UPSC Optional (Maths) Paper-02 Algebra Solution

# 2022

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UPSC Algebra

$$\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}$$

### 1(a) Show that the multiplicative group $$G=\{1,-1, i,-i\}$$, where $$i=\sqrt{-1}$$, is isomorphic to the group $$G'=\{0,1,2,3\}, +_4$$.

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Introduction
In group theory, two groups $G$$G$GG$G$ and ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$ are said to be isomorphic if there exists a bijective function $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$ that preserves the group operation. In this problem, we are asked to show that the multiplicative group $G=\left\{1,-1,i,-i\right\}$$G=\left\{1,-1,i,-i\right\}$G={1,-1,i,-i}G = \{1, -1, i, -i\}$G=\left\{1,-1,i,-i\right\}$, where $i=\sqrt{-1}$$i=\sqrt{-1}$i=sqrt(-1)i = \sqrt{-1}$i=\sqrt{-1}$, is isomorphic to the additive group ${G}^{\prime }=\left\{0,1,2,3\right\}$${G}^{\prime }=\left\{0,1,2,3\right\}$G^(‘)={0,1,2,3}G’ = \{0, 1, 2, 3\}${G}^{\prime }=\left\{0,1,2,3\right\}$ under addition modulo 4, denoted as ${+}_{4}$${+}_{4}$+_(4)+_4${+}_{4}$.
To show that $G$$G$GG$G$ is isomorphic to ${G}^{\prime }$${G}^{\prime }$G^(‘)G’${G}^{\prime }$, we need to:
1. Define a bijective function $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$.
2. Show that $f$$f$ff$f$ preserves the group operation, i.e., $f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$ for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$.
Work/Calculations
Step 1: Define a Bijective Function $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$
Let’s define $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$ as follows:
$f\left(1\right)=0,\phantom{\rule{1em}{0ex}}f\left(-1\right)=2,\phantom{\rule{1em}{0ex}}f\left(i\right)=1,\phantom{\rule{1em}{0ex}}f\left(-i\right)=3$$f\left(1\right)=0,\phantom{\rule{1em}{0ex}}f\left(-1\right)=2,\phantom{\rule{1em}{0ex}}f\left(i\right)=1,\phantom{\rule{1em}{0ex}}f\left(-i\right)=3$f(1)=0,quad f(-1)=2,quad f(i)=1,quad f(-i)=3f(1) = 0, \quad f(-1) = 2, \quad f(i) = 1, \quad f(-i) = 3$f\left(1\right)=0,\phantom{\rule{1em}{0ex}}f\left(-1\right)=2,\phantom{\rule{1em}{0ex}}f\left(i\right)=1,\phantom{\rule{1em}{0ex}}f\left(-i\right)=3$
We can see that $f$$f$ff$f$ is a bijection because it is both injective (one-to-one) and surjective (onto).
Step 2: Show that $f$$f$ff$f$ Preserves the Group Operation
To show that $f$$f$ff$f$ preserves the group operation, we need to verify that $f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$ for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$.
Let’s consider all possible combinations of $a$$a$aa$a$ and $b$$b$bb$b$ in $G$$G$GG$G$ and calculate $f\left(a\cdot b\right)$$f\left(a\cdot b\right)$f(a*b)f(a \cdot b)$f\left(a\cdot b\right)$ and $f\left(a\right){+}_{4}f\left(b\right)$$f\left(a\right){+}_{4}f\left(b\right)$f(a)+_(4)f(b)f(a) +_4 f(b)$f\left(a\right){+}_{4}f\left(b\right)$.
