# 2016

Estimated reading: 48 minutes 74 views

UPSC Algebra

$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$

### 1(a) Let $$\mathbb{K}$$ be a field and $$\mathbb{K}[X]$$ be the ring of polynomials over $$\mathbb{R}$$ in a single variable $$X$$. For a polynomial $$f \in \mathbb{K}[X]$$, let $$(f)$$ denote the ideal in $$\mathbb{K}[X]$$ generated by $$f$$. Show that $$(f)$$ is a maximal ideal in $$\mathbb{K}[X]$$ if and only if $$f$$ is an irreducible polynomial over $$\mathbb{K}$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal in $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$ if and only if $f$$f$ff$f$ is an irreducible polynomial over $\mathbb{K}$$\mathbb{K}$K\mathbb{K}$\mathbb{K}$, we need to prove two directions:
1. If $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal, then $f$$f$ff$f$ is irreducible.
2. If $f$$f$ff$f$ is irreducible, then $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal.
Direction 1: $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal $⇒$$⇒$=>\Rightarrow$⇒$ $f$$f$ff$f$ is irreducible
Step 1: Assume $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal
Let’s assume that $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal in $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$.
Step 2: Show that $f$$f$ff$f$ is irreducible
Suppose, for the sake of contradiction, that $f$$f$ff$f$ is not irreducible. Then $f=gh$$f=gh$f=ghf = gh$f=gh$ for some $g,h\in \mathbb{K}\left[X\right]$$g,h\in \mathbb{K}\left[X\right]$g,h inK[X]g, h \in \mathbb{K}[X]$g,h\in \mathbb{K}\left[X\right]$ with $\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$$\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$deg(g) < deg(f)\deg(g) < \deg(f)$\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$ and $\mathrm{deg}\left(h\right)<\mathrm{deg}\left(f\right)$$\mathrm{deg}\left(h\right)<\mathrm{deg}\left(f\right)$deg(h) < deg(f)\deg(h) < \deg(f)$\mathrm{deg}\left(h\right)<\mathrm{deg}\left(f\right)$.
Let $I=\left(g\right)$$I=\left(g\right)$I=(g)I = (g)$I=\left(g\right)$. Then $f\in I$$f\in I$f in If \in I$f\in I$ because $f=gh$$f=gh$f=ghf = gh$f=gh$, and $I$$I$II$I$ contains $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$. However, $I\ne \mathbb{K}\left[X\right]$$I\ne \mathbb{K}\left[X\right]$I!=K[X]I \neq \mathbb{K}[X]$I\ne \mathbb{K}\left[X\right]$ because $\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$$\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$deg(g) < deg(f)\deg(g) < \deg(f)$\mathrm{deg}\left(g\right)<\mathrm{deg}\left(f\right)$.
This contradicts the assumption that $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal, because we’ve found an ideal $I$$I$II$I$ strictly between $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ and $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$.
Therefore, $f$$f$ff$f$ must be irreducible.
Direction 2: $f$$f$ff$f$ is irreducible $⇒$$⇒$=>\Rightarrow$⇒$ $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal
Step 1: Assume $f$$f$ff$f$ is irreducible
Let’s assume that $f$$f$ff$f$ is an irreducible polynomial in $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$.
Step 2: Show that $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal
Let $I$$I$II$I$ be an ideal in $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$ such that $\left(f\right)\subseteq I\subseteq \mathbb{K}\left[X\right]$$\left(f\right)\subseteq I\subseteq \mathbb{K}\left[X\right]$(f)sube I subeK[X](f) \subseteq I \subseteq \mathbb{K}[X]$\left(f\right)\subseteq I\subseteq \mathbb{K}\left[X\right]$.
Since $f$$f$ff$f$ is in $I$$I$II$I$, there exists a polynomial $g\in I$$g\in I$g in Ig \in I$g\in I$ such that $f=g$$f=g$f=gf = g$f=g$.
Since $f$$f$ff$f$ is irreducible, $g$$g$gg$g$ must either be a unit or associate of $f$$f$ff$f$. If $g$$g$gg$g$ is a unit, then $I=\mathbb{K}\left[X\right]$$I=\mathbb{K}\left[X\right]$I=K[X]I = \mathbb{K}[X]$I=\mathbb{K}\left[X\right]$. If $g$$g$gg$g$ is an associate of $f$$f$ff$f$, then $I=\left(f\right)$$I=\left(f\right)$I=(f)I = (f)$I=\left(f\right)$.
Therefore, $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal.
