are equal, and thus their corresponding coefficients are equal. This yields three equations:

1. $cw = dz$
2. $dy = bw$
3. $bz = cy$

where we consider $y$, $z$, and $w$ to be fixed while $b$, $c$, and $d$ are arbitrary real numbers. Consider the first equation; we may assume that $d = 1$. Then, if $w \neq 0$, we have $c = z/w$. This is a contradiction since $z$ and $w$ are fixed while $c$ is arbitrary. Thus $w = 0$. Similarly, $y = z = 0$, and we have $\beta \in \{ x + 0i + 0j + 0k \ |\ x \in \mathbb{R} \}$.

Now let $S = \{ a + bi \ |\ a,b \in \mathbb{R} \}$. Let $\alpha = a + bi$ and $\beta = c+di$. Since $\alpha - \beta = (a-c) + (b-d)i \in S$ and $0 + 0i \in S$, by the subgroup criterion $S \leq \mathbb{H}$. Since $\alpha\beta = (ac-bd) + (ad + bc)i \in S$, $S$ is a subring of $\mathbb{H}$.

Now if $a + bi \neq 0 + 0i$, we see that $(a+bi)(a-bi)/(a^2 + b^2) = 1$; thus $\alpha$ has an inverse in $S$, specifically $\alpha^{-1} = (a-bi)/(a^2 + b^2)$. Thus $S$ is a division ring. Moreover, $\alpha \beta = (ac - bd) + (ad + bc)i$ $= (ca - db) + (da + bc)i$ $= \beta \alpha$, so that $S$ is a field.

Note, however, that $ij = k$ while $ji = -k$, so that $S \not\subseteq Z(\mathbb{H})$.

### The center of a ring is a subring

The center of a ring $R$ is $Z(R) = \{ z \in R \ |\ zr = rz\ \mathrm{for\ all}\ r \in R \}$. Prove that $Z(R)$ is a subring of $R$ and that if $R$ has a 1, then $1 \in Z(R)$. Prove also that the center of a division ring is a field.

Note first that $0 \in Z(R)$ since $0 \cdot r = 0 = r \cdot 0$ for all $r \in R$; in particular, $Z(R)$ is nonempty. Next, if $x,y \in Z(R)$ and $r \in R$, then $(x-y)r = xr - yr = rx - ry = r(x-y)$. By the subgroup criterion, $Z(R) \leq R$. Moreover, $xyr = xry = rxy$, so that $xy \in Z(R)$; by definition, $Z(R)$ is a subring.

If $R$ has a 1, then by definition, $1 \cdot x = x \dot 1 = x$ for all $x \in R$. Thus $1 \in Z(R)$.

Now let $R$ be a division ring, and consider $Z(R)$. If $x \in Z(R)$, note that by the cancellation law, the inverse of $x$ is unique; denote it by $x^{-1}$. Clearly, $(r^{-1})^{-1} = r$. Since $(ab)(b^{-1}a^{-1}) = 1$, we have $(ab)^{-1} = b^{-1}a^{-1}$. Now let $r \in R$. $x^{-1}r^{-1} = (rx)^{-1}$ $= (xr)^{-1}$ $= r^{-1}x^{-1}$, and since $r^{-1}$ is arbitrary in $R$, $x^{-1} \in Z(R)$. Thus $Z(R)$ is a commutative division ring- that is, a field.

### Decide whether a given subset is a subring

Decide which of the following are subrings of the ring of all functions from the closed interval $[0,1]$ to $\mathbb{R}$.

1. The set of all functions $f(x)$ such that $f(q) = 0$ for all $q \in \mathbb{Q} \cap [0,1]$.
2. The set of all polynomial functions.
3. The set of all functions which have only a finite number of zeros, together with the zero function.
4. The set of all functions which have an infinite number of zeros.
5. The set of all functions such that $\lim_{x \rightarrow 1^-} f(x) = 0$.
6. The set of all rational linear combinations of the functions $\sin mx$ and $\cos nx$, where $m,n \in \mathbb{N}$.