1. $a=1,b=1$$a=1,b=1$a=1,b=1a = 1, b = 1$a=1,b=1$
• $f\left(a\cdot b\right)=f\left(1\cdot 1\right)=f\left(1\right)$$f\left(a\cdot b\right)=f\left(1\cdot 1\right)=f\left(1\right)$f(a*b)=f(1*1)=f(1)f(a \cdot b) = f(1 \cdot 1) = f(1)$f\left(a\cdot b\right)=f\left(1\cdot 1\right)=f\left(1\right)$
• $f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(1\right)=0{+}_{4}0$$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(1\right)=0{+}_{4}0$f(a)+_(4)f(b)=f(1)+_(4)f(1)=0+_(4)0f(a) +_4 f(b) = f(1) +_4 f(1) = 0 +_4 0$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(1\right)=0{+}_{4}0$
After Calculating we get $f\left(a\cdot b\right)=0$$f\left(a\cdot b\right)=0$f(a*b)=0f(a \cdot b) = 0$f\left(a\cdot b\right)=0$ and $f\left(a\right){+}_{4}f\left(b\right)=0$$f\left(a\right){+}_{4}f\left(b\right)=0$f(a)+_(4)f(b)=0f(a) +_4 f(b) = 0$f\left(a\right){+}_{4}f\left(b\right)=0$.
1. $a=1,b=-1$$a=1,b=-1$a=1,b=-1a = 1, b = -1$a=1,b=-1$
• $f\left(a\cdot b\right)=f\left(1\cdot -1\right)=f\left(-1\right)$$f\left(a\cdot b\right)=f\left(1\cdot -1\right)=f\left(-1\right)$f(a*b)=f(1*-1)=f(-1)f(a \cdot b) = f(1 \cdot -1) = f(-1)$f\left(a\cdot b\right)=f\left(1\cdot -1\right)=f\left(-1\right)$
• $f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-1\right)=0{+}_{4}2$$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-1\right)=0{+}_{4}2$f(a)+_(4)f(b)=f(1)+_(4)f(-1)=0+_(4)2f(a) +_4 f(b) = f(1) +_4 f(-1) = 0 +_4 2$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-1\right)=0{+}_{4}2$
After Calculating we get $f\left(a\cdot b\right)=2$$f\left(a\cdot b\right)=2$f(a*b)=2f(a \cdot b) = 2$f\left(a\cdot b\right)=2$ and $f\left(a\right){+}_{4}f\left(b\right)=2$$f\left(a\right){+}_{4}f\left(b\right)=2$f(a)+_(4)f(b)=2f(a) +_4 f(b) = 2$f\left(a\right){+}_{4}f\left(b\right)=2$.
1. $a=1,b=i$$a=1,b=i$a=1,b=ia = 1, b = i$a=1,b=i$
• $f\left(a\cdot b\right)=f\left(1\cdot i\right)=f\left(i\right)$$f\left(a\cdot b\right)=f\left(1\cdot i\right)=f\left(i\right)$f(a*b)=f(1*i)=f(i)f(a \cdot b) = f(1 \cdot i) = f(i)$f\left(a\cdot b\right)=f\left(1\cdot i\right)=f\left(i\right)$
• $f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(i\right)=0{+}_{4}1$$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(i\right)=0{+}_{4}1$f(a)+_(4)f(b)=f(1)+_(4)f(i)=0+_(4)1f(a) +_4 f(b) = f(1) +_4 f(i) = 0 +_4 1$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(i\right)=0{+}_{4}1$
After Calculating we get $f\left(a\cdot b\right)=1$$f\left(a\cdot b\right)=1$f(a*b)=1f(a \cdot b) = 1$f\left(a\cdot b\right)=1$ and $f\left(a\right){+}_{4}f\left(b\right)=1$$f\left(a\right){+}_{4}f\left(b\right)=1$f(a)+_(4)f(b)=1f(a) +_4 f(b) = 1$f\left(a\right){+}_{4}f\left(b\right)=1$.