Summary
We have shown that if $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal, then $f$$f$ff$f$ is irreducible, and conversely, if $f$$f$ff$f$ is irreducible, then $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal. Hence, $\left(f\right)$$\left(f\right)$(f)(f)$\left(f\right)$ is a maximal ideal in $\mathbb{K}\left[X\right]$$\mathbb{K}\left[X\right]$K[X]\mathbb{K}[X]$\mathbb{K}\left[X\right]$ if and only if $f$$f$ff$f$ is an irreducible polynomial over $\mathbb{K}$$\mathbb{K}$K\mathbb{K}$\mathbb{K}$.
5/5
$$b=c\:cos\:A+a\:cos\:C$$
$$\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}$$

### 2(b) Let $$p$$ be a prime number and $$\mathbb{Z}_{p}$$ denote the additive group of integers modulo $$p$$. Show that every non-zero element of $$\mathbb{Z}_{p}$$ generates $$\mathbb{Z}_{p}$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that every non-zero element of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$ generates ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$, we need to prove that for any non-zero element $a$$a$aa$a$ in ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$, the cyclic subgroup generated by $a$$a$aa$a$, denoted $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$, is equal to ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$.
Step 1: Define the Cyclic Subgroup $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$
The cyclic subgroup $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ generated by $a$$a$aa$a$ is defined as:
$⟨a⟩=\left\{{a}^{n}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p:n\in \mathbb{Z}\right\}$$⟨a⟩=\left\{{a}^{n}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p:n\in \mathbb{Z}\right\}$(:a:)={a^(n)quadmodp:n inZ}\langle a \rangle = \{ a^n \mod p : n \in \mathbb{Z} \}$⟨a⟩=\left\{{a}^{n}\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p:n\in \mathbb{Z}\right\}$
Step 2: Show that $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ is a Subset of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$
First, let’s establish that $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ is a subset of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$. This is trivially true since $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ consists of elements that are congruent modulo $p$$p$pp$p$, and therefore, each element in $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ is an element in ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$.
Step 3: Show that $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ Contains $p$$p$pp$p$ Distinct Elements
Let’s assume that $a$$a$aa$a$ is a non-zero element in ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$. We want to show that $a,2a,3a,\dots ,\left(p-1\right)a$$a,2a,3a,\dots ,\left(p-1\right)a$a,2a,3a,dots,(p-1)aa, 2a, 3a, \ldots, (p-1)a$a,2a,3a,\dots ,\left(p-1\right)a$ are all distinct modulo $p$$p$pp$p$.
Suppose, for the sake of contradiction, that $ia\equiv ja\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$$ia\equiv ja\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$ia-=jamodpia \equiv ja \mod p$ia\equiv ja\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$ for some $i\ne j$$i\ne j$i!=ji \neq j$i\ne j$ where $1\le i,j\le p-1$$1\le i,j\le p-1$1 <= i,j <= p-11 \leq i,j \leq p-1$1\le i,j\le p-1$.
This implies $\left(i-j\right)a\equiv 0\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$$\left(i-j\right)a\equiv 0\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$(i-j)a-=0modp(i-j)a \equiv 0 \mod p$\left(i-j\right)a\equiv 0\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$, or $p$$p$pp$p$ divides $\left(i-j\right)a$$\left(i-j\right)a$(i-j)a(i-j)a$\left(i-j\right)a$.
Since $p$$p$pp$p$ is a prime and $p\nmid \left(i-j\right)$$p\nmid \left(i-j\right)$p∤(i-j)p \nmid (i-j)$p\nmid \left(i-j\right)$ (because $i\ne j$$i\ne j$i!=ji \neq j$i\ne j$), it must be the case that $p$$p$pp$p$ divides $a$$a$aa$a$, which contradicts the assumption that $a$$a$aa$a$ is a non-zero element of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$.
Therefore, $a,2a,3a,\dots ,\left(p-1\right)a$$a,2a,3a,\dots ,\left(p-1\right)a$a,2a,3a,dots,(p-1)aa, 2a, 3a, \ldots, (p-1)a$a,2a,3a,\dots ,\left(p-1\right)a$ are all distinct modulo $p$$p$pp$p$, and $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ contains $p$$p$pp$p$ distinct elements.
Step 4: Conclude that $⟨a⟩={\mathbb{Z}}_{p}$$⟨a⟩={\mathbb{Z}}_{p}$(:a:)=Z_(p)\langle a \rangle = \mathbb{Z}_{p}$⟨a⟩={\mathbb{Z}}_{p}$
Since $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ is a subset of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$ and contains $p$$p$pp$p$ distinct elements, it must be the case that $⟨a⟩={\mathbb{Z}}_{p}$$⟨a⟩={\mathbb{Z}}_{p}$(:a:)=Z_(p)\langle a \rangle = \mathbb{Z}_{p}$⟨a⟩={\mathbb{Z}}_{p}$.
