# 2021

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UPSC Algebra

$$a=b\:cos\:C+c\:cos\:B$$

### 1(a) Let $$m_1, m_2, \cdots, m_k$$ be positive integers and $$d>0$$ the greatest common divisor of $$m_1, m_2, \cdots, m_k$$. Show that there exist integers $$x_1, x_2, \cdots, x_k$$ such that $d=x_1 m_1+x_2 m_2+\cdots+x_k m_k$

##### Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Defining a Set S
Let $s=\left\{au+bv\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}u,v$$s=\left\{au+bv\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}u,v$s={au+bv|u,vs=\{a u+b v \,|\, u, v$s=\left\{au+bv\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}u,v$ are integers and $au+bv>0\right\}$$au+bv>0\right\}$au+bv > 0}a u+b v>0\}$au+bv>0\right\}$
Step 2: Investigating Cases for $a$$a$aa$a$ and $b$$b$bb$b$
If $a>0$$a>0$a > 0a>0$a>0$, then $a=a1+b\left(0\right)>0$$a=a1+b\left(0\right)>0$a=a1+b(0) > 0a=a 1+b(0)>0$a=a1+b\left(0\right)>0$, which implies that $a>0$$a>0$a > 0a>0$a>0$.
If $a<0$$a<0$a < 0a<0$a<0$, then $-a=a\left(-1\right)+b\left(0\right)>0$$-a=a\left(-1\right)+b\left(0\right)>0$-a=a(-1)+b(0) > 0-a=a(-1)+b(0)>0$-a=a\left(-1\right)+b\left(0\right)>0$, which implies that $-a\in S$$-a\in S$-a in S-a \in S$-a\in S$.
Similarly, if $b>0$$b>0$b > 0b>0$b>0$, then $b\in S$$b\in S$b in Sb \in S$b\in S$.
If $b<0$$b<0$b < 0b<0$b<0$, then $-b\in S$$-b\in S$-b in S-b \in S$-b\in S$.
Therefore, $S\ne \varphi$$S\ne \varphi$S!=phiS \neq \phi$S\ne \varphi$ and $S$$S$SS$S$ contains positive integers.
By the well-ordering principle, $S$$S$SS$S$ has a least element, say $d$$d$dd$d$.
Step 3: Proving that $d$$d$dd$d$ is the GCD of $a$$a$aa$a$ and $b$$b$bb$b$
Now we have $d\in S$$d\in S$d in Sd \in S$d\in S$ such that $d=ax+by----\left(1\right)$$d=ax+by----\left(1\right)$d=ax+by—-(1)d=a x+b y—-(1)$d=ax+by----\left(1\right)$ for some integers $x,y$$x,y$x,yx, y$x,y$.
Also, $d>0$$d>0$d > 0d>0$d>0$.
To prove that $d$$d$dd$d$ is the greatest common divisor (GCD) of $a$$a$aa$a$ and $b$$b$bb$b$, consider the following:
Let $a=\mathbf{d}q+r----\left(2\right)$$a=\mathbf{d}q+r----\left(2\right)$a=dq+r—-(2)a=\mathbf{d} q+r—-(2)$a=\mathbf{d}q+r----\left(2\right)$ (where $0⩽r$0⩽r0 <= r < d0 \leqslant r < d$0⩽r) be a division of $a$$a$aa$a$ by $d$$d$dd$d$.
If $r\ne 0$$r\ne 0$r!=0r \neq 0$r\ne 0$, then $r=a-\mathbf{d}q$$r=a-\mathbf{d}q$r=a-dqr=a-\mathbf{d} q$r=a-\mathbf{d}q$.