1. Let $A \leq [0,1]$, and let $S$ denote the set of all functions $f : [0,1] \rightarrow \mathbb{R}$ such that $f[A] = 0$. Clearly the zero function is in $S$, and if $f,g \in S$, then $(f-g)(x) = 0$ for all $x \in A$, so that $S$ is a subgroup by the subgroup criterion. Now if $f,g \in S$, then $(fg)(x) = f(x)g(x) = 0$ if $x \in A$. So $S$ is closed under multiplication, and is thus a subring.
2. Since the sum, difference, and product of two polynomials is again a polynomial, this is a subring.
3. Define $f$ and $g$ as follows: $f(x) = 1$ if $0 \leq x < 1/2$, 0 if $x = 1/2$, and $1$ if $1/2 < x \leq 1$, while $g(x) = -1$ if $0 \leq x < 1/2$, 0 if $x = 1/2$, $-1$ if $1/2 < x < 1$, and $0$ if $x = 1$. $f$ and $g$ have finitely many zeroes, but $(f+g)(x) = 0$ if $0 \leq x < 1$ and $1$ if $x = 1$; so $f+g$ has infinitely many zeros and is not itself the zero function. Since this set is not closed under addition, it is not a subring.
4. Consider the functions $f(x)$ and $g(x)$ defined as follows: $f(x) = 0$ if $0 \leq x \leq 1/2$ and 1 otherwise, while $g(x) = 1$ if $1/2 \leq x \leq 1$ and 0 otherwise. Both $f$ and $g$ have infinitely many zeroes. However, $(f+g)(x)$ is 2 at $1/2$ and 1 otherwise, and thus has no zeroes. Hence this set is not closed under addition, and cannot be a subring.
5. First we show that this subset is a subgroup. If $f$ and $g$ approach 0 as $x$ approaches 1, then $\lim_{x \rightarrow 1} (f-g)(x) = \lim_{x \rightarrow 1}f(x) - \lim_{x \rightarrow 1} g(x) = 0$. Since the zero function satisfies this property trivially, by the subgroup criterion this set is a subgroup. Similarly, we know from calculus that if $\lim_{x \rightarrow 1} f(x) = 0$ and $\lim_{x \rightarrow 1} g(x) = 0$, then $\lim_{x \rightarrow 1} (fg)(x) = 0$. Thus this set is closed under multiplication, and is a subring.
6. Recall the product-to-sum formulas for sine and cosine: $\cos x \cos y = (\cos(x-y) + \cos(x+y))/2$, $\cos x \sin y = (\sin(x+y) - \sin(x-y))/2$, $\sin x \cos y = (\sin(x+y) + \sin(x-y))/2$, and $\sin x \sin y = (\cos(x-y) - \cos(x+y))/2$. Due to these identities, the sum, difference, and product of a rational linear combination of sines and cosines of integer multiples of $x$ are again of this type. Thus this set is a subring.

### Decide whether a given set of rationals is a subring

Decide which of the following are subrings of $\mathbb{Q}$.

1. The set of all rational numbers with odd denominators (when written in lowest terms).
2. The set of all rational numbers with even denominators (when written in lowest terms).
3. The set of all nonnegative rational numbers.
4. The set of squares of rational numbers.
5. The set of all rational numbers with odd numerators (when written in lowest terms).
6. The set of all rational numbers with even numerators (when written in lowest terms).

1. In this previous exercise, we showed that this set is a subgroup of $\mathbb{Q}$. Moreover, if $a/b$ and $c/d$ are written in lowest terms and $b$ and $d$ are odd, then when $ac/bd$ is written in lowest terms, its denominator must divide $bd$ and thus is odd. So this subset of $\mathbb{Q}$ is closed under multiplication and is indeed a subring.
2. In this previous exercise, we showed that this set is not a subgroup of $\mathbb{Q}$. Thus it is not a subring of $\mathbb{Q}$.
3. This subset is not a group, since only 0 has an additive inverse. Thus it is not a subgroup of $\mathbb{Q}$, and thus not a subring.
4. Note that $1 = 1^2$ is in this set, but that $1+1 = 2$ is not the square of a rational. To see this, suppose otherwise that $(a/b)^2 = 2$; then $a^2 = 2b^2$. By the fundamental theorem of arithmetic, the number of times 2 divides $a^2$ and $2b^2$ must be the same. However, 2 divides $a^2$ and even number of times, while it divides $2b^2$ an odd number of times, a contradiction. Thus this subset is not a subring, as it is not closed under addition.
5. This set is not closed under addition since $1/3$ is in lowest terms, but $2/3$ is in lowest terms and has an even numerator. Thus this subset is not a subring.
6. First we show that this subset $A_6$ is a subgroup. If $a/b$ and $c/d$ are in lowest terms and $a$ and $c$ are even, then $(a/b) + (-c/d) = (ad-bc)/bd$. Now $b$ and $d$ must be odd, so that when reduced to lowest terms, the numerator of $(ad-bc)/bd$ is even since 2 divides $a$ and $b$. Since $A_6$ contains $2 = 2/1$, $A_6 \leq \mathbb{Q}$ is a subgroup. Now if $a/b$ and $c/d$ are in $A_6$ and in lowest terms, then $b$ and $d$ are odd, so that when expressed in lowest terms, 2 must divide $ac$. Thus $A_6$ is closed under multiplication, and is a subring of $\mathbb{Q}$. Now suppose $A_6$ has an identity element $u/v$. Then for all $a/b$, we have $ua/vb = a/b$, so that $uab = vab$. Thus $u = v$. Since $u$ must be even and $v$ odd, we have a contradiction. So $A_6$ is a ring, but not a unital ring.