Summary
We have shown that for any non-zero element $a$$a$aa$a$ in ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$, the cyclic subgroup $⟨a⟩$$⟨a⟩$(:a:)\langle a \rangle$⟨a⟩$ generated by $a$$a$aa$a$ is equal to ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$. Therefore, every non-zero element of ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$ generates ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_{p}${\mathbb{Z}}_{p}$.
5/5
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$
$$a=b\:cos\:C+c\:cos\:B$$

### 3(a) Let $$K$$ be an extension of a field $$F$$; Prove that the elements of $$K$$, which are algebraic over $$F$$, form a subfield of $$K$$. Further, if $$F \subset K \subset L$$ are fields, $$L$$ is algebraic over $$K$$ and $$K$$ is algebraic over $$F$$, then prove that $$L$$ is algebraic over $$F$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Part 1: Prove that the elements of $K$$K$KK$K$ that are algebraic over $F$$F$FF$F$ form a subfield of $K$$K$KK$K$
To show that the set of elements in $K$$K$KK$K$ that are algebraic over $F$$F$FF$F$ form a subfield, we need to prove that this set is closed under addition, multiplication, and contains multiplicative inverses for non-zero elements. Let’s denote this set as $A$$A$AA$A$.
Let $a,b\in A$$a,b\in A$a,b in Aa, b \in A$a,b\in A$. Then $a$$a$aa$a$ and $b$$b$bb$b$ are roots of some polynomials $f\left(x\right),g\left(x\right)\in F\left[x\right]$$f\left(x\right),g\left(x\right)\in F\left[x\right]$f(x),g(x)in F[x]f(x), g(x) \in F[x]$f\left(x\right),g\left(x\right)\in F\left[x\right]$ respectively. We need to show that $a+b$$a+b$a+ba + b$a+b$ is also algebraic over $F$$F$FF$F$.
Consider the polynomials $f\left(x-b\right)$$f\left(x-b\right)$f(x-b)f(x – b)$f\left(x-b\right)$ and $g\left(x-a\right)$$g\left(x-a\right)$g(x-a)g(x – a)$g\left(x-a\right)$. Their product $h\left(x\right)=f\left(x-b\right)g\left(x-a\right)$$h\left(x\right)=f\left(x-b\right)g\left(x-a\right)$h(x)=f(x-b)g(x-a)h(x) = f(x – b)g(x – a)$h\left(x\right)=f\left(x-b\right)g\left(x-a\right)$ is a polynomial in $F\left[x\right]$$F\left[x\right]$F[x]F[x]$F\left[x\right]$. The root of $h\left(x\right)$$h\left(x\right)$h(x)h(x)$h\left(x\right)$ is $x=a+b$$x=a+b$x=a+bx = a + b$x=a+b$, which means $a+b$$a+b$a+ba + b$a+b$ is algebraic over $F$$F$FF$F$.
Step 2: Closure under Multiplication
Let $a,b\in A$$a,b\in A$a,b in Aa, b \in A$a,b\in A$. Then $a$$a$aa$a$ and $b$$b$bb$b$ are roots of some polynomials $f\left(x\right),g\left(x\right)\in F\left[x\right]$$f\left(x\right),g\left(x\right)\in F\left[x\right]$f(x),g(x)in F[x]f(x), g(x) \in F[x]$f\left(x\right),g\left(x\right)\in F\left[x\right]$ respectively. We need to show that $ab$$ab$abab$ab$ is also algebraic over $F$$F$FF$F$.
Consider the polynomial $h\left(x\right)=f\left(x/b\right)g\left(x/a\right)$$h\left(x\right)=f\left(x/b\right)g\left(x/a\right)$h(x)=f(x//b)g(x//a)h(x) = f(x/b)g(x/a)$h\left(x\right)=f\left(x/b\right)g\left(x/a\right)$. This is a polynomial in $F\left[x\right]$$F\left[x\right]$F[x]F[x]$F\left[x\right]$. The root of $h\left(x\right)$$h\left(x\right)$h(x)h(x)$h\left(x\right)$ is $x=ab$$x=ab$x=abx = ab$x=ab$, which means $ab$$ab$abab$ab$ is algebraic over $F$$F$FF$F$.