$\begin{array}{rl}& =a-\left(ax+by\right)q\phantom{\rule{1em}{0ex}}\text{(as}d=ax+by\text{)}\\ & =a\left(1-\lambda q\right)+b\left(-yz\right)>0\phantom{\rule{1em}{0ex}}\text{(since}1-\lambda q,-yq\text{are integers}\right)\\ & r>0,r\in S\text{if}r{:[=a-(ax+by)q quad(as d=ax+by”)”],[=a(1-lambda q)+b(-yz) > 0quad(since 1-lambda q”,”-yq” are integers”)],[r > 0″,”r in S” if “r < d” and “r in S]:}\begin{aligned} &=a-(a x+b y) q \quad \text{(as } d=a x+b y\text{)} \\ &=a(1-\lambda q)+b(-y z)>0 \quad \text{(since } 1-\lambda q, -yq\text{ are integers}) \\ &r>0, r \in S \text{ if } r<d \text{ and } r \in S \end{aligned}
This leads to a contradiction since $d$$d$dd$d$ is the least element of $S$$S$SS$S$.
Therefore, $r=0$$r=0$r=0r=0$r=0$, and we have $a=\mathbf{d}q$$a=\mathbf{d}q$a=dqa=\mathbf{d} q$a=\mathbf{d}q$.
This implies that $\frac{a}{d}=q$$\frac{a}{d}=q$(a)/(d)=q\frac{a}{d}=q$\frac{a}{d}=q$ and $\frac{d}{a}$$\frac{d}{a}$(d)/(a)\frac{d}{a}$\frac{d}{a}$ are integers.
Similarly, $\frac{d}{b}$$\frac{d}{b}$(d)/(b)\frac{d}{b}$\frac{d}{b}$ is also an integer.
Step 4: Proving Uniqueness of GCD
Suppose $\frac{C}{c},\frac{c}{b}⇒\frac{c}{ax+by}⇒\frac{c}{d}$$\frac{C}{c},\frac{c}{b}⇒\frac{c}{ax+by}⇒\frac{c}{d}$(C)/(c),(c)/(b)=>(c)/(ax+by)=>(c)/(d)\frac{C}{c}, \frac{c}{b} \Rightarrow \frac{c}{a x+b y} \Rightarrow \frac{c}{d}$\frac{C}{c},\frac{c}{b}⇒\frac{c}{ax+by}⇒\frac{c}{d}$
Therefore $\frac{\mathbf{d}}{a},\frac{d}{b}$$\frac{\mathbf{d}}{a},\frac{d}{b}$(d)/(a),(d)/(b)\frac{\mathbf{d}}{a}, \frac{d}{b}$\frac{\mathbf{d}}{a},\frac{d}{b}$ also if $\frac{c}{a},\frac{c}{b}⇒\frac{c}{d}$$\frac{c}{a},\frac{c}{b}⇒\frac{c}{d}$(c)/(a),(c)/(b)=>(c)/(d)\frac{c}{a}, \frac{c}{b} \Rightarrow \frac{c}{d}$\frac{c}{a},\frac{c}{b}⇒\frac{c}{d}$
$d$$d$dd$d$ is gcd of $a$$a$aa$a$ and $b$$b$bb$b$
If possible ${d}^{\prime }$${d}^{\prime }$d^(‘)d’${d}^{\prime }$ is also gcd of $a$$a$aa$a$ and $b$$b$bb$b$
Then $\frac{{d}^{\mathrm{\prime }}}{a},\frac{{d}^{\mathrm{\prime }}}{b}⇒\frac{d}{{d}^{\mathrm{\prime }}}\to$$\frac{{d}^{\mathrm{\prime }}}{a},\frac{{d}^{\mathrm{\prime }}}{b}⇒\frac{d}{{d}^{\mathrm{\prime }}}\to$(d^(‘))/(a),(d^(‘))/(b)=>(d)/(d^(‘))rarr\frac{d^{\prime}}{a}, \frac{d^{\prime}}{b} \Rightarrow \frac{d}{d^{\prime}} \rightarrow$\frac{{d}^{\mathrm{\prime }}}{a},\frac{{d}^{\mathrm{\prime }}}{b}⇒\frac{d}{{d}^{\mathrm{\prime }}}\to$ (3)
Similarly, $\frac{\mathbf{d}}{c},\frac{d}{b}⇒\frac{{d}^{\mathrm{\prime }}}{d}\to$$\frac{\mathbf{d}}{c},\frac{d}{b}⇒\frac{{d}^{\mathrm{\prime }}}{d}\to$(d)/(c),(d)/(b)=>(d^(‘))/(d)rarr\frac{\mathbf{d}}{c}, \frac{d}{b} \Rightarrow \frac{d^{\prime}}{d} \rightarrow$\frac{\mathbf{d}}{c},\frac{d}{b}⇒\frac{{d}^{\mathrm{\prime }}}{d}\to$ (4)
This implies $\frac{d}{{d}^{\mathrm{\prime }}}$$\frac{d}{{d}^{\mathrm{\prime }}}$(d)/(d^(‘))\frac{d}{d^{\prime}}$\frac{d}{{d}^{\mathrm{\prime }}}$ (from the divisibility relation) which, in turn, implies $d={d}^{\mathrm{\prime }}$$d={d}^{\mathrm{\prime }}$d=d^(‘)d=d^{\prime}$d={d}^{\mathrm{\prime }}$ (from the uniqueness of GCD).
