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PPSC Lecturer Math Ring Theory Solved MCQs & Quiz
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PPSC Lecturer Math Ring Theory Solved MCQs & Quiz
Ring Theory forms the backbone of advanced Abstract Algebra questions in competitive exams such as PPSC, FPSC, and SPSC Lecturer Mathematics papers. Understanding structural properties of algebraic rings, ideals, quotient structures, and polynomial rings is vital for candidate success.
| Concept / Structure | Mathematical Definition | Key Exam Property / Result |
|---|---|---|
| Subring Test | $S \subseteq R$ is a subring if $\forall a,b \in S$, $a - b \in S$ and $ab \in S$. | Non-empty subset closed under subtraction and multiplication. |
| Two-Sided Ideal | $I \subseteq R$ such that $(I,+)$ is a subgroup and $\forall r \in R, a \in I \implies ra \in I \text{ and } ar \in I$. | Kernel of any ring homomorphism is always a two-sided ideal. |
| Quotient Ring | $R/I = \{r + I \mid r \in R\}$ with operations $(a+I)+(b+I) = (a+b)+I$ and $(a+I)(b+I) = ab+I$. | $R/I$ is a field iff $I$ is a maximal ideal (for commutative ring $R$ with unity). |
| Prime Ideal | $P \subsetneq R$ where $ab \in P \implies a \in P \text{ or } b \in P$. | $R/P$ is an integral domain iff $P$ is a prime ideal. |
| Maximal Ideal | $M \subsetneq R$ such that if $M \subseteq I \subseteq R$, then $I = M$ or $I = R$. | Every maximal ideal in a commutative ring with $1$ is a prime ideal. |
| Characteristic | Smallest positive integer $n$ such that $n \cdot 1 = 0$; if no such $n$ exists, $\text{char}(R) = 0$. | Characteristic of an integral domain is either $0$ or a prime $p$. |
Question 1 (Part i): Prove whether the subset $I = \{f(x) \in \mathbb{Z}[x] \mid f(0) \in 2\mathbb{Z}\}$ forms a two-sided ideal of the polynomial ring $\mathbb{Z}[x]$.
Solution:
To prove that $I$ is an ideal of $\mathbb{Z}[x]$, we apply the Subring/Ideal Test:
1. Non-emptiness & Subgroup Test: Let $f(x), g(x) \in I$. Then $f(0) = 2k$ and $g(0) = 2m$ for some $k, m \in \mathbb{Z}$.
Consider $(f - g)(x) = f(x) - g(x)$.
Evaluating at $0$: $(f - g)(0) = f(0) - g(0) = 2k - 2m = 2(k - m) \in 2\mathbb{Z}$.
Hence, $(I, +)$ is an additive subgroup of $\mathbb{Z}[x]$.
2. Ideal Absorption Property: Let $h(x) \in \mathbb{Z}[x]$ be an arbitrary polynomial, and $f(x) \in I$.
Evaluate the product $(h \cdot f)(x) = h(x)f(x)$ at $x = 0$:
$(h \cdot f)(0) = h(0) \cdot f(0) = h(0) \cdot (2k) = 2(h(0) \cdot k)$.
Since $h(0) \in \mathbb{Z}$ and $k \in \mathbb{Z}$, the product $h(0)k \in \mathbb{Z}$. Thus $(h \cdot f)(0) \in 2\mathbb{Z}$.
Because $\mathbb{Z}[x]$ is a commutative ring, $h(x)f(x) = f(x)h(x) \in I$.
Therefore, $I$ satisfies all requirements of a two-sided ideal generated by $\langle 2, x \rangle$.
Answer: $$I = \langle 2, x \rangle \text{ is a valid two-sided ideal of } \mathbb{Z}[x]$$
Question 1 (Part ii): Let $S = \left\{ \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} : a,b,c \in \mathbb{R} \right\}$ be a subset of the matrix ring $M_2(\mathbb{R})$. Determine whether $S$ is a subring, a left ideal, or a right ideal of $M_2(\mathbb{R})$.
