Ring Theory MCQs – Daily Mathematics Practice

Detailed Solution

We are given that $u$ is a unit, $n$ is nilpotent, and $$ un=nu. $$ Since $u$ is a unit, there exists $u^{-1}\in R$ such that $$ uu^{-1}=u^{-1}u=1. $$ Factor $u+n$ as $$ u+n=u(1+u^{-1}n). $$ We will show that both factors on the right are units. Step 1: Show that $u^{-1}n$ is nilpotent. Since $un=nu$, the elements $u^{-1}$ and $n$ also commute. Therefore, $$ (u^{-1}n)^k=u^{-k}n^k=0. $$ Thus $u^{-1}n$ is nilpotent. Step 2: Show that $1+u^{-1}n$ is a unit. Let $a=u^{-1}n$. Since $a^k=0$, the finite geometric series gives $$ (1+a)^{-1}=1-a+a^2-\cdots+(-1)^{k-1}a^{k-1}. $$ Indeed, $$ (1+a)(1-a+a^2-\cdots+(-1)^{k-1}a^{k-1})=1, $$ because all intermediate terms cancel and $a^k=0$. The same product in the reverse order is also $1$. Hence $1+a=1+u^{-1}n$ is a unit. Step 3: Conclude that $u+n$ is a unit. We have expressed $u+n$ as a product of two units: $$ u+n=u(1+u^{-1}n). $$ Therefore, $u+n$ is a unit. An explicit inverse is $$ \boxed{ (u+n)^{-1} =\left(\sum_{j=0}^{k-1}(-u^{-1}n)^j\right)u^{-1}. } $$ Thus option A is correct. Why the other options are incorrect: Option B: A ring need not be free of zero divisors for this result to hold. The proof uses only the fact that $u$ is a unit, $n$ is nilpotent, and $u$ commutes with $n$. Option C: The sum need not be nilpotent. For example, in $\mathbb Z$, take $u=1$ and $n=0$. Then $u+n=1$, which is not nilpotent. Option D: False under the stated hypotheses. The sum $u+n$ is necessarily a unit. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

Let $$ f(x)=a_mx^m+\cdots+a_1x+a_0 $$ and $$ g(x)=b_nx^n+\cdots+b_1x+b_0, $$ where $a_m\neq0$ and $b_n\neq0$ are the leading coefficients of $f$ and $g$. When we multiply $f(x)$ and $g(x)$, the coefficient of the highest possible power $x^{m+n}$ is $$ a_mb_n. $$ Since $R$ is an integral domain, it has no zero divisors. Therefore, the product of two nonzero elements is nonzero: $$ a_m\neq0,\quad b_n\neq0 \quad\Longrightarrow\quad a_mb_n\neq0. $$ Consequently, the term $a_mb_nx^{m+n}$ cannot vanish. No term of degree greater than $m+n$ can occur in the product. Hence $$ \boxed{\deg(fg)=m+n}. $$ Thus option A is correct. Why the other options are incorrect: Option B: The degree formula does not hold in every polynomial ring. For example, consider $R=\mathbb Z_4$, which has a nonzero zero divisor because $2\cdot2=0$. In $\mathbb Z_4[x]$, let $$ f(x)=2x+1,\qquad g(x)=2x+1. $$ Then $$ \begin{aligned} f(x)g(x) &=4x^2+4x+1\\ &=1 \end{aligned} $$ in $\mathbb Z_4[x]$. Therefore, $$ \deg(fg)=0\neq2=\deg f+\deg g. $$ Option C: False. If $R$ is an integral domain, then $R[x]$ is also an integral domain. Indeed, the product of any two nonzero polynomials has a nonzero leading coefficient. Option D: False. Zero divisors in the coefficient ring can cause the leading coefficient of a product to vanish, reducing its degree. The example over $\mathbb Z_4[x]$ above demonstrates this. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$ Key takeaway: For nonzero polynomials $f,g\in R[x]$, $$ \deg(fg)=\deg f+\deg g $$ holds whenever $R$ is an integral domain. The absence of zero divisors guarantees that the product of the leading coefficients is nonzero.

Detailed Solution

We are given that $$ x^3=x $$ for every $x\in R$. We will prove that $$ 6\cdot1_R=0, $$ which implies that the characteristic of $R$ divides $6$. Step 1: Substitute $x+1$. Since every element satisfies the given identity, $$ (x+1)^3=x+1. $$ Expanding and using the fact that $1$ commutes with every element, $$ x^3+3x^2+3x+1=x+1. $$ Since $x^3=x$, we obtain $$ 3x^2+3x=0. $$ Thus, $$ 3x^2+3x=0. \tag{1} $$ Step 2: Substitute $x-1$. Similarly, $$ (x-1)^3=x-1. $$ Expanding gives $$ x^3-3x^2+3x-1=x-1. $$ Using $x^3=x$, we get $$ -3x^2+3x=0. \tag{2} $$ Step 3: Add equations (1) and (2). Adding yields $$ 6x=0 $$ for every $x\in R$. In particular, taking $x=1_R$, we obtain $$ 6\cdot1_R=0. $$ The characteristic of a ring with unity is the least positive integer $n$ such that $$ n\cdot1_R=0, $$ provided such an integer exists. Since $6\cdot1_R=0$, the characteristic must divide $6$. Therefore, $$ \boxed{\operatorname{char}(R)\mid6}. $$ Hence option B is correct. Why the other options are incorrect: Option A: A Boolean ring must satisfy $x^2=x$ for every element. The identity $x^3=x$ does not imply this. For example, in $\mathbb Z_3$, every element satisfies $x^3=x$, but $2^2=1\neq2$. Option C: The ring need not be a field. For example, $\mathbb Z_6$ satisfies $x^3=x$ for every element but is not a field. Option D: A definite conclusion follows from the identity: the characteristic divides $6$. Thus the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