Step 5: Extending the Argument
The above argument can be extended to more than two integers.
If $d=\mathrm{gcd}\left({m}_{1},{m}_{2},\dots ,{m}_{k}\right)$$d=\mathrm{gcd}\left({m}_{1},{m}_{2},\dots ,{m}_{k}\right)$d=gcd(m_(1),m_(2),dots,m_(k))d=\operatorname{gcd}\left(m_1, m_2, \ldots, m_k\right)$d=\mathrm{gcd}\left({m}_{1},{m}_{2},\dots ,{m}_{k}\right)$, there exist integers ${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{k}$${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{k}$lambda_(1),lambda_(2),dots,lambda _(k)\lambda_1, \lambda_2, \ldots, \lambda_k${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{k}$ such that
$\begin{array}{rl}& d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{m}_{k}{\lambda }_{k}\\ & ⇒d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{\lambda }_{k}{m}_{k}\end{array}$$\begin{array}{r}d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{m}_{k}{\lambda }_{k}\\ ⇒d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{\lambda }_{k}{m}_{k}\end{array}${:[d=lambda_(1)m_(1)+lambda_(2)m_(2)+dots+m_(k)lambda _(k)],[=>d=lambda_(1)m_(1)+lambda_(2)m_(2)+dots+lambda _(k)m_(k)]:}\begin{aligned} & d=\lambda_1 m_1+\lambda_2 m_2+\ldots+m_k \lambda_k \\ & \Rightarrow d=\lambda_1 m_1+\lambda_2 m_2+\ldots+\lambda_k m_k \end{aligned}$\begin{array}{rl}& d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{m}_{k}{\lambda }_{k}\\ & ⇒d={\lambda }_{1}{m}_{1}+{\lambda }_{2}{m}_{2}+\dots +{\lambda }_{k}{m}_{k}\end{array}$
This concludes the proof.
Verified Answer
5/5
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$
$$\sec ^2 \theta=1+\tan ^2 \theta$$

### 2(b) Let $$F$$ be a field and $$f(x) \in F[x]$$, a polynomial of degree $$>0$$ over $$F$$. Show that there is a field $$F^{\prime}$$ and an imbedding $$q: F \rightarrow F^{\prime}$$ such that the polynomial $$f^q \in F^{\prime}[x]$$ has a root in $$F^{\prime}$$, where $$f^q$$ is obtained by replacing each coefficient $$a$$ of $$f$$ by $$q(a)$$.

##### Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Problem Statement
We are given a field $F$$F$FF$F$ and a polynomial $f\left(x\right)\in F\left[x\right]$$f\left(x\right)\in F\left[x\right]$f(x)in F[x]f(x) \in F[x]$f\left(x\right)\in F\left[x\right]$ of degree greater than 0. The goal is to show the existence of a field ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$ and an embedding $q:F\to {F}^{\mathrm{\prime }}$$q:F\to {F}^{\mathrm{\prime }}$q:F rarrF^(‘)q: F \rightarrow F^{\prime}$q:F\to {F}^{\mathrm{\prime }}$ such that the polynomial ${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$ has a root in ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$. Here, ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$ is obtained by replacing each coefficient $a$$a$aa$a$ of $f$$f$ff$f$ by $q\left(a\right)$$q\left(a\right)$q(a)q(a)$q\left(a\right)$.