Step 3: Existence of Multiplicative Inverses for Non-zero Elements
Let $a\in A$$a\in A$a in Aa \in A$a\in A$ and $a\ne 0$$a\ne 0$a!=0a \neq 0$a\ne 0$. Then $a$$a$aa$a$ is a root of some polynomial $f\left(x\right)\in F\left[x\right]$$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]$. We need to show that ${a}^{-1}$${a}^{-1}$a^(-1)a^{-1}${a}^{-1}$ is also algebraic over $F$$F$FF$F$.
Consider the polynomial $g\left(x\right)={x}^{n}f\left(1/x\right)$$g\left(x\right)={x}^{n}f\left(1/x\right)$g(x)=x^(n)f(1//x)g(x) = x^n f(1/x)$g\left(x\right)={x}^{n}f\left(1/x\right)$, where $n$$n$nn$n$ is the degree of $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$. This is a polynomial in $F\left[x\right]$$F\left[x\right]$F[x]F[x]$F\left[x\right]$. The root of $g\left(x\right)$$g\left(x\right)$g(x)g(x)$g\left(x\right)$ is $x={a}^{-1}$$x={a}^{-1}$x=a^(-1)x = a^{-1}$x={a}^{-1}$, which means ${a}^{-1}$${a}^{-1}$a^(-1)a^{-1}${a}^{-1}$ is algebraic over $F$$F$FF$F$.
Step 4: $0$$0$00$0$ and $1$$1$11$1$ are in $A$$A$AA$A$
The zero polynomial and the constant polynomial $x-1$$x-1$x-1x – 1$x-1$ show that $0$$0$00$0$ and $1$$1$11$1$ are algebraic over $F$$F$FF$F$.
Summary for Part 1
We have shown that $A$$A$AA$A$ is closed under addition, multiplication, and contains multiplicative inverses for non-zero elements. Also, $0$$0$00$0$ and $1$$1$11$1$ are in $A$$A$AA$A$. Therefore, $A$$A$AA$A$ is a subfield of $K$$K$KK$K$.
Part 2: If $F\subset K\subset L$$F\subset K\subset L$F sub K sub LF \subset K \subset L$F\subset K\subset L$ and $L$$L$LL$L$ is algebraic over $K$$K$KK$K$ and $K$$K$KK$K$ is algebraic over $F$$F$FF$F$, then $L$$L$LL$L$ is algebraic over $F$$F$FF$F$
Step 1: Take an element $a\in L$$a\in L$a in La \in L$a\in L$
Let $a$$a$aa$a$ be an element in $L$$L$LL$L$.
Step 2: Show that $a$$a$aa$a$ is algebraic over $K$$K$KK$K$
Since $L$$L$LL$L$ is algebraic over $K$$K$KK$K$, $a$$a$aa$a$ is a root of some polynomial $f\left(x\right)\in K\left[x\right]$$f\left(x\right)\in K\left[x\right]$f(x)in K[x]f(x) \in K[x]$f\left(x\right)\in K\left[x\right]$.
Step 3: Show that $a$$a$aa$a$ is algebraic over $F$$F$FF$F$
The coefficients of $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ are in $K$$K$KK$K$ and are algebraic over $F$$F$FF$F$. Let these coefficients be ${k}_{1},{k}_{2},\dots ,{k}_{n}$${k}_{1},{k}_{2},\dots ,{k}_{n}$k_(1),k_(2),dots,k_(n)k_1, k_2, \ldots, k_n${k}_{1},{k}_{2},\dots ,{k}_{n}$. Each ${k}_{i}$${k}_{i}$k_(i)k_i${k}_{i}$ is a root of some polynomial ${g}_{i}\left(x\right)\in F\left[x\right]$${g}_{i}\left(x\right)\in F\left[x\right]$g_(i)(x)in F[x]g_i(x) \in F[x]${g}_{i}\left(x\right)\in F\left[x\right]$.
Consider the polynomial $h\left(x\right)=f\left(x,{k}_{1},{k}_{2},\dots ,{k}_{n}\right)$$h\left(x\right)=f\left(x,{k}_{1},{k}_{2},\dots ,{k}_{n}\right)$h(x)=f(x,k_(1),k_(2),dots,k_(n))h(x) = f(x, k_1, k_2, \ldots, k_n)$h\left(x\right)=f\left(x,{k}_{1},{k}_{2},\dots ,{k}_{n}\right)$ where we treat ${k}_{1},{k}_{2},\dots ,{k}_{n}$${k}_{1},{k}_{2},\dots ,{k}_{n}$k_(1),k_(2),dots,k_(n)k_1, k_2, \ldots, k_n${k}_{1},{k}_{2},\dots ,{k}_{n}$ as variables. The polynomial $h\left(x\right)$$h\left(x\right)$h(x)h(x)$h\left(x\right)$ has $a$$a$aa$a$ as a root and its coefficients are in $F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$$F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$F(k_(1),k_(2),dots,k_(n))F(k_1, k_2, \ldots, k_n)$F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$.