Every element of $R$ has the form $$ A=\begin{pmatrix} a&b\\ 0&c \end{pmatrix}, \qquad a,b,c\in\mathbb Z_2. $$ Since each of $a,b,c$ has two possible values, the ring contains $$ 2^3=8 $$ elements. A square matrix over a field is invertible if and only if its determinant is nonzero. Here, $\mathbb Z_2$ is a field, and $$ \det(A)=ac. $$ Therefore, $A$ is invertible precisely when $$ ac=1. $$ In $\mathbb Z_2$, this requires $$ a=1,\qquad c=1. $$ The entry $b$ can be either $0$ or $1$. Thus the only units are $$ \begin{pmatrix}1&0\\0&1\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}1&1\\0&1\end{pmatrix}. $$ Consequently, the ring has exactly $$ 2 $$ units. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Since $e$ is idempotent, we have $$ e^2=e. $$ Let $f=1-e$. We prove that $f$ is also idempotent and that $ef=fe=0$. First, calculate $f^2$: $$ \begin{aligned} f^2&=(1-e)^2\\ &=1-e-e+e^2\\ &=1-2e+e\\ &=1-e\\ &=f. \end{aligned} $$ Here we used $e^2=e$. Thus $f$ is idempotent. Next, using the distributive law, $$ \begin{aligned} ef&=e(1-e)\\ &=e-e^2\\ &=e-e\\ &=0. \end{aligned} $$ Similarly, $$ \begin{aligned} fe&=(1-e)e\\ &=e-e^2\\ &=e-e\\ &=0. \end{aligned} $$ Therefore, $$ \boxed{f^2=f,\qquad ef=fe=0}. $$ These identities require neither commutativity nor any additional assumptions beyond the existence of unity and the idempotency of $e$. Hence option A is correct. Option B is incorrect because the proof of $f^2=f$ uses only distributivity and $e^2=e$, not commutativity. Option C is incorrect. Nontrivial idempotents can also satisfy these identities. For example, in $M_2(\mathbb R)$, take $$ e=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad f=I-e=\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$ Then $e\neq0$, $e\neq I$, and $ef=fe=0$. Option D is incorrect because $f=1-e$ need not be a unit. For example, in $M_2(\mathbb R)$, the matrix $$ f=\begin{pmatrix}0&0\\0&1\end{pmatrix} $$ is idempotent but singular, so it is not invertible. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

An element $e$ of a ring is called idempotent if $$ e^2=e. $$ In the product ring $R=\mathbb Z_3\times\mathbb Z_3$, multiplication is defined componentwise: $$ (a,b)(c,d)=(ac,bd), $$ where all operations are performed modulo $3$. For $e=(1,0)$, we obtain $$ e^2=(1,0)(1,0)=(1^2,0^2)=(1,0)=e. $$ Thus $e$ is idempotent. Moreover, $$ e=(1,0)\neq(0,0) $$ and $$ e=(1,0)\neq(1,1)=1_R. $$ Therefore, $e$ is a nontrivial idempotent, proving that option A is correct. We examine the remaining options. Option B is incorrect. Although $$ e(1,1)-e=(1,0)-(1,0)=(0,0), $$ this only shows that $(1,1)$ is the multiplicative identity. It does not establish nilpotency. In fact, since $e^2=e$, we have $$ e^n=e\neq0 $$ for every positive integer $n$. Hence $e$ is not nilpotent. Option C is incorrect. A nonzero element need not be a unit. If $e=(1,0)$ had an inverse $(a,b)$, then $$ (1,0)(a,b)=(1,1). $$ But $$ (1,0)(a,b)=(a,0), $$ whose second component is always $0$. Therefore, no inverse exists, and $e$ is not a unit. Option D is incorrect. Although the only idempotents in $\mathbb Z_3$ are $0$ and $1$, the product ring contains the idempotents $$ (0,0),\quad(1,0),\quad(0,1),\quad(1,1). $$ The two idempotents $(1,0)$ and $(0,1)$ are nontrivial. Thus the correct option is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

We work in the polynomial ring $$ R=\mathbb Z_4[x], $$ where every coefficient is calculated modulo $4$. Given $$ f(x)=2x+1,\qquad g(x)=2x-1, $$ their product is $$ \begin{aligned} f(x)g(x) &=(2x+1)(2x-1)\\ &=4x^2-2x+2x-1\\ &=4x^2-1. \end{aligned} $$ Since $4\equiv0\pmod4$, we obtain $$ 4x^2-1\equiv-1\equiv3\pmod4. $$ Therefore, $$ \boxed{f(x)g(x)=3}. $$ This proves that option B is correct. We examine the remaining options. Option A is false. Both $f$ and $g$ have degree $1$, but their product is the nonzero constant polynomial $3$, which has degree $0$. Thus $$ \deg(fg)=0\neq2=\deg f+\deg g. $$ The usual degree formula can fail when the coefficient ring has zero divisors. Option C is false. In fact, both $f$ and $g$ are units. We have $$ \begin{aligned} f(x)^2 &=(1+2x)^2\\ &=1+4x+4x^2\\ &=1 \end{aligned} $$ in $\mathbb Z_4[x]$. Hence $f^{-1}=f$. Similarly, $$ \begin{aligned} g(x)^2 &=(2x-1)^2\\ &=4x^2-4x+1\\ &=1. \end{aligned} $$ Hence $g^{-1}=g$. Therefore, both polynomials are units, not nonunits. Option D is false. The coefficient ring $\mathbb Z_4$ has a nonzero zero divisor because $$ 2\cdot2=4\equiv0\pmod4. $$ The same relation holds in $\mathbb Z_4[x]$, where the constant polynomial $2$ is nonzero but satisfies $2\cdot2=0$. Thus $\mathbb Z_4[x]$ has zero divisors. Consequently, the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

