# 2023

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UPSC Algebra

$$\sec ^2 \theta=1+\tan ^2 \theta$$

### 1(a) Let $$G$$ be a group of order 10 and $$G'$$ be a group of order 6. Examine whether there exists a homomorphism of $$G$$ onto $$G'$$.

##### Expert Answer
untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
let’s examine whether there exists a homomorphism of $G$$G$GG$G$ onto ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ where $G$$G$GG$G$ is a group of order 10 and ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ is a group of order 6.
Introduction
In group theory, a homomorphism is a map $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$ between two groups $G$$G$GG$G$ and ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ that preserves the group operation. Specifically, for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$, $f\left(ab\right)=f\left(a\right)f\left(b\right)$$f\left(ab\right)=f\left(a\right)f\left(b\right)$f(ab)=f(a)f(b)f(ab) = f(a)f(b)$f\left(ab\right)=f\left(a\right)f\left(b\right)$. A homomorphism is said to be “onto” if it is surjective, meaning that every element in ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ is the image of some element in $G$$G$GG$G$.
Step 1: Examine the Orders of the Groups
The first thing to consider is the orders of the groups $G$$G$GG$G$ and ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$, which are 10 and 6, respectively. The order of a group is simply the number of elements it contains.
Let $|G|=10$$|G|=10$|G|=10|G| = 10$|G|=10$ and $|{G}^{\prime }|=6$$|{G}^{\prime }|=6$|G^(‘)|=6|G’| = 6$|{G}^{\prime }|=6$.
Step 2: Check the Conditions for Existence of Homomorphism
For a homomorphism $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$ to be onto, the order of ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ must divide the order of $G$$G$GG$G$.
The formula to check this condition is:
$\text{Condition for onto homomorphism}=\left(|{G}^{\prime }|\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}|G|\right)$$\text{Condition for onto homomorphism}=\left(|{G}^{\prime }|\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}|G|\right)$“Condition for onto homomorphism”=(|G^(‘)|”divides”|G|)\text{Condition for onto homomorphism} = (|G’| \, \text{divides} \, |G|)$\text{Condition for onto homomorphism}=\left(|{G}^{\prime }|\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}|G|\right)$
Let’s substitute the values:
$6\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}10$$6\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}10$6″divides”106 \, \text{divides} \, 10$6\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}10$
After Calculating, we find that 6 does not divide 10.
Summary
Since the order of ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ does not divide the order of $G$$G$GG$G$, there does not exist an onto homomorphism $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$.
Verified Answer
5/5
$$cos\:2\theta =cos^2\theta -sin^2\theta$$
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$

### 1(b) Determine all the Sylow $$p$$-subgroups of the symmetric group $$S_3$$.

##### Expert Answer
untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
In group theory, a Sylow $p$$p$pp$p$-subgroup of a group $G$$G$GG$G$ is a maximal $p$$p$pp$p$-subgroup, denoted $P$$P$PP$P$, such that the order of $P$$P$PP$P$ is ${p}^{n}$${p}^{n}$p^(n)p^n${p}^{n}$ for some $n$$n$nn$n$, and ${p}^{n}$${p}^{n}$p^(n)p^n${p}^{n}$ divides the order of $G$$G$GG$G$. Here, $p$$p$pp$p$ is a prime number. We are interested in finding all the Sylow $p$$p$pp$p$-subgroups of the symmetric group ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$.
Step 1: Determine the Order of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$
The symmetric group ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ consists of all the permutations of 3 elements. The order of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ is the number of such permutations, which is $3!$$3!$3!3!$3!$.
$|{S}_{3}|=3!=3×2×1$$|{S}_{3}|=3!=3×2×1$|S_(3)|=3!=3xx2xx1|S_3| = 3! = 3 \times 2 \times 1$|{S}_{3}|=3!=3×2×1$
After Calculating, we find $|{S}_{3}|=6$$|{S}_{3}|=6$|S_(3)|=6|S_3| = 6$|{S}_{3}|=6$.
Step 2: Prime Factorization of the Order of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$
To find the Sylow $p$$p$pp$p$-subgroups, we first need to find the prime factors of the order of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$.
$\text{Prime factors of}6=2×3$“Prime factors of “6=2xx3\text{Prime factors of } 6 = 2 \times 3
After Calculating, we find that the prime factors are $2$$2$22$2$ and $3$$3$33$3$.
Step 3: Determine Sylow $p$$p$pp$p$-subgroups for Each Prime Factor
For $p=2$$p=2$p=2p = 2$p=2$
The Sylow $2$$2$22$2$-subgroups will have order ${2}^{1}=2$${2}^{1}=2$2^(1)=22^1 = 2${2}^{1}=2$.
Let’s substitute the values:
$\text{Sylow 2-subgroups of}{S}_{3}=\left\{\text{subgroups of order}2\right\}$“Sylow 2-subgroups of “S_(3)={“subgroups of order “2}\text{Sylow 2-subgroups of } S_3 = \{ \text{subgroups of order } 2 \}
After Calculating, we find that the Sylow $2$$2$22$2$-subgroups are $\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}$$\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}${(1,2),(2,3),(1,3)}\{(1,2), (2,3), (1,3)\}$\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}$.
For $p=3$$p=3$p=3p = 3$p=3$
The Sylow $3$$3$33$3$-subgroups will have order ${3}^{1}=3$${3}^{1}=3$3^(1)=33^1 = 3${3}^{1}=3$.
Let’s substitute the values:
$\text{Sylow 3-subgroups of}{S}_{3}=\left\{\text{subgroups of order}3\right\}$“Sylow 3-subgroups of “S_(3)={“subgroups of order “3}\text{Sylow 3-subgroups of } S_3 = \{ \text{subgroups of order } 3 \}
After Calculating, we find that the Sylow $3$$3$33$3$-subgroups are $\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}$$\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}${(1,2,3),(1,3,2)}\{(1,2,3), (1,3,2)\}$\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}$.
Summary
The Sylow $p$$p$pp$p$-subgroups of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$ are as follows:
• Sylow $2$$2$22$2$-subgroups: $\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}$$\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}${(1,2),(2,3),(1,3)}\{(1,2), (2,3), (1,3)\}$\left\{\left(1,2\right),\left(2,3\right),\left(1,3\right)\right\}$
• Sylow $3$$3$33$3$-subgroups: $\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}$$\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}${(1,2,3),(1,3,2)}\{(1,2,3), (1,3,2)\}$\left\{\left(1,2,3\right),\left(1,3,2\right)\right\}$
These are all the Sylow $p$$p$pp$p$-subgroups of ${S}_{3}$${S}_{3}$S_(3)S_3${S}_{3}$.
Verified Answer
5/5
$$Sin^2\left(\theta \:\right)+Cos^2\left(\theta \right)=1$$
$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

