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 = { a u + b v | u , v s = { a u + b v | u , v s={au+bv|u,vs=\{a u+b v \,|\, u, vs={au+bv|u,v are integers and a u + b v > 0 } a u + b v > 0 } au+bv > 0}a u+b v>0\}au+bv>0}
Step 2: Investigating Cases for a a aaa and b b bbb
If a > 0 a > 0 a > 0a>0a>0, then a = a 1 + b ( 0 ) > 0 a = a 1 + b ( 0 ) > 0 a=a1+b(0) > 0a=a 1+b(0)>0a=a1+b(0)>0, which implies that a > 0 a > 0 a > 0a>0a>0.
If a < 0 a < 0 a < 0a<0a<0, then a = a ( 1 ) + b ( 0 ) > 0 a = a ( 1 ) + b ( 0 ) > 0 -a=a(-1)+b(0) > 0-a=a(-1)+b(0)>0a=a(1)+b(0)>0, which implies that a S a S -a in S-a \in SaS.
Similarly, if b > 0 b > 0 b > 0b>0b>0, then b S b S b in Sb \in SbS.
If b < 0 b < 0 b < 0b<0b<0, then b S b S -b in S-b \in SbS.
Therefore, S ϕ S ϕ S!=phiS \neq \phiSϕ and S S SSS contains positive integers.
By the well-ordering principle, S S SSS has a least element, say d d ddd.
Step 3: Proving that d d ddd is the GCD of a a aaa and b b bbb
Now we have d S d S d in Sd \in SdS such that d = a x + b y ( 1 ) d = a x + b y ( 1 ) d=ax+by—-(1)d=a x+b y—-(1)d=ax+by(1) for some integers x , y x , y x,yx, yx,y.
Also, d > 0 d > 0 d > 0d>0d>0.
To prove that d d ddd is the greatest common divisor (GCD) of a a aaa and b b bbb, consider the following:
Let a = d q + r ( 2 ) a = d q + r ( 2 ) a=dq+r—-(2)a=\mathbf{d} q+r—-(2)a=dq+r(2) (where 0 r < d 0 r < d 0 <= r < d0 \leqslant r < d0r<d) be a division of a a aaa by d d ddd.
If r 0 r 0 r!=0r \neq 0r0, then r = a d q r = a d q r=a-dqr=a-\mathbf{d} qr=adq.
= a ( a x + b y ) q (as d = a x + b y ) = a ( 1 λ q ) + b ( y z ) > 0 (since 1 λ q , y q are integers ) r > 0 , r S if r < d and r S = a ( a x + b y ) q (as  d = a x + b y ) = a ( 1 λ q ) + b ( y z ) > 0 (since  1 λ q , y q  are integers ) r > 0 , r S  if  r < d  and  r S {:[=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}=a(ax+by)q(as d=ax+by)=a(1λq)+b(yz)>0(since 1λq,yq are integers)r>0,rS if r<d and rS
This leads to a contradiction since d d ddd is the least element of S S SSS.
Therefore, r = 0 r = 0 r=0r=0r=0, and we have a = d q a = d q a=dqa=\mathbf{d} qa=dq.
This implies that a d = q a d = q (a)/(d)=q\frac{a}{d}=qad=q and d a d a (d)/(a)\frac{d}{a}da are integers.
Similarly, d b d b (d)/(b)\frac{d}{b}db is also an integer.
Step 4: Proving Uniqueness of GCD
Suppose C c , c b c a x + b y c d C c , c b c a x + b y 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}Cc,cbcax+bycd
Therefore d a , d b d a , d b (d)/(a),(d)/(b)\frac{\mathbf{d}}{a}, \frac{d}{b}da,db also if c a , c b c d c a , c b c d (c)/(a),(c)/(b)=>(c)/(d)\frac{c}{a}, \frac{c}{b} \Rightarrow \frac{c}{d}ca,cbcd
d d ddd is gcd of a a aaa and b b bbb
If possible d d d^(‘)d’d is also gcd of a a aaa and b b bbb
Then d a , d b d d d a , d b d d (d^(‘))/(a),(d^(‘))/(b)=>(d)/(d^(‘))rarr\frac{d^{\prime}}{a}, \frac{d^{\prime}}{b} \Rightarrow \frac{d}{d^{\prime}} \rightarrowda,dbdd (3)
Similarly, d c , d b d d d c , d b d d (d)/(c),(d)/(b)=>(d^(‘))/(d)rarr\frac{\mathbf{d}}{c}, \frac{d}{b} \Rightarrow \frac{d^{\prime}}{d} \rightarrowdc,dbdd (4)
This implies d d d d (d)/(d^(‘))\frac{d}{d^{\prime}}dd (from the divisibility relation) which, in turn, implies d = d d = d d=d^(‘)d=d^{\prime}d=d (from the uniqueness of GCD).