Step 2: Maximal Ideal and Field Extension
Suppose $f\left(x\right)\in F\left[x\right]$$f\left(x\right)\in F\left[x\right]$f(x)in F[x]f(x) \in F[x]$f\left(x\right)\in F\left[x\right]$ is a polynomial of degree greater than 0 over a field $F$$F$FF$F$. We consider the ideal $M=⟨f\left(x\right)⟩$$M=⟨f\left(x\right)⟩$M=(:f(x):)M = \langle f(x) \rangle$M=⟨f\left(x\right)⟩$ in $F\left[x\right]$$F\left[x\right]$F[x]F[x]$F\left[x\right]$. Since $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ is irreducible, $M$$M$MM$M$ is a maximal ideal in $F\left[x\right]$$F\left[x\right]$F[x]F[x]$F\left[x\right]$.
Step 3: Field Extension
We create a field extension by taking $F$$F$FF$F$ modulo this maximal ideal, denoted as $\frac{F\left[x\right]}{M}$$\frac{F\left[x\right]}{M}$(F[x])/(M)\frac{F[x]}{M}$\frac{F\left[x\right]}{M}$, which is itself a field.
Step 4: Define Embedding $q$$q$qq$q$
Define the embedding $q:F\to \frac{F\left[x\right]}{M}$$q:F\to \frac{F\left[x\right]}{M}$q:F rarr(F[x])/(M)q: F \rightarrow \frac{F[x]}{M}$q:F\to \frac{F\left[x\right]}{M}$ as $q\left(a\right)=a+M$$q\left(a\right)=a+M$q(a)=a+Mq(a) = a + M$q\left(a\right)=a+M$. This embedding $q$$q$qq$q$ is a homomorphism.
Step 5: Ker $q$$q$qq$q$ and Injectiveness
We consider the kernel of $q$$q$qq$q$, i.e., elements $a\in F$$a\in F$a in Fa \in F$a\in F$ such that $q\left(a\right)=0+M$$q\left(a\right)=0+M$q(a)=0+Mq(a) = 0 + M$q\left(a\right)=0+M$. This implies $a+M=M$$a+M=M$a+M=Ma + M = M$a+M=M$, which further implies $a\in M=⟨f\left(x\right)⟩$$a\in M=⟨f\left(x\right)⟩$a in M=(:f(x):)a \in M = \langle f(x) \rangle$a\in M=⟨f\left(x\right)⟩$. Therefore, $a=f\left(x\right)q\left(x\right)$$a=f\left(x\right)q\left(x\right)$a=f(x)q(x)a = f(x)q(x)$a=f\left(x\right)q\left(x\right)$ for some $q\left(x\right)\in F\left[x\right]$$q\left(x\right)\in F\left[x\right]$q(x)in F[x]q(x) \in F[x]$q\left(x\right)\in F\left[x\right]$. Since $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ is irreducible and $a$$a$aa$a$ is a polynomial of degree greater than 0, this relation holds only when $a=0$$a=0$a=0a = 0$a=0$. Consequently, $\text{ker}q=\left\{0\right\}$“ker “q={0}\text{ker } q = \{0\} or $q$$q$qq$q$ is injective.
Step 6: Isomorphism and Subfield
Hence, $F$$F$FF$F$ is isomorphic to $q\left(F\right)$$q\left(F\right)$q(F)q(F)$q\left(F\right)$. We can view ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$ as containing $F$$F$FF$F$ by identifying $a\in F$$a\in F$a in Fa \in F$a\in F$ with $q\left(a\right)$$q\left(a\right)$q(a)q(a)$q\left(a\right)$ and vice versa.