1. $a=1,b=-i$$a=1,b=-i$a=1,b=-ia = 1, b = -i$a=1,b=-i$
• $f\left(a\cdot b\right)=f\left(1\cdot -i\right)=f\left(-i\right)$$f\left(a\cdot b\right)=f\left(1\cdot -i\right)=f\left(-i\right)$f(a*b)=f(1*-i)=f(-i)f(a \cdot b) = f(1 \cdot -i) = f(-i)$f\left(a\cdot b\right)=f\left(1\cdot -i\right)=f\left(-i\right)$
• $f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-i\right)=0{+}_{4}3$$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-i\right)=0{+}_{4}3$f(a)+_(4)f(b)=f(1)+_(4)f(-i)=0+_(4)3f(a) +_4 f(b) = f(1) +_4 f(-i) = 0 +_4 3$f\left(a\right){+}_{4}f\left(b\right)=f\left(1\right){+}_{4}f\left(-i\right)=0{+}_{4}3$
After Calculating we get $f\left(a\cdot b\right)=3$$f\left(a\cdot b\right)=3$f(a*b)=3f(a \cdot b) = 3$f\left(a\cdot b\right)=3$ and $f\left(a\right){+}_{4}f\left(b\right)=3$$f\left(a\right){+}_{4}f\left(b\right)=3$f(a)+_(4)f(b)=3f(a) +_4 f(b) = 3$f\left(a\right){+}_{4}f\left(b\right)=3$.
We can continue this for all combinations of $a$$a$aa$a$ and $b$$b$bb$b$ in $G$$G$GG$G$. For brevity, we can summarize that for all combinations, $f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$f(a*b)=f(a)+_(4)f(b)f(a \cdot b) = f(a) +_4 f(b)$f\left(a\cdot b\right)=f\left(a\right){+}_{4}f\left(b\right)$.
Conclusion
We have defined a bijective function $f:G\to {G}^{\prime }$$f:G\to {G}^{\prime }$f:G rarrG^(‘)f: G \to G’$f:G\to {G}^{\prime }$ and verified that it preserves the group operation. Therefore, the multiplicative group $G=\left\{1,-1,i,-i\right\}$$G=\left\{1,-1,i,-i\right\}$G={1,-1,i,-i}G = \{1, -1, i, -i\}$G=\left\{1,-1,i,-i\right\}$ is isomorphic to the additive group ${G}^{\prime }=\left\{0,1,2,3\right\}$${G}^{\prime }=\left\{0,1,2,3\right\}$G^(‘)={0,1,2,3}G’ = \{0, 1, 2, 3\}${G}^{\prime }=\left\{0,1,2,3\right\}$ under addition modulo 4.
5/5
$$cos\:2\theta =1-2\:sin^2\theta$$
$$b=c\:cos\:A+a\:cos\:C$$

### 2(b) Prove that every homomorphic image of a group $$G$$ is isomorphic to some quotient group of $$G$$.

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Introduction: In this problem, we are asked to prove that every homomorphic image of a group $G$$G$GG$G$ is isomorphic to some quotient group of $G$$G$GG$G$.
Definition:
1. Group homomorphism: A group homomorphism is a map $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \rightarrow H$\varphi :G\to H$ between two groups, $G$$G$GG$G$ and $H$$H$HH$H$, such that for all elements $a,b\in G,\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)$$a,b\in G,\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)$a,b in G,phi(ab)=phi(a)phi(b)a, b \in G, \phi(a b)=\phi(a) \phi(b)$a,b\in G,\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)$.
2. Homomorphic image: The homomorphic image of a group $G$$G$GG$G$ under a group homomorphism $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \rightarrow H$\varphi :G\to H$ is the set $\varphi \left(G\right)=\left\{\varphi \left(g\right)\mid g\in G\right\}$$\varphi \left(G\right)=\left\{\varphi \left(g\right)\mid g\in G\right\}$phi(G)={phi(g)∣g in G}\phi(G)=\{\phi(g) \mid g \in G\}$\varphi \left(G\right)=\left\{\varphi \left(g\right)\mid g\in G\right\}$, which is a subgroup of $H$$H$HH$H$.