Step 5: Extending the Argument
The above argument can be extended to more than two integers.
If d = gcd ( m 1 , m 2 , , m k ) d = gcd m 1 , m 2 , , m k d=gcd(m_(1),m_(2),dots,m_(k))d=\operatorname{gcd}\left(m_1, m_2, \ldots, m_k\right)d=gcd(m1,m2,,mk), there exist integers λ 1 , λ 2 , , λ k λ 1 , λ 2 , , λ k lambda_(1),lambda_(2),dots,lambda _(k)\lambda_1, \lambda_2, \ldots, \lambda_kλ1,λ2,,λk such that
d = λ 1 m 1 + λ 2 m 2 + + m k λ k d = λ 1 m 1 + λ 2 m 2 + + λ k m k d = λ 1 m 1 + λ 2 m 2 + + m k λ k d = λ 1 m 1 + λ 2 m 2 + + λ k m k {:[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}d=λ1m1+λ2m2++mkλkd=λ1m1+λ2m2++λkmk
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 FFF and a polynomial f ( x ) F [ x ] f ( x ) F [ x ] f(x)in F[x]f(x) \in F[x]f(x)F[x] of degree greater than 0. The goal is to show the existence of a field F F F^(‘)F^{\prime}F and an embedding q : F F q : F F q:F rarrF^(‘)q: F \rightarrow F^{\prime}q:FF such that the polynomial f q F [ x ] f q F [ x ] f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]fqF[x] has a root in F F F^(‘)F^{\prime}F. Here, f q f q f^(q)f^qfq is obtained by replacing each coefficient a a aaa of f f fff by q ( a ) q ( a ) q(a)q(a)q(a).
Step 2: Maximal Ideal and Field Extension
Suppose f ( x ) F [ x ] f ( x ) F [ x ] f(x)in F[x]f(x) \in F[x]f(x)F[x] is a polynomial of degree greater than 0 over a field F F FFF. We consider the ideal M = f ( x ) M = f ( x ) M=(:f(x):)M = \langle f(x) \rangleM=f(x) in F [ x ] F [ x ] F[x]F[x]F[x]. Since f ( x ) f ( x ) f(x)f(x)f(x) is irreducible, M M MMM is a maximal ideal in F [ x ] F [ x ] F[x]F[x]F[x].
Step 3: Field Extension
We create a field extension by taking F F FFF modulo this maximal ideal, denoted as F [ x ] M F [ x ] M (F[x])/(M)\frac{F[x]}{M}F[x]M, which is itself a field.
Step 4: Define Embedding q q qqq
Define the embedding q : F F [ x ] M q : F F [ x ] M q:F rarr(F[x])/(M)q: F \rightarrow \frac{F[x]}{M}q:FF[x]M as q ( a ) = a + M q ( a ) = a + M q(a)=a+Mq(a) = a + Mq(a)=a+M. This embedding q q qqq is a homomorphism.
Step 5: Ker q q qqq and Injectiveness
We consider the kernel of q q qqq, i.e., elements a F a F a in Fa \in FaF such that q ( a ) = 0 + M q ( a ) = 0 + M q(a)=0+Mq(a) = 0 + Mq(a)=0+M. This implies a + M = M a + M = M a+M=Ma + M = Ma+M=M, which further implies a M = f ( x ) a M = f ( x ) a in M=(:f(x):)a \in M = \langle f(x) \rangleaM=f(x). Therefore, a = f ( x ) q ( x ) a = f ( x ) q ( x ) a=f(x)q(x)a = f(x)q(x)a=f(x)q(x) for some q ( x ) F [ x ] q ( x ) F [ x ] q(x)in F[x]q(x) \in F[x]q(x)F[x]. Since f ( x ) f ( x ) f(x)f(x)f(x) is irreducible and a a aaa is a polynomial of degree greater than 0, this relation holds only when a = 0 a = 0 a=0a = 0a=0. Consequently, ker q = { 0 } ker  q = { 0 } “ker “q={0}\text{ker } q = \{0\}ker q={0} or q q qqq is injective.
