2016

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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 \(\mathbb{K}\).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that ( f ) ( f ) (f)(f)(f) is a maximal ideal in K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X] if and only if f f fff is an irreducible polynomial over K K K\mathbb{K}K, we need to prove two directions:
  1. If ( f ) ( f ) (f)(f)(f) is a maximal ideal, then f f fff is irreducible.
  2. If f f fff is irreducible, then ( f ) ( f ) (f)(f)(f) is a maximal ideal.
Direction 1: ( f ) ( f ) (f)(f)(f) is a maximal ideal =>\Rightarrow f f fff is irreducible
Step 1: Assume ( f ) ( f ) (f)(f)(f) is a maximal ideal
Let’s assume that ( f ) ( f ) (f)(f)(f) is a maximal ideal in K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X].
Step 2: Show that f f fff is irreducible
Suppose, for the sake of contradiction, that f f fff is not irreducible. Then f = g h f = g h f=ghf = ghf=gh for some g , h K [ X ] g , h K [ X ] g,h inK[X]g, h \in \mathbb{K}[X]g,hK[X] with deg ( g ) < deg ( f ) deg ( g ) < deg ( f ) deg(g) < deg(f)\deg(g) < \deg(f)deg(g)<deg(f) and deg ( h ) < deg ( f ) deg ( h ) < deg ( f ) deg(h) < deg(f)\deg(h) < \deg(f)deg(h)<deg(f).
Let I = ( g ) I = ( g ) I=(g)I = (g)I=(g). Then f I f I f in If \in IfI because f = g h f = g h f=ghf = ghf=gh, and I I III contains ( f ) ( f ) (f)(f)(f). However, I K [ X ] I K [ X ] I!=K[X]I \neq \mathbb{K}[X]IK[X] because deg ( g ) < deg ( f ) deg ( g ) < deg ( f ) deg(g) < deg(f)\deg(g) < \deg(f)deg(g)<deg(f).
This contradicts the assumption that ( f ) ( f ) (f)(f)(f) is a maximal ideal, because we’ve found an ideal I I III strictly between ( f ) ( f ) (f)(f)(f) and K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X].
Therefore, f f fff must be irreducible.
Direction 2: f f fff is irreducible =>\Rightarrow ( f ) ( f ) (f)(f)(f) is a maximal ideal
Step 1: Assume f f fff is irreducible
Let’s assume that f f fff is an irreducible polynomial in K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X].
Step 2: Show that ( f ) ( f ) (f)(f)(f) is a maximal ideal
Let I I III be an ideal in K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X] such that ( f ) I K [ X ] ( f ) I K [ X ] (f)sube I subeK[X](f) \subseteq I \subseteq \mathbb{K}[X](f)IK[X].
Since f f fff is in I I III, there exists a polynomial g I g I g in Ig \in IgI such that f = g f = g f=gf = gf=g.
Since f f fff is irreducible, g g ggg must either be a unit or associate of f f fff. If g g ggg is a unit, then I = K [ X ] I = K [ X ] I=K[X]I = \mathbb{K}[X]I=K[X]. If g g ggg is an associate of f f fff, then I = ( f ) I = ( f ) I=(f)I = (f)I=(f).
Therefore, ( f ) ( f ) (f)(f)(f) is a maximal ideal.
Summary
We have shown that if ( f ) ( f ) (f)(f)(f) is a maximal ideal, then f f fff is irreducible, and conversely, if f f fff is irreducible, then ( f ) ( f ) (f)(f)(f) is a maximal ideal. Hence, ( f ) ( f ) (f)(f)(f) is a maximal ideal in K [ X ] K [ X ] K[X]\mathbb{K}[X]K[X] if and only if f f fff is an irreducible polynomial over K K K\mathbb{K}K.
Verified Answer
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 \( \mathbb{Z}_{p} \).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that every non-zero element of Z p Z p Z_(p)\mathbb{Z}_{p}Zp generates Z p Z p Z_(p)\mathbb{Z}_{p}Zp, we need to prove that for any non-zero element a a aaa in Z p Z p Z_(p)\mathbb{Z}_{p}Zp, the cyclic subgroup generated by a a aaa, denoted a a (:a:)\langle a \ranglea, is equal to Z p Z p Z_(p)\mathbb{Z}_{p}Zp.