3. Quotient group: Let $G$$G$GG$G$ be a group, and let $N$$N$NN$N$ be a normal subgroup of $G$$G$GG$G$. The quotient group of $G$$G$GG$G$ by $N$$N$NN$N$, denoted as $G/N$$G/N$G//NG / N$G/N$, is the set of cosets of $N$$N$NN$N$ in $G$$G$GG$G$ with the group operation defined as $\left(aN\right)\left(bN\right)=\left(ab\right)N$$\left(aN\right)\left(bN\right)=\left(ab\right)N$(aN)(bN)=(ab)N(a N)(b N)=(a b) N$\left(aN\right)\left(bN\right)=\left(ab\right)N$ for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$.
Theorem: Every homomorphic image of a group $G$$G$GG$G$ is isomorphic to some quotient group of $G$$G$GG$G$.
Proof: Let $\varphi :G\to H$$\varphi :G\to H$phi:G rarr H\phi: G \rightarrow H$\varphi :G\to H$ be a group homomorphism, and let $N=\mathrm{ker}\varphi$$N=\mathrm{ker}\varphi$N=ker phiN=\operatorname{ker} \phi$N=\mathrm{ker}\varphi$ be the kernel of $\varphi$$\varphi$phi\phi$\varphi$, where $\mathrm{ker}\varphi =\left\{g\in G\mid \varphi \left(g\right)={e}_{H}\right\}$$\mathrm{ker}\varphi =\left\{g\in G\mid \varphi \left(g\right)={e}_{H}\right\}$ker phi={g in G∣phi(g)=e_(H)}\operatorname{ker} \phi=\left\{g \in G \mid \phi(g)=e_H\right\}$\mathrm{ker}\varphi =\left\{g\in G\mid \varphi \left(g\right)={e}_{H}\right\}$ and ${e}_{H}$${e}_{H}$e_(H)e_H${e}_{H}$ is the identity element of the group $H$$H$HH$H$. We claim that the homomorphic image $\varphi \left(G\right)$$\varphi \left(G\right)$phi(G)\phi(G)$\varphi \left(G\right)$ is isomorphic to the quotient group $G/N$$G/N$G//NG / N$G/N$.
First, note that the kernel $N$$N$NN$N$ is a normal subgroup of $G$$G$GG$G$. Now, we define a map $\psi :G/N\to$$\psi :G/N\to$psi:G//N rarr\psi: G / N \rightarrow$\psi :G/N\to$ $\varphi \left(G\right)$$\varphi \left(G\right)$phi(G)\phi(G)$\varphi \left(G\right)$ by $\psi \left(aN\right)=\varphi \left(a\right)$$\psi \left(aN\right)=\varphi \left(a\right)$psi(aN)=phi(a)\psi(a N)=\phi(a)$\psi \left(aN\right)=\varphi \left(a\right)$ for all $a\in G$$a\in G$a in Ga \in G$a\in G$. We need to show that $\psi$$\psi$psi\psi$\psi$ is a well-defined isomorphism.
1. Well-defined: If $aN=bN$$aN=bN$aN=bNa N=b N$aN=bN$, then ${b}^{-1}a\in N$${b}^{-1}a\in N$b^(-1)a in Nb^{-1} a \in N${b}^{-1}a\in N$, so $\varphi \left({b}^{-1}a\right)={e}_{H}$$\varphi \left({b}^{-1}a\right)={e}_{H}$phi(b^(-1)a)=e_(H)\phi\left(b^{-1} a\right)=e_H$\varphi \left({b}^{-1}a\right)={e}_{H}$, which implies $\varphi \left(b{\right)}^{-1}\varphi \left(a\right)={e}_{H}$$\varphi \left(b{\right)}^{-1}\varphi \left(a\right)={e}_{H}$phi(b)^(-1)phi(a)=e_(H)\phi(b)^{-1} \phi(a)=e_H$\varphi \left(b{\right)}^{-1}\varphi \left(a\right)={e}_{H}$, or equivalently, $\varphi \left(b\right)=\varphi \left(a\right)$$\varphi \left(b\right)=\varphi \left(a\right)$phi(b)=phi(a)\phi(b)=\phi(a)$\varphi \left(b\right)=\varphi \left(a\right)$. Thus, $\psi \left(aN\right)=\psi \left(bN\right)$$\psi \left(aN\right)=\psi \left(bN\right)$psi(aN)=psi(bN)\psi(a N)=\psi(b N)$\psi \left(aN\right)=\psi \left(bN\right)$, and $\psi$$\psi$psi\psi$\psi$ is well-defined.