An element $u=a+b\sqrt{-5}$ is a unit if and only if its norm is $1$. Thus we need to solve $$ a^2+5b^2=1, \qquad a,b\in\mathbb Z. $$ Since $a^2\geq0$ and $5b^2\geq0$, we have $$ 0\leq 5b^2\leq1. $$ Because $b$ is an integer, this forces $b=0$. Substituting $b=0$ into the norm equation gives $$ a^2=1, $$ so $$ a=\pm1. $$ Therefore, the only elements of norm $1$ are $$ u=1\quad\text{and}\quad u=-1. $$ Both are units because $$ 1^{-1}=1,\qquad (-1)^{-1}=-1. $$ Hence the complete set of units is $$ \boxed{\mathbb Z[\sqrt{-5}]^\times=\{1,-1\}}. $$ Thus option A is correct. Option B is incorrect because $$ N(\sqrt{-5})=0^2+5(1)^2=5\neq1. $$ Therefore, neither $\sqrt{-5}$ nor $-\sqrt{-5}$ is a unit. Option C is incorrect. For example, the element $1+\sqrt{-5}$ has norm $$ N(1+\sqrt{-5})=1^2+5(1)^2=6. $$ Since $6$ divides $6$ but the norm is not $1$, this element is not a unit. Option D is incorrect because $1$ and $-1$ are units. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

The multiplication in this ring is defined by intersection: $$ A\otimes B=A\cap B. $$ For every subset $A\subseteq X$, the intersection of $A$ with itself is $A$. Therefore, $$ A\otimes A=A\cap A=A. $$ Hence every element of the ring is idempotent under multiplication. This proves that option B is correct. We verify the other options. Option A is false. The multiplicative identity is $X$, because $$ A\cap X=A $$ for every $A\subseteq X$. A nonempty subset $A$ is a unit only if there exists $B\subseteq X$ such that $$ A\cap B=X. $$ This requires $A=X$. Thus $X$ is the only unit, so not every nonempty subset is a unit. Option C is false. Intersection is commutative: $$ A\otimes B=A\cap B=B\cap A=B\otimes A. $$ Option D is false. The characteristic of a ring is the least positive integer $n$ for which $n\cdot 1_R=0_R$. Here, the additive identity is $\varnothing$, and the multiplicative identity is $X$. Since addition is symmetric difference, $$ X\oplus X=\varnothing. $$ Thus $2\cdot 1_R=0_R$, while $X\neq\varnothing$. Hence the characteristic is $2$, not $3$. Therefore, the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

In the quotient ring $$ R=\mathbb Z_2[x]/(x^2), $$ the polynomial $x^2$ is identified with $0$. Thus, every element of $R$ has a unique representative of the form $$ a+bx,\qquad a,b\in\mathbb Z_2. $$ Consequently, the four elements of $R$ are $$ R=\{0,1,x,1+x\}. $$ Hence $|R|=4$. The characteristic of a ring with unity is the least positive integer $n$ such that $$ n\cdot 1_R=0. $$ Since $\mathbb Z_2$ has characteristic $2$, we have $$ 1_R+1_R=0. $$ Therefore, $$ 2\cdot 1_R=0, $$ while $1_R\neq0$. Hence the characteristic of $R$ is exactly $$ \operatorname{char}(R)=2. $$ Thus option B is correct. We also verify the remaining options. Option A is incorrect because the characteristic of a ring need not equal its number of elements. Here, $|R|=4$, but $\operatorname{char}(R)=2$. Option C is incorrect because $x\neq0$ in the quotient ring, while $$ x^2=0. $$ Therefore, $R$ contains a nonzero nilpotent element. A field has no nonzero nilpotent elements, so $R$ cannot be a field and is not isomorphic to $\mathbb F_4$. Option D is incorrect for the same reason: $x$ is a nonzero nilpotent element because $x^2=0$. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

A matrix in $M_2(\mathbb Z_2)$ is a **unit** if and only if it is invertible. Since $\mathbb Z_2=\{0,1\}$, there are $$ 2^4=16 $$ matrices in $M_2(\mathbb Z_2)$ altogether. To count the invertible matrices, we can choose their columns successively. For an invertible $2\times2$ matrix over $\mathbb Z_2$: The first column can be any nonzero vector in $\mathbb Z_2^2$. There are $$ 2^2-1=3 $$ choices: $$ \begin{pmatrix}1\\0\end{pmatrix}, \quad \begin{pmatrix}0\\1\end{pmatrix}, \quad \begin{pmatrix}1\\1\end{pmatrix}. $$ The second column must be linearly independent of the first column. The vector space $\mathbb Z_2^2$ has $4$ vectors in total. Once a nonzero first column is chosen, its span contains $$ 2 $$ vectors: $$ \{0,v\}. $$ Therefore, there are $$ 4-2=2 $$ choices for the second column. Hence the number of invertible matrices is $$ (2^2-1)(2^2-2) =3\cdot2 =6. $$ Therefore, $$ \boxed{|R^\times|=6}. $$ Thus the correct answer is $$ \boxed{\mathrm{A}}. $$ For $$ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}, $$ the matrix is invertible over $\mathbb Z_2$ precisely when $$ \det(A)=ad-bc\neq0. $$ Since the only nonzero element of $\mathbb Z_2$ is $1$, this means $$ \det(A)=1. $$ Exactly $6$ of the $16$ matrices satisfy this condition. **Option B:** $8$ counts too many matrices; not every matrix with nonzero entries is invertible. **Option C:** $15$ is the number of nonzero matrices, but a nonzero matrix need not be invertible. For example, $$ \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} $$ is nonzero but singular. **Option D:** $16$ is the total number of elements of $M_2(\mathbb Z_2)$, not the number of units. Therefore, $$ \boxed{\text{The ring }M_2(\mathbb Z_2)\text{ has exactly }6\text{ units}.} $$