Step 7: Construct $\mathrm{\Psi }$$\mathrm{\Psi }$Psi\Psi$\mathrm{\Psi }$
Define $\mathrm{\Psi }:F\left[x\right]\to \frac{F\left[x\right]}{M}$$\mathrm{\Psi }:F\left[x\right]\to \frac{F\left[x\right]}{M}$Psi:F[x]rarr(F[x])/(M)\Psi: F[x] \rightarrow \frac{F[x]}{M}$\mathrm{\Psi }:F\left[x\right]\to \frac{F\left[x\right]}{M}$ such that $\mathrm{\Psi }\left(f\left(x\right)\right)=f\left(x\right)+M$$\mathrm{\Psi }\left(f\left(x\right)\right)=f\left(x\right)+M$Psi(f(x))=f(x)+M\Psi(f(x)) = f(x) + M$\mathrm{\Psi }\left(f\left(x\right)\right)=f\left(x\right)+M$. $\mathrm{\Psi }$$\mathrm{\Psi }$Psi\Psi$\mathrm{\Psi }$ is a natural homomorphism.
Step 8: Show $\alpha$$\alpha$alpha\alpha$\alpha$ as a Root
Let $\alpha =\mathrm{\Psi }\left(x\right)=x+M$$\alpha =\mathrm{\Psi }\left(x\right)=x+M$alpha=Psi(x)=x+M\alpha = \Psi(x) = x + M$\alpha =\mathrm{\Psi }\left(x\right)=x+M$. We claim that $\alpha$$\alpha$alpha\alpha$\alpha$ is a root of ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$ in $F$$F$FF$F$.
Step 9: Evaluate ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$
Evaluate ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$ as follows:
$\begin{array}{rl}\mathrm{\Psi }\left({f}^{q}\right)& =\mathrm{\Psi }\left({\alpha }_{0}+{\alpha }_{1}x+{\alpha }_{2}{x}^{2}+\dots +{\alpha }_{n}{x}^{n}\right)\\ & =\mathrm{\Psi }\left({\alpha }_{0}\right)+\mathrm{\Psi }\left({\alpha }_{1}\right)\mathrm{\Psi }\left(x\right)+\dots +\mathrm{\Psi }\left({\alpha }_{n}\right)\mathrm{\Psi }\left({x}^{n}\right)\end{array}$$\begin{array}{r}\mathrm{\Psi }\left({f}^{q}\right)=\mathrm{\Psi }\left({\alpha }_{0}+{\alpha }_{1}x+{\alpha }_{2}{x}^{2}+\dots +{\alpha }_{n}{x}^{n}\right)\\ =\mathrm{\Psi }\left({\alpha }_{0}\right)+\mathrm{\Psi }\left({\alpha }_{1}\right)\mathrm{\Psi }\left(x\right)+\dots +\mathrm{\Psi }\left({\alpha }_{n}\right)\mathrm{\Psi }\left({x}^{n}\right)\end{array}${:[Psi(f^(q))=Psi(alpha_(0)+alpha_(1)x+alpha_(2)x^(2)+dots+alpha _(n)x^(n))],[=Psi(alpha_(0))+Psi(alpha_(1))Psi(x)+dots+Psi(alpha _(n))Psi(x^(n))]:}\begin{aligned} \Psi(f^q) &= \Psi(\alpha_0 + \alpha_1 x + \alpha_2 x^2 + \ldots + \alpha_n x^n) \\ &= \Psi(\alpha_0) + \Psi(\alpha_1)\Psi(x) + \ldots + \Psi(\alpha_n)\Psi(x^n) \end{aligned}$\begin{array}{rl}\mathrm{\Psi }\left({f}^{q}\right)& =\mathrm{\Psi }\left({\alpha }_{0}+{\alpha }_{1}x+{\alpha }_{2}{x}^{2}+\dots +{\alpha }_{n}{x}^{n}\right)\\ & =\mathrm{\Psi }\left({\alpha }_{0}\right)+\mathrm{\Psi }\left({\alpha }_{1}\right)\mathrm{\Psi }\left(x\right)+\dots +\mathrm{\Psi }\left({\alpha }_{n}\right)\mathrm{\Psi }\left({x}^{n}\right)\end{array}$