2. Homomorphism: Let $aN,bN\in G/N$$aN,bN\in G/N$aN,bN in G//Na N, b N \in G / N$aN,bN\in G/N$. Then,
$\psi \left(\left(aN\right)\left(bN\right)\right)=\psi \left(abN\right)=\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)=\psi \left(aN\right)\psi \left(bN\right)$$\psi \left(\left(aN\right)\left(bN\right)\right)=\psi \left(abN\right)=\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)=\psi \left(aN\right)\psi \left(bN\right)$psi((aN)(bN))=psi(abN)=phi(ab)=phi(a)phi(b)=psi(aN)psi(bN)\psi((a N)(b N))=\psi(a b N)=\phi(a b)=\phi(a) \phi(b)=\psi(a N) \psi(b N)$\psi \left(\left(aN\right)\left(bN\right)\right)=\psi \left(abN\right)=\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)=\psi \left(aN\right)\psi \left(bN\right)$
which shows that $\psi$$\psi$psi\psi$\psi$ is a group homomorphism.
1. Injective: Suppose $\psi \left(aN\right)=\psi \left(bN\right)$$\psi \left(aN\right)=\psi \left(bN\right)$psi(aN)=psi(bN)\psi(a N)=\psi(b N)$\psi \left(aN\right)=\psi \left(bN\right)$, then $\varphi \left(a\right)=\varphi \left(b\right)$$\varphi \left(a\right)=\varphi \left(b\right)$phi(a)=phi(b)\phi(a)=\phi(b)$\varphi \left(a\right)=\varphi \left(b\right)$, which implies $\varphi \left({b}^{-1}a\right)={e}_{H}$$\varphi \left({b}^{-1}a\right)={e}_{H}$phi(b^(-1)a)=e_(H)\phi\left(b^{-1} a\right)=e_H$\varphi \left({b}^{-1}a\right)={e}_{H}$ . Thus, ${b}^{-1}a\in N$${b}^{-1}a\in N$b^(-1)a in Nb^{-1} a \in N${b}^{-1}a\in N$, which means $aN=bN$$aN=bN$aN=bNa N=b N$aN=bN$. Therefore, $\psi$$\psi$psi\psi$\psi$ is injective.
2. Surjective : For any element $h\in \varphi \left(G\right)$$h\in \varphi \left(G\right)$h in phi(G)h \in \phi(G)$h\in \varphi \left(G\right)$, there exists an element $a\in G$$a\in G$a in Ga \in G$a\in G$ such that $\varphi \left(a\right)=h$$\varphi \left(a\right)=h$phi(a)=h\phi(a)=h$\varphi \left(a\right)=h$. Then, $\psi \left(aN\right)=\varphi \left(a\right)=h$$\psi \left(aN\right)=\varphi \left(a\right)=h$psi(aN)=phi(a)=h\psi(a N)=\phi(a)=h$\psi \left(aN\right)=\varphi \left(a\right)=h$, so $\psi$$\psi$psi\psi$\psi$ is surjective.
Since $\psi$$\psi$psi\psi$\psi$ is a well-defined group homomorphism that is both injective and surjective, it is an isomorphism. Therefore, the homomorphic image $\varphi \left(G\right)$$\varphi \left(G\right)$phi(G)\phi(G)$\varphi \left(G\right)$ is isomorphic to the quotient group $G/N$$G/N$G//NG / N$G/N$
Conclusion: Every homomorphic image of a group $G$$G$GG$G$ is isomorphic to some quotient group of $G$$G$GG$G$.
5/5
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$

### 4(a) Let $$R$$ be a field of real numbers and $$S$$, the field of all those polynomials $$f(x) \in R[x]$$ such that $$f(0)=0=f(1)$$. Prove that $$S$$ is an ideal of $$R[x]$$. Is the residue class ring $$R[x] / S$$ an integral domain? Give justification for your answer.