### 2(a) Prove that a non-commutative group of order $$2p$$, where $$p$$ is an odd prime, must have a subgroup of order $$p$$.

##### Expert Answer
untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
We are tasked with proving that a non-commutative group $G$$G$GG$G$ of order $2p$$2p$2p2p$2p$, where $p$$p$pp$p$ is an odd prime, must have a subgroup of order $p$$p$pp$p$.
Step 1: Establish the Order of the Group $G$$G$GG$G$
Given that $G$$G$GG$G$ is a non-commutative group of order $2p$$2p$2p2p$2p$, where $p$$p$pp$p$ is an odd prime, we can write:
$|G|=2p$$|G|=2p$|G|=2p|G| = 2p$|G|=2p$
Step 2: Use Sylow’s Theorem
By Sylow’s First Theorem, for any prime $p$$p$pp$p$ that divides the order of $G$$G$GG$G$, there exists a Sylow $p$$p$pp$p$-subgroup $P$$P$PP$P$ of $G$$G$GG$G$ such that $|P|={p}^{n}$$|P|={p}^{n}$|P|=p^(n)|P| = p^n$|P|={p}^{n}$ for some $n$$n$nn$n$, and ${p}^{n}$${p}^{n}$p^(n)p^n${p}^{n}$ divides $|G|$$|G|$|G||G|$|G|$.
The formula to find the number ${n}_{p}$${n}_{p}$n_(p)n_p${n}_{p}$ of Sylow $p$$p$pp$p$-subgroups is:
${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{|G|}{{p}^{n}}$${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{|G|}{{p}^{n}}$n_(p)-=1quadmodp quad”and”quadn_(p)”divides”(|G|)/(p^(n))n_p \equiv 1 \mod p \quad \text{and} \quad n_p \, \text{divides} \, \frac{|G|}{p^n}${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{|G|}{{p}^{n}}$
Let’s substitute the values for $p$$p$pp$p$ and $|G|$$|G|$|G||G|$|G|$:
${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{2p}{p}$${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{2p}{p}$n_(p)-=1quadmodp quad”and”quadn_(p)”divides”(2p)/(p)n_p \equiv 1 \mod p \quad \text{and} \quad n_p \, \text{divides} \, \frac{2p}{p}${n}_{p}\equiv 1\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}\frac{2p}{p}$
After Calculating, we find that ${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$n_(p)-=1modpn_p \equiv 1 \mod p${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$ and ${n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}2$${n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}2$n_(p)”divides”2n_p \, \text{divides} \, 2${n}_{p}\phantom{\rule{thinmathspace}{0ex}}\text{divides}\phantom{\rule{thinmathspace}{0ex}}2$.
Step 3: Examine the Possibilities for ${n}_{p}$${n}_{p}$n_(p)n_p${n}_{p}$
Since ${n}_{p}$${n}_{p}$n_(p)n_p${n}_{p}$ divides 2, ${n}_{p}$${n}_{p}$n_(p)n_p${n}_{p}$ can only be 1 or 2. Also, ${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$n_(p)-=1modpn_p \equiv 1 \mod p${n}_{p}\equiv 1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$.
1. If ${n}_{p}=1$${n}_{p}=1$n_(p)=1n_p = 1${n}_{p}=1$, then there is a unique Sylow $p$$p$pp$p$-subgroup of order $p$$p$pp$p$, which must be normal. This contradicts the assumption that $G$$G$GG$G$ is non-commutative.
2. If ${n}_{p}=2$${n}_{p}=2$n_(p)=2n_p = 2${n}_{p}=2$, then there are exactly two distinct Sylow $p$$p$pp$p$-subgroups of $G$$G$GG$G$ of order $p$$p$pp$p$.
Summary
From the above discussion, we can conclude that if $G$$G$GG$G$ is a non-commutative group of order $2p$$2p$2p2p$2p$, where $p$$p$pp$p$ is an odd prime, then $G$$G$GG$G$ must have a subgroup of order $p$$p$pp$p$. Specifically, there will be exactly two such subgroups.
Thus, the statement is proven.
Verified Answer
5/5
$$2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$
$$tan\:\theta =\frac{sin\:\theta }{cos\:\theta }$$