Detailed Solution

First, $\mathbb Z$ is indeed a subring of $\mathbb Q$ because it is closed under addition, subtraction, and multiplication, and contains the multiplicative identity $1$. However, being a **subring** does not imply being an **ideal**. For $\mathbb Z$ to be an ideal of $\mathbb Q$, we would need $$ qz\in\mathbb Z $$ for every $q\in\mathbb Q$ and every $z\in\mathbb Z$. Take $$ q=\frac12\in\mathbb Q \qquad\text{and}\qquad z=1\in\mathbb Z. $$ Then $$ qz=\frac12\notin\mathbb Z. $$ Therefore, $\mathbb Z$ is **not** an ideal of $\mathbb Q$. Hence, $$ \boxed{\mathbb Z\text{ is a subring of }\mathbb Q\text{ but not an ideal of }\mathbb Q.} $$ **Option A:** False. Although $\mathbb Z$ is a subring of $\mathbb Q$, it fails the absorption property required for an ideal. **Option C:** False. $\mathbb Z$ is naturally a subring of $\mathbb Q$. **Option D:** False. A subring of a commutative ring need not be an ideal. The example $$ \mathbb Z\subset\mathbb Q $$ is a direct counterexample. Thus the correct answer is $$ \boxed{\mathrm{B}}. $$ **Key distinction:** A **subring** is closed under the ring operations internally, whereas an **ideal** must additionally satisfy the absorption property $$ rI\subseteq I \qquad\text{for every }r\in R. $$

Detailed Solution

The opposite ring $R^{\mathrm{op}}$ has exactly the same elements and addition as $R$, but its multiplication is reversed: $$ A*B=BA. $$ Consider the transpose map $$ \Phi:R\longrightarrow R^{\mathrm{op}}, \qquad \Phi(A)=A^T. $$ We show that $\Phi$ is a ring isomorphism. For $A,B\in R$, $$ \Phi(A+B) =(A+B)^T =A^T+B^T =\Phi(A)+\Phi(B). $$ Thus $\Phi$ is additive. In the ordinary matrix ring $R$, $$ \Phi(AB)=(AB)^T=B^TA^T. $$ But multiplication in $R^{\mathrm{op}}$ is defined by $$ X*Y=YX. $$ Therefore, $$ \Phi(A)*\Phi(B) =A^T*B^T =B^TA^T. $$ Hence $$ \Phi(AB)=\Phi(A)*\Phi(B). $$ Thus the transpose map is a ring homomorphism $$ R\longrightarrow R^{\mathrm{op}}. $$ Since $$ (A^T)^T=A, $$ the transpose map is its own inverse. Hence $\Phi$ is bijective. Therefore, $$ \boxed{\Phi(A)=A^T} $$ is a ring isomorphism from $M_2(\mathbb R)$ onto its opposite ring. Hence the correct answer is $$ \boxed{\mathrm{B}}. $$ **Option A:** False. The opposite multiplication $$ a*b=ba $$ still satisfies associativity and the ring axioms. Thus $R^{\mathrm{op}}$ is indeed a ring. **Option C:** False in general. Although $R$ and $R^{\mathrm{op}}$ have the same underlying set and additive structure, their multiplications are reversed. Since $M_2(\mathbb R)$ is noncommutative, the two ring structures are not literally the same. For example, if $$ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, $$ then $$ AB= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad BA= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Thus $AB\neq BA$. **Option D:** False. By definition, $R^{\mathrm{op}}$ has **exactly the same additive structure** as $R$. Only the multiplication is reversed. Therefore, $$ \boxed{\text{The transpose map }A\mapsto A^T\text{ gives }R\cong R^{\mathrm{op}}.} $$

Detailed Solution

First observe that both $f$ and $g$ are nonzero continuous functions. For $f$, $$ f(x)= \begin{cases} 0, & 0\le x\le \frac12,\\[2mm] x-\frac12, & \frac12<x\le1. \end{cases} $$ Thus, for example, $$ f(1)=\frac12\neq0. $$ Hence $f$ is a nonzero element of $C[0,1]$. Similarly, $$ g(x)= \begin{cases} \frac12-x, & 0\le x<\frac12,\\[2mm] 0, & \frac12\le x\le1. \end{cases} $$ and therefore $$ g(0)=\frac12\neq0. $$ So $g$ is also nonzero. Now consider their pointwise product. Here, $$ f(x)=0, $$ so $$ f(x)g(x)=0. $$ Here, $$ g(x)=0, $$ so again $$ f(x)g(x)=0. $$ Therefore, $$ f(x)g(x)=0 \qquad\text{for every }x\in[0,1]. $$ Hence $$ fg=0 $$ in the ring $C[0,1]$, even though $$ f\neq0,\qquad g\neq0. $$ This is exactly the definition of zero divisors. Therefore, $f$ and $g$ are nonzero zero divisors, and consequently $C[0,1]$ **has zero divisors**. Thus, $$ \boxed{\text{A: }C[0,1]\text{ has zero divisors}.} $$ **Option B:** False. In fact, $f$ and $g$ **are** zero divisors because $$ f\neq0,\quad g\neq0,\quad fg=0. $$ **Option C:** False. The explicit pair $f,g$ provides nonzero zero divisors in $C[0,1]$. **Option D:** False. A unit in $C[0,1]$ must be nonzero at every point of $[0,1]$. But $$ f(x)=0\quad\text{for }x\le\frac12, $$ and $$ g(x)=0\quad\text{for }x\ge\frac12. $$ Therefore neither $f$ nor $g$ can have a multiplicative inverse in $C[0,1]$. Hence the correct answer is $$ \boxed{\mathrm{A}}. $$