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Introduction: In this problem, we are asked to prove that the set $S$$S$SS$S$ of polynomials in $R\left[x\right]$$R\left[x\right]$R[x]R[x]$R\left[x\right]$ satisfying $f\left(0\right)=0=f\left(1\right)$$f\left(0\right)=0=f\left(1\right)$f(0)=0=f(1)f(0)=0=f(1)$f\left(0\right)=0=f\left(1\right)$ is an ideal of $R\left[x\right]$$R\left[x\right]$R[x]R[x]$R\left[x\right]$, and to determine whether the residue class ring $R\left[x\right]/S$$R\left[x\right]/S$R[x]//SR[x] / S$R\left[x\right]/S$ is an integral domain. We will show that $S$$S$SS$S$ satisfies the properties of an ideal and then analyze the structure of the quotient ring.
Assumptions: We assume that $R$$R$RR$R$ is a field of real numbers.
Definitions:
1. An ideal of a ring is a non-empty subset that is closed under addition and multiplication by any element of the ring.
2. A residue class ring (or quotient ring) is the set of equivalence classes of a ring modulo an ideal.
3. An integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero.
Method/Approach: We will first prove that $S$$S$SS$S$ is an ideal by showing that it is closed under addition and multiplication by elements of $R\left[x\right]$$R\left[x\right]$R[x]R[x]$R\left[x\right]$. Then, we will examine the structure of the residue class ring $R\left[x\right]/S$$R\left[x\right]/S$R[x]//SR[x] / S$R\left[x\right]/S$ to determine whether it is an integral domain.
Work/Calculations:
1. Prove that $S$$S$SS$S$ is an ideal:
a) Closure under addition: Let $f\left(x\right),g\left(x\right)\in S$$f\left(x\right),g\left(x\right)\in S$f(x),g(x)in Sf(x), g(x) \in S$f\left(x\right),g\left(x\right)\in S$. Then, $f\left(0\right)=0=g\left(0\right)$$f\left(0\right)=0=g\left(0\right)$f(0)=0=g(0)f(0)=0=g(0)$f\left(0\right)=0=g\left(0\right)$ and $f\left(1\right)=0=$$f\left(1\right)=0=$f(1)=0=f(1)=0=$f\left(1\right)=0=$ $g\left(1\right)$$g\left(1\right)$g(1)g(1)$g\left(1\right)$. Consider the polynomial $h\left(x\right)=f\left(x\right)+g\left(x\right)$$h\left(x\right)=f\left(x\right)+g\left(x\right)$h(x)=f(x)+g(x)h(x)=f(x)+g(x)$h\left(x\right)=f\left(x\right)+g\left(x\right)$. We have:
$\begin{array}{rl}& h\left(0\right)=f\left(0\right)+g\left(0\right)=0+0=0\\ & h\left(1\right)=f\left(1\right)+g\left(1\right)=0+0=0\end{array}$$\begin{array}{r}h\left(0\right)=f\left(0\right)+g\left(0\right)=0+0=0\\ h\left(1\right)=f\left(1\right)+g\left(1\right)=0+0=0\end{array}${:[h(0)=f(0)+g(0)=0+0=0],[h(1)=f(1)+g(1)=0+0=0]:}\begin{aligned} & h(0)=f(0)+g(0)=0+0=0 \\ & h(1)=f(1)+g(1)=0+0=0 \end{aligned}$\begin{array}{rl}& h\left(0\right)=f\left(0\right)+g\left(0\right)=0+0=0\\ & h\left(1\right)=f\left(1\right)+g\left(1\right)=0+0=0\end{array}$
Thus, $h\left(x\right)\in S$$h\left(x\right)\in S$h(x)in Sh(x) \in S$h\left(x\right)\in S$, and $S$$S$SS$S$ is closed under addition.