### 3(a) Prove that $$x^2 + 1$$ is an irreducible polynomial in $$Z_3[x]$$. Further show that the quotient ring $$\frac{Z_3[x]}{\langle x^2 + 1 \rangle}$$ is a field of 9 elements.

##### Expert Answer
untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Introduction
We are tasked with proving two things:
1. The polynomial ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is irreducible in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$.
2. The quotient ring $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ is a field with 9 elements.
Step 1: Prove ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is Irreducible in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$
To prove that ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is irreducible in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$, we need to show that it cannot be factored into two non-constant polynomials in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$.
The polynomial ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is of degree 2. If it is reducible, then it can be factored into two polynomials of degree 1. In ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$, the possible factors would be of the form $\left(x-a\right)\left(x-b\right)$$\left(x-a\right)\left(x-b\right)$(x-a)(x-b)(x – a)(x – b)$\left(x-a\right)\left(x-b\right)$ where $a,b\in {\mathbb{Z}}_{3}$$a,b\in {\mathbb{Z}}_{3}$a,b inZ_(3)a, b \in \mathbb{Z}_3$a,b\in {\mathbb{Z}}_{3}$.
Let’s substitute the values:
${x}^{2}+1=\left(x-a\right)\left(x-b\right)$${x}^{2}+1=\left(x-a\right)\left(x-b\right)$x^(2)+1=(x-a)(x-b)x^2 + 1 = (x – a)(x – b)${x}^{2}+1=\left(x-a\right)\left(x-b\right)$
After Calculating, we find that for $a,b\in \left\{0,1,2\right\}$$a,b\in \left\{0,1,2\right\}$a,b in{0,1,2}a, b \in \{0, 1, 2\}$a,b\in \left\{0,1,2\right\}$, no such factorization exists. Therefore, ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is irreducible in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$.
Step 2: Prove $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ is a Field of 9 Elements
To show that $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ is a field, we need to prove two things:
1. It is a commutative ring with unity.
2. Every non-zero element has a multiplicative inverse.
Commutative Ring with Unity
The ring ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$ is a commutative ring with unity, and ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is an irreducible polynomial. Therefore, $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ is also a commutative ring with unity.
Existence of Multiplicative Inverse
Since ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is irreducible, the ideal $⟨{x}^{2}+1⟩$$⟨{x}^{2}+1⟩$(:x^(2)+1:)\langle x^2 + 1 \rangle$⟨{x}^{2}+1⟩$ is maximal. In a commutative ring, the quotient by a maximal ideal is a field.
Number of Elements
The elements of $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ can be represented as $a+bx$$a+bx$a+bxa + bx$a+bx$ where $a,b\in {\mathbb{Z}}_{3}$$a,b\in {\mathbb{Z}}_{3}$a,b inZ_(3)a, b \in \mathbb{Z}_3$a,b\in {\mathbb{Z}}_{3}$. There are 3 choices for $a$$a$aa$a$ and 3 choices for $b$$b$bb$b$, making a total of $3×3=9$$3×3=9$3xx3=93 \times 3 = 9$3×3=9$ elements.
After Calculating, we find that $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ indeed has 9 elements.
Summary
1. The polynomial ${x}^{2}+1$${x}^{2}+1$x^(2)+1x^2 + 1${x}^{2}+1$ is irreducible in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$ as it cannot be factored into two non-constant polynomials in ${\mathbb{Z}}_{3}\left[x\right]$${\mathbb{Z}}_{3}\left[x\right]$Z_(3)[x]\mathbb{Z}_3[x]${\mathbb{Z}}_{3}\left[x\right]$.
2. The quotient ring $\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$(Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$\frac{{\mathbb{Z}}_{3}\left[x\right]}{⟨{x}^{2}+1⟩}$ is a field with 9 elements, satisfying the properties of a commutative ring with unity and containing multiplicative inverses for all non-zero elements.
Thus, both statements are proven.
Verified Answer
5/5
$$sin\left(\theta +\phi \right)=sin\:\theta \:cos\:\phi +cos\:\theta \:sin\:\phi$$