Detailed Solution

The abelian group $\mathbb Z\times\mathbb Z$ is a free abelian group of rank $2$. Every group endomorphism of $\mathbb Z\times\mathbb Z$ is therefore determined by the images of the standard generators $$ e_1=(1,0),\qquad e_2=(0,1). $$ Thus every endomorphism can be represented by a $2\times2$ integer matrix $$ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}, \qquad a,b,c,d\in\mathbb Z. $$ Under this correspondence, addition of endomorphisms corresponds to matrix addition, while composition corresponds to matrix multiplication. Hence $$ R\cong M_2(\mathbb Z), $$ the ring of $2\times2$ matrices with integer entries. Matrix multiplication is not commutative in general. For example, consider $$ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad B= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}. $$ Then $$ AB= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, $$ whereas $$ BA= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Therefore, $$ AB\neq BA. $$ Thus $M_2(\mathbb Z)$, and hence $R$, is noncommutative. Option C is false because $R$ does have nonzero zero divisors. For the matrices above, $$ A\neq0,\qquad B\neq0,\qquad AB\neq0, $$ so these particular matrices do not give zero divisors in that order. However, taking $$ C= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \qquad D= \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix}, $$ gives $$ CD= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, $$ so this pair also does not give zero. Instead, take $$ E= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad F= \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}. $$ These are both nonzero, but $$ EF= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}. $$ Hence $R$ has nonzero zero divisors. Option D is false because not every nonzero matrix is invertible. For example, $$ \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} \neq0 $$ has determinant $0$, so it is not a unit in $M_2(\mathbb Z)$. Therefore, the correct statement is that $R$ is a noncommutative ring. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

By definition, a zero divisor in this question must be a nonzero element $x\in\mathbb Z_8$ for which there exists a nonzero $y\in\mathbb Z_8$ such that $$ xy\equiv0\pmod 8. $$ We examine the nonzero elements $1,2,3,4,5,6,7$. The elements $2$, $4$, and $6$ are zero divisors because $$ 2\cdot4=8\equiv0\pmod8, $$ so $2$ is a zero divisor; $$ 4\cdot2=8\equiv0\pmod8, $$ so $4$ is a zero divisor; and $$ 6\cdot4=24\equiv0\pmod8, $$ so $6$ is a zero divisor. Now consider $1,3,5,7$. Each of these is relatively prime to $8$: $$ \gcd(1,8)=\gcd(3,8)=\gcd(5,8)=\gcd(7,8)=1. $$ Therefore, each of $1,3,5,7$ is a unit in $\mathbb Z_8$. A unit cannot be a zero divisor: if $xy\equiv0\pmod8$ and $x$ has a multiplicative inverse, multiplying by $x^{-1}$ gives $$ y\equiv0\pmod8, $$ contradicting the requirement that $y\neq0$. Hence the nonzero zero divisors are exactly $$ \{2,4,6\}. $$ Notice that $0$ is not counted because the definition explicitly requires $x\neq0$. Therefore, there are exactly $3$ zero divisors. Option B is incorrect because it includes $0$, which is excluded by the definition given in the question. Option C is incorrect because not every nonzero element of $\mathbb Z_8$ is a zero divisor. In particular, $1,3,5,$ and $7$ are units and hence are not zero divisors. Option D is incorrect because $2$, $4$, and $6$ explicitly satisfy the zero-divisor condition. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

For an element $u=a+b\omega\in\mathbb Z[\omega]$, the norm is \[ N(u)=a^2-ab+b^2. \]An element is a unit if and only if its norm is $1$. Therefore, we need to determine all integer pairs $(a,b)$ satisfying \[ a^2-ab+b^2=1. \]Complete the square: \[ a^2-ab+b^2 =\left(a-\frac b2\right)^2+\frac{3b^2}{4}. \]Since this expression is equal to $1$, we must have \[ \frac{3b^2}{4}\leq 1, \]which gives \[ b^2\leq\frac43. \]Because $b\in\mathbb Z$, we obtain \[ b\in\{-1,0,1\}. \]If $b=0$, then \[ a^2=1, \]so $a=\pm1$. This gives the two units \[ 1,\quad -1. \]If $b=1$, then \[ a^2-a+1=1, \]so \[ a(a-1)=0. \]Hence $a=0$ or $a=1$, giving \[ \omega,\quad 1+\omega. \]Since $\omega$ is a primitive cube root of unity, \[ 1+\omega=-\omega^2. \]Thus these two elements are $\omega$ and $-\omega^2$. If $b=-1$, then \[ a^2+a+1=1, \]so \[ a(a+1)=0. \]Hence $a=0$ or $a=-1$, giving \[ -\omega,\quad -1-\omega. \]Using $1+\omega=-\omega^2$, we have \[ -1-\omega=\omega^2. \]Thus these two elements are $-\omega$ and $\omega^2$. Therefore, the complete set of elements having norm $1$ is \[ \{1,-1,\omega,-\omega,\omega^2,-\omega^2\} = \{\pm1,\pm\omega,\pm\omega^2\}. \]Hence $\mathbb Z[\omega]$ has exactly six units. Option A is incorrect because ${1,-1,i,-i}$ are the units of the Gaussian integers $\mathbb Z[i]$, not the Eisenstein integers. Option C is incorrect because a nonzero Eisenstein integer need not have norm $1$. For example, \[ N(2)=4\neq1, \]so $2$ is not a unit. Option D is incorrect because irrationality of $\omega$ does not prevent it from being a unit. In fact, \[ N(\omega)=0^2-0(1)+1^2=1, \]so $\omega$ is a unit. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Since $1$ is the (two-sided) multiplicative identity of $R$, it satisfies $x\cdot1=x$ for every $x\in R$, in particular for $x=e$: $$ e\cdot1=e. $$ On the other hand, the hypothesis on $e$ says $e\cdot a=a$ for every $a\in R$, so taking $a=1$ gives $$ e\cdot1=1. $$ Combining these two equalities, $$ e=e\cdot1=1. $$ So $e=1$ is forced, and this argument used only the defining property $x\cdot1=x$ of the unity together with the hypothesis on $e$ — commutativity of $R$ was never needed. This rules out option B. Option C fails because if $e=0$, then $e\cdot a=0\neq a$ for any nonzero $a$ (recalling $1\neq0$ guarantees $R$ has nonzero elements), contradicting the hypothesis. Option D is also false, since the argument above works in any ring with unity, commutative or not, and does not require $R$ to be a field. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