Step 1: Define the Cyclic Subgroup a a (:a:)\langle a \ranglea
The cyclic subgroup a a (:a:)\langle a \ranglea generated by a a aaa is defined as:
a = { a n mod p : n Z } a = { a n mod p : n Z } (:a:)={a^(n)quadmodp:n inZ}\langle a \rangle = \{ a^n \mod p : n \in \mathbb{Z} \}a={anmodp:nZ}
Step 2: Show that a a (:a:)\langle a \ranglea is a Subset of Z p Z p Z_(p)\mathbb{Z}_{p}Zp
First, let’s establish that a a (:a:)\langle a \ranglea is a subset of Z p Z p Z_(p)\mathbb{Z}_{p}Zp. This is trivially true since a a (:a:)\langle a \ranglea consists of elements that are congruent modulo p p ppp, and therefore, each element in a a (:a:)\langle a \ranglea is an element in Z p Z p Z_(p)\mathbb{Z}_{p}Zp.
Step 3: Show that a a (:a:)\langle a \ranglea Contains p p ppp Distinct Elements
Let’s assume that a a aaa is a non-zero element in Z p Z p Z_(p)\mathbb{Z}_{p}Zp. We want to show that a , 2 a , 3 a , , ( p 1 ) a a , 2 a , 3 a , , ( p 1 ) a a,2a,3a,dots,(p-1)aa, 2a, 3a, \ldots, (p-1)aa,2a,3a,,(p1)a are all distinct modulo p p ppp.
Suppose, for the sake of contradiction, that i a j a mod p i a j a mod p ia-=jamodpia \equiv ja \mod piajamodp for some i j i j i!=ji \neq jij where 1 i , j p 1 1 i , j p 1 1 <= i,j <= p-11 \leq i,j \leq p-11i,jp1.
This implies ( i j ) a 0 mod p ( i j ) a 0 mod p (i-j)a-=0modp(i-j)a \equiv 0 \mod p(ij)a0modp, or p p ppp divides ( i j ) a ( i j ) a (i-j)a(i-j)a(ij)a.
Since p p ppp is a prime and p ( i j ) p ( i j ) p∤(i-j)p \nmid (i-j)p(ij) (because i j i j i!=ji \neq jij), it must be the case that p p ppp divides a a aaa, which contradicts the assumption that a a aaa is a non-zero element of Z p Z p Z_(p)\mathbb{Z}_{p}Zp.
Therefore, a , 2 a , 3 a , , ( p 1 ) a a , 2 a , 3 a , , ( p 1 ) a a,2a,3a,dots,(p-1)aa, 2a, 3a, \ldots, (p-1)aa,2a,3a,,(p1)a are all distinct modulo p p ppp, and a a (:a:)\langle a \ranglea contains p p ppp distinct elements.
Step 4: Conclude that a = Z p a = Z p (:a:)=Z_(p)\langle a \rangle = \mathbb{Z}_{p}a=Zp
Since a a (:a:)\langle a \ranglea is a subset of Z p Z p Z_(p)\mathbb{Z}_{p}Zp and contains p p ppp distinct elements, it must be the case that a = Z p a = Z p (:a:)=Z_(p)\langle a \rangle = \mathbb{Z}_{p}a=Zp.
Summary
We have shown that for any non-zero element a a aaa in Z p Z p Z_(p)\mathbb{Z}_{p}Zp, the cyclic subgroup a a (:a:)\langle a \ranglea generated by a a aaa is equal to Z p Z p Z_(p)\mathbb{Z}_{p}Zp. Therefore, every non-zero element of Z p Z p Z_(p)\mathbb{Z}_{p}Zp generates Z p Z p Z_(p)\mathbb{Z}_{p}Zp.
Verified Answer
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 \( F \).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Part 1: Prove that the elements of K K KKK that are algebraic over F F FFF form a subfield of K K KKK
To show that the set of elements in K K KKK that are algebraic over F F FFF 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 AAA.