Step 10: Use ${p}^{\left(x\right)}$${p}^{\left(x\right)}$p^((x))p^{(x)}${p}^{\left(x\right)}$
Note that $\mathrm{\Psi }\left({p}^{\left(x\right)}\right)={p}^{\left(x\right)}+M={p}^{\left(x\right)}+⟨{p}^{\left(x\right)}⟩=M=$$\mathrm{\Psi }\left({p}^{\left(x\right)}\right)={p}^{\left(x\right)}+M={p}^{\left(x\right)}+⟨{p}^{\left(x\right)}⟩=M=$Psi(p^((x)))=p^((x))+M=p^((x))+(:p^((x)):)=M=\Psi(p^{(x)}) = p^{(x)} + M = p^{(x)} + \langle p^{(x)} \rangle = M =$\mathrm{\Psi }\left({p}^{\left(x\right)}\right)={p}^{\left(x\right)}+M={p}^{\left(x\right)}+⟨{p}^{\left(x\right)}⟩=M=$ zero of $F$$F$FF$F$.
Step 11: Zero of ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$
Thus, the zero of ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$ is:
$\begin{array}{rl}& \text{Zero of}{F}^{\mathrm{\prime }}=\mathrm{\Psi }\left({\alpha }_{0}\right)+\mathrm{\Psi }\left({\alpha }_{1}\right)\mathrm{\Psi }\left(x\right)+\dots +\mathrm{\Psi }\left({\alpha }_{n}\right)\mathrm{\Psi }\left({x}^{n}\right)\\ & =q\left({a}_{0}\right)+q\left({a}_{1}\right)\mathrm{\Psi }\left(x\right)+\dots +q\left({a}_{n}\right)\mathrm{\Psi }\left({x}^{n}\right)\end{array}${:[“Zero of “F^(‘)=Psi(alpha_(0))+Psi(alpha_(1))Psi(x)+dots+Psi(alpha _(n))Psi(x^(n))],[=q(a_(0))+q(a_(1))Psi(x)+dots+q(a_(n))Psi(x^(n))]:}\begin{aligned} & \text{Zero of } F^{\prime} = \Psi(\alpha_0) + \Psi(\alpha_1)\Psi(x) + \ldots + \Psi(\alpha_n)\Psi(x^n) \\ & = q(a_0) + q(a_1)\Psi(x) + \ldots + q(a_n)\Psi(x^n) \end{aligned}
Step 12: Zero is a Root
Since $F\cong q\left(F\right)$$F\cong q\left(F\right)$F~=q(F)F \cong q(F)$F\cong q\left(F\right)$, the zero is also ${a}_{0}+{a}_{1}\alpha +{a}_{2}{\alpha }^{2}+\dots +{a}_{n}{\alpha }^{n}$${a}_{0}+{a}_{1}\alpha +{a}_{2}{\alpha }^{2}+\dots +{a}_{n}{\alpha }^{n}$a_(0)+a_(1)alpha+a_(2)alpha^(2)+dots+a_(n)alpha ^(n)a_0 + a_1\alpha + a_2\alpha^2 + \ldots + a_n\alpha^n${a}_{0}+{a}_{1}\alpha +{a}_{2}{\alpha }^{2}+\dots +{a}_{n}{\alpha }^{n}$, which is equal to $f\left(a\right)={f}^{q}$$f\left(a\right)={f}^{q}$f(a)=f^(q)f(a) = f^q$f\left(a\right)={f}^{q}$.
Step 13: Conclusion
Hence, $a$$a$aa$a$ is a root of $f\left(a\right)$$f\left(a\right)$f(a)f(a)$f\left(a\right)$ in $F$$F$FF$F$. By replacing each coefficient $a$$a$aa$a$ of $f$$f$ff$f$ by $q\left(a\right)$$q\left(a\right)$q(a)q(a)$q\left(a\right)$, we obtain ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$.