By definition, $\varepsilon\neq0$ but $\varepsilon^2=0$, so $\varepsilon$ is a nonzero nilpotent element of $R[\varepsilon]$. This immediately rules out option A: a field can have no nonzero nilpotents, since if $x\neq0$ were nilpotent with $x^n=0$, then multiplying by $x^{-1}$ repeatedly would force $x=0$, a contradiction. It also rules out option C, since $\varepsilon\cdot\varepsilon=0$ with $\varepsilon\neq0$ is precisely a zero-divisor relation. For option D, check whether the proposed map $\phi(a+b\varepsilon)=(a,a+b)$ is a ring homomorphism. Compute the product in $R[\varepsilon]$: $$ (a+b\varepsilon)(c+d\varepsilon)=ac+(ad+bc)\varepsilon, $$ using $\varepsilon^2=0$. So $$ \phi\big((a+b\varepsilon)(c+d\varepsilon)\big)=(ac,\;ac+ad+bc). $$ But the componentwise product in $R\times R$ gives $$ \phi(a+b\varepsilon)\cdot\phi(c+d\varepsilon)=(a,a+b)\cdot(c,c+d)=(ac,\;(a+b)(c+d))=(ac,\;ac+ad+bc+bd). $$ These two results differ by the extra term $bd$, so $\phi$ is not multiplicative and hence not a ring isomorphism. In fact $R[\varepsilon]$ cannot be isomorphic to $R\times R$ at all, since $R\times R$ has no nonzero nilpotent elements while $R[\varepsilon]$ does (namely $\varepsilon$). Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

Since $R$ is commutative, the binomial theorem applies: $$ (a+b)^p=\sum_{k=0}^{p}\binom{p}{k}a^{p-k}b^k=a^p+\sum_{k=1}^{p-1}\binom{p}{k}a^{p-k}b^k+b^p. $$ For each $k$ with $0<k<p$, the binomial coefficient $\binom{p}{k}=\dfrac{p!}{k!(p-k)!}$ has $p$ (a prime) dividing the numerator $p!$ but not dividing $k!$ or $(p-k)!$, since all factors in those are strictly smaller than $p$. Hence $p\mid\binom{p}{k}$ for every such $k$. Since $R$ has characteristic $p$, multiplying any element of $R$ by $p$ gives $0$. Therefore every middle term $\binom{p}{k}a^{p-k}b^k$ vanishes in $R$, leaving $$ (a+b)^p=a^p+b^p. $$ This identity requires only that $R$ be commutative (so the binomial theorem is valid) and have characteristic $p$; it does not require $R$ to be a field, an integral domain, or free of zero divisors. It also fails in general for noncommutative rings, since the binomial expansion itself is invalid there without commutativity. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Write $x=e-g$ and $y=e+g$. Multiply using distributivity and the relations $e\cdot e=e$, $e\cdot g=g\cdot e=g$, and $g\cdot g=e$ (since $g^2=e$): $$ xy=(e-g)(e+g)=e\cdot e+e\cdot g-g\cdot e-g\cdot g. $$ Substituting the relations, $$ xy=e+g-g-e=0. $$ So $xy=0$ even though neither $x$ nor $y$ is the zero element of $\mathbb Z[C_2]$ (their coefficient vectors are $(1,-1)$ and $(1,1)$, both nonzero). This shows that $\mathbb Z[C_2]$ has genuine zero divisors, despite the fact that the coefficient ring $\mathbb Z$ itself is an integral domain. This is a standard illustration of why group rings of groups with torsion (elements of finite order greater than $1$, like $g$ here) typically fail to be integral domains: the relation $g^2=e$ forces $(e-g)(e+g)=e^2-g^2=e-e=0$, mirroring a difference-of-squares factorization of zero. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

In the polynomial ring $R[x]$ (with $R$ a field), the units are exactly the nonzero constant polynomials, since the degree of a product of nonzero polynomials adds and a unit must have degree $0$. In the power series ring $R[[x]]$, however, a series is a unit if and only if its constant term is nonzero. This is because if $$ f(x)=1-x, $$ we can check directly that $$ f(x)\cdot(1+x+x^2+\cdots)=(1-x)(1+x+x^2+\cdots)=1, $$ using the fact that the series multiplication is well-defined term by term and the terms telescope. So $f$ has a genuine inverse inside $R[[x]]$, even though $1-x$ is certainly not invertible as a polynomial in $R[x]$ (its only polynomial "inverse" would have to be an infinite series, which isn't a polynomial). This shows units in $R[[x]]$ are far more plentiful than units in $R[x]$: any series with nonzero constant term is invertible, not just the nonzero constants. Hence option A is false, C is false since $f$ is invertible in $R[[x]]$, and D is false since $R[[x]]$ is in fact an integral domain when $R$ is a field (no zero divisors at all), while this has nothing to do with matching $R[x]$'s zero divisor structure trivially — the real distinguishing feature here is the units, not zero divisors. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