Step 1: Closure under Addition
Let a , b A a , b A a,b in Aa, b \in Aa,bA. Then a a aaa and b b bbb are roots of some polynomials f ( x ) , g ( x ) F [ x ] f ( x ) , g ( x ) F [ x ] f(x),g(x)in F[x]f(x), g(x) \in F[x]f(x),g(x)F[x] respectively. We need to show that a + b a + b a+ba + ba+b is also algebraic over F F FFF.
Consider the polynomials f ( x b ) f ( x b ) f(x-b)f(x – b)f(xb) and g ( x a ) g ( x a ) g(x-a)g(x – a)g(xa). Their product h ( x ) = f ( x b ) g ( x a ) h ( x ) = f ( x b ) g ( x a ) h(x)=f(x-b)g(x-a)h(x) = f(x – b)g(x – a)h(x)=f(xb)g(xa) is a polynomial in F [ x ] F [ x ] F[x]F[x]F[x]. The root of h ( x ) h ( x ) h(x)h(x)h(x) is x = a + b x = a + b x=a+bx = a + bx=a+b, which means a + b a + b a+ba + ba+b is algebraic over F F FFF.
Step 2: Closure under Multiplication
Let a , b A a , b A a,b in Aa, b \in Aa,bA. Then a a aaa and b b bbb are roots of some polynomials f ( x ) , g ( x ) F [ x ] f ( x ) , g ( x ) F [ x ] f(x),g(x)in F[x]f(x), g(x) \in F[x]f(x),g(x)F[x] respectively. We need to show that a b a b ababab is also algebraic over F F FFF.
Consider the polynomial h ( x ) = f ( x / b ) g ( x / a ) h ( x ) = f ( x / b ) g ( x / a ) h(x)=f(x//b)g(x//a)h(x) = f(x/b)g(x/a)h(x)=f(x/b)g(x/a). This is a polynomial in F [ x ] F [ x ] F[x]F[x]F[x]. The root of h ( x ) h ( x ) h(x)h(x)h(x) is x = a b x = a b x=abx = abx=ab, which means a b a b ababab is algebraic over F F FFF.
Step 3: Existence of Multiplicative Inverses for Non-zero Elements
Let a A a A a in Aa \in AaA and a 0 a 0 a!=0a \neq 0a0. Then a a aaa is a root of some polynomial f ( x ) F [ x ] f ( x ) F [ x ] f(x)in F[x]f(x) \in F[x]f(x)F[x]. We need to show that a 1 a 1 a^(-1)a^{-1}a1 is also algebraic over F F FFF.
Consider the polynomial g ( x ) = x n f ( 1 / x ) g ( x ) = x n f ( 1 / x ) g(x)=x^(n)f(1//x)g(x) = x^n f(1/x)g(x)=xnf(1/x), where n n nnn is the degree of f ( x ) f ( x ) f(x)f(x)f(x). This is a polynomial in F [ x ] F [ x ] F[x]F[x]F[x]. The root of g ( x ) g ( x ) g(x)g(x)g(x) is x = a 1 x = a 1 x=a^(-1)x = a^{-1}x=a1, which means a 1 a 1 a^(-1)a^{-1}a1 is algebraic over F F FFF.
Step 4: 0 0 000 and 1 1 111 are in A A AAA
The zero polynomial and the constant polynomial x 1 x 1 x-1x – 1x1 show that 0 0 000 and 1 1 111 are algebraic over F F FFF.
Summary for Part 1
We have shown that A A AAA is closed under addition, multiplication, and contains multiplicative inverses for non-zero elements. Also, 0 0 000 and 1 1 111 are in A A AAA. Therefore, A A AAA is a subfield of K K KKK.
Part 2: If F K L F K L F sub K sub LF \subset K \subset LFKL and L L LLL is algebraic over K K KKK and K K KKK is algebraic over F F FFF, then L L LLL is algebraic over F F FFF
Step 1: Take an element a L a L a in La \in LaL
Let a a aaa be an element in L L LLL.
Step 2: Show that a a aaa is algebraic over K K KKK
Since L L LLL is algebraic over K K KKK, a a aaa is a root of some polynomial f ( x ) K [ x ] f ( x ) K [ x ] f(x)in K[x]f(x) \in K[x]f(x)K[x].