Since each ${k}_{i}$${k}_{i}$k_(i)k_i${k}_{i}$ is algebraic over $F$$F$FF$F$, $F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$$F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$F(k_(1),k_(2),dots,k_(n))F(k_1, k_2, \ldots, k_n)$F\left({k}_{1},{k}_{2},\dots ,{k}_{n}\right)$ is a finite extension of $F$$F$FF$F$ and hence is algebraic over $F$$F$FF$F$. Therefore, the coefficients of $h\left(x\right)$$h\left(x\right)$h(x)h(x)$h\left(x\right)$ are algebraic over $F$$F$FF$F$, and $a$$a$aa$a$ is algebraic over $F$$F$FF$F$.
Summary for Part 2
We have shown that any element $a\in L$$a\in L$a in La \in L$a\in L$ is algebraic over $F$$F$FF$F$. Therefore, $L$$L$LL$L$ is algebraic over $F$$F$FF$F$.
Final Summary
We have proven that the elements of $K$$K$KK$K$ that are algebraic over $F$$F$FF$F$ form a subfield of $K$$K$KK$K$ and that if $F\subset K\subset L$$F\subset K\subset L$F sub K sub LF \subset K \subset L$F\subset K\subset L$ and $L$$L$LL$L$ is algebraic over $K$$K$KK$K$ and $K$$K$KK$K$ is algebraic over $F$$F$FF$F$, then $L$$L$LL$L$ is algebraic over $F$$F$FF$F$.
5/5
$$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$$
$$sin\:3\theta =3\:sin\:\theta -4\:sin^3\:\theta$$

### 4(a) Show that every algebraically closed field is infinite.

To show that every algebraically closed field $K$$K$KK$K$ is infinite, we will proceed by contradiction. We will assume that $K$$K$KK$K$ is a finite field and algebraically closed, and then show that this leads to a contradiction.
Step 1: Assume $K$$K$KK$K$ is a Finite Algebraically Closed Field
Let’s assume that $K$$K$KK$K$ is a finite field with $q$$q$qq$q$ elements and is algebraically closed. Being algebraically closed means that every non-constant polynomial $f\left(x\right)\in K\left[x\right]$$f\left(x\right)\in K\left[x\right]$f(x)in K[x]f(x) \in K[x]$f\left(x\right)\in K\left[x\right]$ has a root in $K$$K$KK$K$.
Step 2: Consider a Polynomial of Degree $q$$q$qq$q$
Consider the polynomial $f\left(x\right)=\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\cdots \left(x-{a}_{q}\right)$$f\left(x\right)=\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\cdots \left(x-{a}_{q}\right)$f(x)=(x-a_(1))(x-a_(2))cdots(x-a_(q))f(x) = (x – a_1)(x – a_2) \cdots (x – a_q)$f\left(x\right)=\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\cdots \left(x-{a}_{q}\right)$, where ${a}_{1},{a}_{2},\dots ,{a}_{q}$${a}_{1},{a}_{2},\dots ,{a}_{q}$a_(1),a_(2),dots,a_(q)a_1, a_2, \ldots, a_q${a}_{1},{a}_{2},\dots ,{a}_{q}$ are all the elements of $K$$K$KK$K$.
Step 3: Show that $f\left(x\right)$$f\left(x\right)$f(x)f(x)$f\left(x\right)$ has No Root in $K$$K$KK$K$
Notice that $f\left(a\right)$$f\left(a\right)$f(a)f(a)$f\left(a\right)$ for any $a\in K$$a\in K$a in Ka \in K$a\in K$ is given by:
$f\left(a\right)=\left(a-{a}_{1}\right)\left(a-{a}_{2}\right)\cdots \left(a-{a}_{q}\right)$$f\left(a\right)=\left(a-{a}_{1}\right)\left(a-{a}_{2}\right)\cdots \left(a-{a}_{q}\right)$f(a)=(a-a_(1))(a-a_(2))cdots(a-a_(q))f(a) = (a – a_1)(a – a_2) \cdots (a – a_q)$f\left(a\right)=\left(a-{a}_{1}\right)\left(a-{a}_{2}\right)\cdots \left(a-{a}_{q}\right)$
Since $a$$a$aa$a$ is one of the ${a}_{i}$${a}_{i}$a_(i)a_i${a}_{i}$‘s, one of the factors in this product is zero, making $f\left(a\right)=0$$f\left(a\right)=0$f(a)=0f(a) = 0