For a quaternion $$ q=a+bi+cj+dk, $$ define its conjugate by $$ \overline q=a-bi-cj-dk. $$ Then $$ q\overline q=\overline q q=a^2+b^2+c^2+d^2=N(q). $$ If $q\neq0$, then $$ N(q)=a^2+b^2+c^2+d^2>0. $$ Hence $$ q^{-1}=\frac{\overline q}{N(q)} $$ exists in $\mathbb H$. Therefore every nonzero quaternion is invertible, so $\mathbb H$ is a division ring. Moreover, $\mathbb H$ is not commutative because $$ ij=k,\qquad ji=-k, $$ and hence $ij\neq ji$. Thus $\mathbb H$ is a division ring but not a field. It also has no zero divisors. Indeed, if $pq=0$, then using multiplicativity of the norm, $$ N(pq)=N(p)N(q)=0. $$ Since $N(p),N(q)\ge0$, this implies $N(p)=0$ or $N(q)=0$, and hence $p=0$ or $q=0$. Therefore the correct option is $$ \boxed{\mathrm{B}}. $$

Detailed Solution

An element $x$ is nilpotent if $x^k\equiv0\pmod9$ for some positive integer $k$. Checking each residue: $3^2=9\equiv0$, so $3$ is nilpotent; $6^2=36\equiv0\pmod9$ (since $36=4\times9$), so $6$ is nilpotent; and $0$ is trivially nilpotent. Every other nonzero residue ($1,2,4,5,7,8$) is coprime to $9$, hence a unit in $\mathbb Z_9$, and a unit can never be nilpotent (if $x^k=0$ and $x$ were invertible, multiplying by $x^{-k}$ would give $1=0$, absurd). So the complete set of nilpotents is $\{0,3,6\}$. Thus the correct option is $\boxed{\mathrm{C}}$.

Detailed Solution

In $R=\{0\}$, the single element $0$ serves as both the additive identity and the multiplicative identity, so $0=1$ in this ring. In any ring with unity where $0=1$, every element $x$ satisfies $x=x\cdot1=x\cdot0=0$, forcing the ring to be trivial; so $\{0\}$ is the *only* ring where the two identities coincide, proving (A). $R$ does satisfy all ring axioms (associativity, distributivity, etc.) trivially with just one element, ruling out (B) and (C). By convention, a field is required to have $0\neq1$ (a nontrivial multiplicative identity distinct from $0$), so the trivial ring is explicitly excluded from being a field, ruling out (D). Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

A unit $u=a+bi$ must satisfy $N(u)=a^2+b^2=1$ with $a,b\in\mathbb Z$. The only integer solutions are $(a,b)\in\{(1,0),(-1,0),(0,1),(0,-1)\}$, giving exactly the four units $\{1,-1,i,-i\}$ — this proves (A). $\mathbb Z[i]$ is not a field: for instance $N(2)=4\neq1$, so $2$ has no inverse in $\mathbb Z[i]$, ruling out (B). Option (C) overcounts: it would include $1+i$, but $N(1+i)=1+1=2\neq1$, so $1+i$ is not a unit. Option (D) is false: $N(2+i)=4+1=5\neq1$, so $2+i$ is not a unit. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Using only associativity, $(ba)^2=(ba)(ba)=b(ab)a=b(0)a=0$. Since $R$ has the property that squaring to zero forces the element to be zero, $(ba)^2=0$ immediately gives $ba=0$ — and this argument never used commutativity, only that $ab=0$ was given, proving (A) and ruling out (B) and (C). Option (D) is false: the hypothesis (no nonzero square-zero elements) does not rule out zero divisors in general — for instance $R=\mathbb Z\times\mathbb Z$ has this property (no nonzero nilpotent squares to $0$) yet $(1,0)(0,1)=(0,0)$ is a genuine zero-divisor pair with neither factor zero. Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Define $\varphi:\mathbb Z\to\mathbb Z$ by $\varphi(x)=1-x$; this is a bijection (it is its own inverse). Check $\varphi(a\oplus b)=1-(a+b-1)=2-a-b=(1-a)+(1-b)=\varphi(a)+\varphi(b)$, and $\varphi(a\otimes b)=1-(a+b-ab)=1-a-b+ab=(1-a)(1-b)=\varphi(a)\varphi(b)$. So $\varphi$ is a ring isomorphism from $(\mathbb Z,\oplus,\otimes)$ onto the ordinary ring $(\mathbb Z,+,\times)$, which is an integral domain; hence so is $(\mathbb Z,\oplus,\otimes)$, proving (B) and simultaneously ruling out (C) (an integral domain has no zero divisors). Under this correspondence, the $\oplus$-identity is $\varphi^{-1}(0)=1$, not $0$ (ruling out (D)), and the $\otimes$-identity is $\varphi^{-1}(1)=0$, not $1$ (so (A), which swaps the two, is false). Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

$R$ is a commutative ring with pointwise operations, and the constant function $1(x)=1$ is a multiplicative identity, so (A) is true. Let $f=\chi_{[0,1]}$ and $g=\chi_{(1,2]}$; both are nonzero (e.g. $f(0)=1$, $g(1.5)=1$), but since $[0,1]$ and $(1,2]$ are disjoint, $f(x)g(x)=0$ for every $x\in\mathbb R$, so $fg=0$. Hence $f$ and $g$ are nonzero zero divisors, so (B) is true. If $f(x)\neq 0$ for every $x$, then $g(x)=1/f(x)$ defines $g\in R$ with $fg=1$, so $f$ is a unit; conversely if $fg=1$ then $f(x)g(x)=1$ forces $f(x)\neq 0$ for every $x$. So (C) is true. Since (B) exhibits an explicit pair of nonzero zero divisors, statement (D) is FALSE. Thus the correct option is $\boxed{\mathrm{D}}$.