Step 3: Show that a a aaa is algebraic over F F FFF
The coefficients of f ( x ) f ( x ) f(x)f(x)f(x) are in K K KKK and are algebraic over F F FFF. Let these coefficients be k 1 , k 2 , , k n k 1 , k 2 , , k n k_(1),k_(2),dots,k_(n)k_1, k_2, \ldots, k_nk1,k2,,kn. Each k i k i k_(i)k_iki is a root of some polynomial g i ( x ) F [ x ] g i ( x ) F [ x ] g_(i)(x)in F[x]g_i(x) \in F[x]gi(x)F[x].
Consider the polynomial h ( x ) = f ( x , k 1 , k 2 , , k n ) h ( x ) = f ( x , k 1 , k 2 , , k n ) h(x)=f(x,k_(1),k_(2),dots,k_(n))h(x) = f(x, k_1, k_2, \ldots, k_n)h(x)=f(x,k1,k2,,kn) where we treat k 1 , k 2 , , k n k 1 , k 2 , , k n k_(1),k_(2),dots,k_(n)k_1, k_2, \ldots, k_nk1,k2,,kn as variables. The polynomial h ( x ) h ( x ) h(x)h(x)h(x) has a a aaa as a root and its coefficients are in F ( k 1 , k 2 , , k n ) F ( k 1 , k 2 , , k n ) F(k_(1),k_(2),dots,k_(n))F(k_1, k_2, \ldots, k_n)F(k1,k2,,kn).
Since each k i k i k_(i)k_iki is algebraic over F F FFF, F ( k 1 , k 2 , , k n ) F ( k 1 , k 2 , , k n ) F(k_(1),k_(2),dots,k_(n))F(k_1, k_2, \ldots, k_n)F(k1,k2,,kn) is a finite extension of F F FFF and hence is algebraic over F F FFF. Therefore, the coefficients of h ( x ) h ( x ) h(x)h(x)h(x) are algebraic over F F FFF, and a a aaa is algebraic over F F FFF.
Summary for Part 2
We have shown that any element a L a L a in La \in LaL is algebraic over F F FFF. Therefore, L L LLL is algebraic over F F FFF.
Final Summary
We have proven that the elements of K K KKK that are algebraic over F F FFF form a subfield of K K KKK and that if F K L F K L F sub K sub LF \subset K \subset LFKL and L L LLL is algebraic over K K KKK and K K KKK is algebraic over F F FFF, then L L LLL is algebraic over F F FFF.
Verified Answer
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.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that every algebraically closed field K K KKK is infinite, we will proceed by contradiction. We will assume that K K KKK is a finite field and algebraically closed, and then show that this leads to a contradiction.
Step 1: Assume K K KKK is a Finite Algebraically Closed Field
Let’s assume that K K KKK is a finite field with q q qqq elements and is algebraically closed. Being algebraically closed means that every non-constant polynomial f ( x ) K [ x ] f ( x ) K [ x ] f(x)in K[x]f(x) \in K[x]f(x)K[x] has a root in K K KKK.
Step 2: Consider a Polynomial of Degree q q qqq
Consider the polynomial f ( x ) = ( x a 1 ) ( x a 2 ) ( x a q ) f ( x ) = ( x a 1 ) ( x a 2 ) ( x a q ) 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(x)=(xa1)(xa2)(xaq), where a 1 , a 2 , , a q a 1 , a 2 , , a q a_(1),a_(2),dots,a_(q)a_1, a_2, \ldots, a_qa1,a2,,aq are all the elements of K K KKK.
Step 3: Show that f ( x ) f ( x ) f(x)f(x)f(x) has No Root in K K KKK
Notice that f ( a ) f ( a ) f(a)f(a)f(a) for any a K a K a in Ka \in KaK is given by:
f ( a ) = ( a a 1 ) ( a a 2 ) ( a a q ) f ( a ) = ( a a 1 ) ( a a 2 ) ( a a q ) 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(a)=(aa1)(aa2)(aaq)
Since a a aaa is one of the a i a i a_(i)a_iai‘s, one of the factors in this product is zero, making f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0