b) Closure under multiplication: Let $f\left(x\right)\in S$$f\left(x\right)\in S$f(x)in Sf(x) \in S$f\left(x\right)\in S$ and $g\left(x\right)\in R\left[x\right]$$g\left(x\right)\in R\left[x\right]$g(x)in R[x]g(x) \in R[x]$g\left(x\right)\in R\left[x\right]$. Consider the polynomial $h\left(x\right)=f\left(x\right)g\left(x\right)$$h\left(x\right)=f\left(x\right)g\left(x\right)$h(x)=f(x)g(x)h(x)=f(x) g(x)$h\left(x\right)=f\left(x\right)g\left(x\right)$. We have:
$\begin{array}{rl}& h\left(0\right)=f\left(0\right)g\left(0\right)=0\cdot g\left(0\right)=0\\ & h\left(1\right)=f\left(1\right)g\left(1\right)=0\cdot g\left(1\right)=0\end{array}$$\begin{array}{r}h\left(0\right)=f\left(0\right)g\left(0\right)=0\cdot g\left(0\right)=0\\ h\left(1\right)=f\left(1\right)g\left(1\right)=0\cdot g\left(1\right)=0\end{array}${:[h(0)=f(0)g(0)=0*g(0)=0],[h(1)=f(1)g(1)=0*g(1)=0]:}\begin{aligned} & h(0)=f(0) g(0)=0 \cdot g(0)=0 \\ & h(1)=f(1) g(1)=0 \cdot g(1)=0 \end{aligned}$\begin{array}{rl}& h\left(0\right)=f\left(0\right)g\left(0\right)=0\cdot g\left(0\right)=0\\ & h\left(1\right)=f\left(1\right)g\left(1\right)=0\cdot g\left(1\right)=0\end{array}$
Thus, $h\left(x\right)\in S$$h\left(x\right)\in S$h(x)in Sh(x) \in S$h\left(x\right)\in S$, and $S$$S$SS$S$ is closed under multiplication by elements of $R\left[x\right]$$R\left[x\right]$R[x]R[x]$R\left[x\right]$.
Since $S$$S$SS$S$ is closed under addition and multiplication by elements of $R\left[x\right],S$$R\left[x\right],S$R[x],SR[x], S$R\left[x\right],S$ is an ideal of $R\left[x\right]$$R\left[x\right]$R[x]R[x]$R\left[x\right]$.
1. Determine whether $R\left[x\right]/S$$R\left[x\right]/S$R[x]//SR[x] / S$R\left[x\right]/S$ is an integral domain:
Recall that an integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero. To check whether $R\left[x\right]/S$$R\left[x\right]/S$R[x]//SR[x] / S$R\left[x\right]/S$ is an integral domain, we will see if there exist any nonzero elements in $R\left[x\right]/S$$R\left[x\right]/S$R[x]//SR[x] / S$R\left[x\right]/S$ whose product is zero.
Consider the polynomials $f\left(x\right)=x\left(x-1\right)$$f\left(x\right)=x\left(x-1\right)$f(x)=x(x-1)f(x)=x(x-1)$f\left(x\right)=x\left(x-1\right)$ and $g\left(x\right)=x$$g\left(x\right)=x$g(x)=xg(x)=x$g\left(x\right)=x$. Both $f\left(x\right),g\left(x\right)\in S$$f\left(x\right),g\left(x\right)\in S$f(x),g(x)in Sf(x), g(x) \in S$f\left(x\right),g\left(x\right)\in S$. Their product is $h\left(x\right)=f\left(x\right)g\left(x\right)={x}^{2}\left(x-1\right)$$h\left(x\right)=f\left(x\right)g\left(x\right)={x}^{2}\left(x-1\right)$h(x)=f(x)g(x)=x^(2)(x-1)h(x)=f(x) g(x)=x^2(x-1)$h\left(x\right)=f\left(x\right)g\left(x\right)={x}^{2}\left(x-1\right)$, which also belongs to $S$$S$SS$S$. However, the equivalence classes $\left[f\left(x\right)\right]$$\left[f\left(x\right)\right]$[f(x)][f(x)]