Solution:
1. Subring Check:
Take $A = \begin{pmatrix} a_1 & 0 \\ b_1 & c_1 \end{pmatrix}$ and $B = \begin{pmatrix} a_2 & 0 \\ b_2 & c_2 \end{pmatrix} \in S$.
$A - B = \begin{pmatrix} a_1-a_2 & 0 \\ b_1-b_2 & c_1-c_2 \end{pmatrix} \in S$.
$AB = \begin{pmatrix} a_1 a_2 & 0 \\ b_1 a_2 + c_1 b_2 & c_1 c_2 \end{pmatrix} \in S$.
Thus, $S$ is a subring of $M_2(\mathbb{R})$.
2. Ideal Check:
Let $R = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \in M_2(\mathbb{R})$ and $A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \in S$.
Product $RA = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \notin S$ (since top-right entry $1 \neq 0$).
Product $AR = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \notin S$.
Hence, $S$ is neither a left ideal nor a right ideal.
Answer: $$S \text{ is a subring, but NEITHER a left nor a right ideal of } M_2(\mathbb{R})$$
Question 2 (Part i): Consider the evaluation homomorphism $\phi: \mathbb{Z}[x] \to \mathbb{R}$ defined by $\phi(f(x)) = f(\sqrt{2})$. Find the kernel $\ker(\phi)$ and determine the quotient structure $\mathbb{Z}[x]/\ker(\phi)$.
Solution:
1. Kernel Determination:
$\ker(\phi) = \{f(x) \in \mathbb{Z}[x] \mid f(\sqrt{2}) = 0\}$.
The minimal polynomial of $\sqrt{2}$ over $\mathbb{Z}$ is $x^2 - 2$.
By polynomial division algorithm in $\mathbb{Q}[x]$ and Gauss's Lemma for $\mathbb{Z}[x]$, any polynomial $f(x) \in \mathbb{Z}[x]$ having $f(\sqrt{2}) = 0$ must be a multiple of $x^2 - 2$.
Thus, $\ker(\phi) = \langle x^2 - 2 \rangle = (x^2 - 2)\mathbb{Z}[x]$.
2. Image and Isomorphism:
The image $\text{Im}(\phi) = \{f(\sqrt{2}) \mid f(x) \in \mathbb{Z}[x]\} = \{a + b\sqrt{2} \mid a, b \in \mathbb{Z}\} = \mathbb{Z}[\sqrt{2}]$.
Applying the First Isomorphism Theorem for Rings:
$\mathbb{Z}[x]/\ker(\phi) \cong \text{Im}(\phi) \implies \mathbb{Z}[x]/\langle x^2 - 2 \rangle \cong \mathbb{Z}[\sqrt{2}]$.
Answer: $$\ker(\phi) = \langle x^2 - 2 \rangle \quad \text{and} \quad \mathbb{Z}[x]/\langle x^2 - 2 \rangle \cong \mathbb{Z}[\sqrt{2}]$$
Question 2 (Part ii): Calculate the exact number of ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}_{20}$.
Solution:
A ring homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}_{20}$ is completely determined by the image of $1$, say $\phi(1) = e \in \mathbb{Z}_{20}$.
Since $1 \cdot 1 = 1$, we must have $\phi(1 \cdot 1) = \phi(1) \cdot \phi(1) \implies e^2 \equiv e \pmod{20}$.
Thus, $e$ must be an idempotent element of $\mathbb{Z}_{20}$.
We solve $e^2 - e \equiv 0 \pmod{20} \implies e(e - 1) \equiv 0 \pmod{20}$.
Since $20 = 4 \times 5$:
- $e \equiv 0 \text{ or } 1 \pmod 4$
- $e \equiv 0 \text{ or } 1 \pmod 5$
By Chinese Remainder Theorem, there are $2^2 = 4$ idempotent solutions in $\mathbb{Z}_{20}$:
1. $e \equiv 0 \pmod 4, e \equiv 0 \pmod 5 \implies e = 0$