Detailed Solution

For any $x\in R$, apply idempotency to $x+x$: $(x+x)^2=x+x$. Expanding, $(x+x)(x+x)=4x^2=4x$ (using $x^2=x$), so $4x=2x$, giving $2x=0$ for every $x$ — hence $R$ has characteristic $2$. For commutativity, expand $(x+y)^2=x+y$ for any $x,y\in R$: $x^2+xy+yx+y^2=x+y$, and since $x^2=x$, $y^2=y$, this simplifies to $xy+yx=0$, i.e. $xy=-yx$. Since the characteristic is $2$, $-1=1$, so $xy=yx$ — proving $R$ is commutative. This confirms (C) and rules out (A) (characteristic is $2$, not $3$) and (B) (commutativity is forced, not merely "need not" hold). Option (D) is false: $R=\mathbb Z_2\times\mathbb Z_2$ is a Boolean ring (every element satisfies $x^2=x$) but is not a field, since $(1,0)(0,1)=(0,0)$ shows it has zero divisors. Thus the correct option is $\boxed{\mathrm{C}}$.

Detailed Solution

Since $a$ and $b$ commute, the binomial theorem applies: $(a+b)^{m+n-1}=\sum_{k=0}^{m+n-1}\binom{m+n-1}{k}a^k b^{m+n-1-k}$. For every term, either $k\ge m$ (making $a^k=0$) or $k\le m-1$, which forces the exponent of $b$ to satisfy $m+n-1-k\ge n$ (making $b^{m+n-1-k}=0$). So every term vanishes, giving $(a+b)^{m+n-1}=0$ — proving (A). This argument relies essentially on $ab=ba$ to justify the binomial expansion; without commutativity, sums of nilpotents need not be nilpotent in general noncommutative rings, so (B) is false. Since $a,b$ commute, $(ab)^{\max(m,n)}=a^{\max(m,n)}b^{\max(m,n)}=0$, so $ab$ is always nilpotent here, ruling out (C). There is no general reason $a-b$ must be a unit (e.g. taking $a=b$ gives $a-b=0$, never a unit in a nontrivial ring), ruling out (D). Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

The characteristic of $R$ is the smallest positive integer $k$ such that $k\cdot(1,1)=(0,0)$, i.e. $k\cdot1=0$ in both $\mathbb Z_4$ and $\mathbb Z_6$ simultaneously. This requires $k$ to be a multiple of $\operatorname{char}(\mathbb Z_4)=4$ and of $\operatorname{char}(\mathbb Z_6)=6$, so the smallest such $k$ is $\operatorname{lcm}(4,6)=12$. Thus the correct option is $\boxed{\mathrm{B}}$.

Detailed Solution

$2\mathbb Z$ is a commutative ring, inheriting associativity, commutativity and distributivity from $\mathbb Z$, so (A) is true. For $2\mathbb Z$ to have a multiplicative identity $e=2k$, we would need $(2k)(2m)=2m$ for every $m\in\mathbb Z$, i.e. $4km=2m$ for all $m$, forcing $k=1/2$ — not an integer. So no element of $2\mathbb Z$ can serve as unity, confirming (B), and $1\notin2\mathbb Z$ directly confirms (C). Option (D) is the false statement: subrings do not automatically inherit the unity of their parent ring, and $2\mathbb Z$ is precisely a classic counterexample — a "rng" (ring without unity) sitting inside a ring that does have one. Thus the correct option is $\boxed{\mathrm{D}}$.

Detailed Solution

Expanding directly using the distributive law (without assuming commutativity): $(A+B)^2=(A+B)(A+B)=A\cdot A+A\cdot B+B\cdot A+B\cdot B=A^2+AB+BA+B^2$. Since $M_2(\mathbb R)$ is generally noncommutative ($AB\neq BA$ for most matrices), the cross terms $AB$ and $BA$ cannot be combined into $2AB$ or $2BA$, ruling out (B) and (D); they also do not cancel, ruling out (C). Thus the correct option is $\boxed{\mathrm{A}}$.

Detailed Solution

Checking $x^2\bmod12$ against $x$ for each residue: $0^2=0$, $1^2=1$, $4^2=16\equiv4$, and $9^2=81\equiv9\pmod{12}$ all satisfy $x^2=x$. Direct computation shows none of the remaining residues ($2,3,5,6,7,8,10,11$) satisfy the condition. So the complete set of idempotents is $\{0,1,4,9\}$, exactly $4$ elements. Thus the correct option is $\boxed{\mathrm{C}}$.

Detailed Solution

For $a+b\sqrt2,\ c+d\sqrt2\in S$, the product $(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2$ has integer coefficients, so $S$ is closed under multiplication, ruling out (A). $S$ is a commutative ring with unity $1=1+0\sqrt2$, inheriting these properties from $\mathbb R$. However, $S$ is not a field: for a generic nonzero element such as $2=2+0\sqrt2$, its inverse in $\mathbb R$ is $1/2$, which is not of the form $a+b\sqrt2$ with integer $a,b$, so $2$ has no inverse within $S$, ruling out (C). Since $S$ is a subring of the field $\mathbb R$, it inherits the "no zero divisors" property (a product of two nonzero real numbers is never zero), ruling out (D). Thus the correct option is $\boxed{\mathrm{B}}$.