Therefore, there exists a field ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$ and an embedding $q:F\to {F}^{\mathrm{\prime }}$$q:F\to {F}^{\mathrm{\prime }}$q:F rarrF^(‘)q: F \rightarrow F^{\prime}$q:F\to {F}^{\mathrm{\prime }}$ such that the polynomial ${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]${f}^{q}\in {F}^{\mathrm{\prime }}\left[x\right]$ has a root in ${F}^{\mathrm{\prime }}$${F}^{\mathrm{\prime }}$F^(‘)F^{\prime}${F}^{\mathrm{\prime }}$, where ${f}^{q}$${f}^{q}$f^(q)f^q${f}^{q}$ is obtained by replacing each coefficient $a$$a$aa$a$ of $f$$f$ff$f$ by $q\left(a\right)$$q\left(a\right)$q(a)q(a)$q\left(a\right)$.
Verified Answer
5/5
$$Sin^2\left(\theta \:\right)+Cos^2\left(\theta \right)=1$$
$$cos^2\left(\frac{\theta }{2}\right)=\frac{1+cos\:\theta }{2}$$

### 4(a) Show that there are infinitely many subgroups of the additive group $$\mathbb{Q}$$ of rational numbers.

##### Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
The additive group of rational numbers, denoted by $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$, consists of all rational numbers under the operation of addition. To show that there are infinitely many subgroups of $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$, we can construct an infinite family of subgroups.
Construction of Subgroups:
Consider the set $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$, where $n$$n$nn$n$ is any positive integer. This set consists of all rational numbers that can be written in the form $nq$$nq$nqnq$nq$, where $q$$q$qq$q$ is any rational number. Formally, $n\mathbb{Q}=\left\{nq\mid q\in \mathbb{Q}\right\}$$n\mathbb{Q}=\left\{nq\mid q\in \mathbb{Q}\right\}$nQ={nq∣q inQ}n\mathbb{Q} = \{nq \mid q \in \mathbb{Q}\}$n\mathbb{Q}=\left\{nq\mid q\in \mathbb{Q}\right\}$.
Properties:
1. Identity Element: $0$$0$00$0$ is in $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$ because $0=n×0$$0=n×0$0=n xx00 = n \times 0$0=n×0$.
2. Closure under Addition: If $a,b\in n\mathbb{Q}$$a,b\in n\mathbb{Q}$a,b in nQa, b \in n\mathbb{Q}$a,b\in n\mathbb{Q}$, then $a+b=n{q}_{1}+n{q}_{2}=n\left({q}_{1}+{q}_{2}\right)$$a+b=n{q}_{1}+n{q}_{2}=n\left({q}_{1}+{q}_{2}\right)$a+b=nq_(1)+nq_(2)=n(q_(1)+q_(2))a + b = nq_1 + nq_2 = n(q_1 + q_2)$a+b=n{q}_{1}+n{q}_{2}=n\left({q}_{1}+{q}_{2}\right)$ is also in $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$.
3. Closure under Inverse: If $a\in n\mathbb{Q}$$a\in n\mathbb{Q}$a in nQa \in n\mathbb{Q}$a\in n\mathbb{Q}$, then $-a=-nq=n\left(-q\right)$$-a=-nq=n\left(-q\right)$-a=-nq=n(-q)-a = -nq = n(-q)$-a=-nq=n\left(-q\right)$ is also in $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$.
Therefore, $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$ is a subgroup of $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ for each positive integer $n$$n$nn$n$.
Infinitude of Subgroups:
Since $n$$n$nn$n$ can be any positive integer, and there are infinitely many positive integers, we have constructed an infinite family of subgroups $n\mathbb{Q}$$n\mathbb{Q}$nQn\mathbb{Q}$n\mathbb{Q}$ of $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$.
Conclusion:
Hence, we have shown that there are infinitely many subgroups of the additive group $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ of rational numbers.
Verified Answer
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