Step 6: Isomorphism and Subfield
Hence, F F FFF is isomorphic to q ( F ) q ( F ) q(F)q(F)q(F). We can view F F F^(‘)F^{\prime}F as containing F F FFF by identifying a F a F a in Fa \in FaF with q ( a ) q ( a ) q(a)q(a)q(a) and vice versa.
Step 7: Construct Ψ Ψ Psi\PsiΨ
Define Ψ : F [ x ] F [ x ] M Ψ : F [ x ] F [ x ] M Psi:F[x]rarr(F[x])/(M)\Psi: F[x] \rightarrow \frac{F[x]}{M}Ψ:F[x]F[x]M such that Ψ ( f ( x ) ) = f ( x ) + M Ψ ( f ( x ) ) = f ( x ) + M Psi(f(x))=f(x)+M\Psi(f(x)) = f(x) + MΨ(f(x))=f(x)+M. Ψ Ψ Psi\PsiΨ is a natural homomorphism.
Step 8: Show α α alpha\alphaα as a Root
Let α = Ψ ( x ) = x + M α = Ψ ( x ) = x + M alpha=Psi(x)=x+M\alpha = \Psi(x) = x + Mα=Ψ(x)=x+M. We claim that α α alpha\alphaα is a root of f q f q f^(q)f^qfq in F F FFF.
Step 9: Evaluate f q f q f^(q)f^qfq
Evaluate f q f q f^(q)f^qfq as follows:
Ψ ( f q ) = Ψ ( α 0 + α 1 x + α 2 x 2 + + α n x n ) = Ψ ( α 0 ) + Ψ ( α 1 ) Ψ ( x ) + + Ψ ( α n ) Ψ ( x n ) Ψ ( f q ) = Ψ ( α 0 + α 1 x + α 2 x 2 + + α n x n ) = Ψ ( α 0 ) + Ψ ( α 1 ) Ψ ( x ) + + Ψ ( α n ) Ψ ( x n ) {:[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}Ψ(fq)=Ψ(α0+α1x+α2x2++αnxn)=Ψ(α0)+Ψ(α1)Ψ(x)++Ψ(αn)Ψ(xn)
Step 10: Use p ( x ) p ( x ) p^((x))p^{(x)}p(x)
Note that Ψ ( p ( x ) ) = p ( x ) + M = p ( x ) + p ( x ) = M = Ψ ( p ( x ) ) = p ( x ) + M = p ( x ) + p ( x ) = 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 =Ψ(p(x))=p(x)+M=p(x)+p(x)=M= zero of F F FFF.
Step 11: Zero of F F F^(‘)F^{\prime}F
Thus, the zero of F F F^(‘)F^{\prime}F is:
Zero of F = Ψ ( α 0 ) + Ψ ( α 1 ) Ψ ( x ) + + Ψ ( α n ) Ψ ( x n ) = q ( a 0 ) + q ( a 1 ) Ψ ( x ) + + q ( a n ) Ψ ( x n ) Zero of  F = Ψ ( α 0 ) + Ψ ( α 1 ) Ψ ( x ) + + Ψ ( α n ) Ψ ( x n ) = q ( a 0 ) + q ( a 1 ) Ψ ( x ) + + q ( a n ) Ψ ( x n ) {:[“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}Zero of F=Ψ(α0)+Ψ(α1)Ψ(x)++Ψ(αn)Ψ(xn)=q(a0)+q(a1)Ψ(x)++q(an)Ψ(xn)
Step 12: Zero is a Root
Since F q ( F ) F q ( F ) F~=q(F)F \cong q(F)Fq(F), the zero is also a 0 + a 1 α + a 2 α 2 + + a n α n a 0 + a 1 α + a 2 α 2 + + a n α 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^na0+a1α+a2α2++anαn, which is equal to f ( a ) = f q f ( a ) = f q f(a)=f^(q)f(a) = f^qf(a)=fq.
Step 13: Conclusion
Hence, a a aaa is a root of f ( a ) f ( a ) f(a)f(a)f(a) in F F FFF. By replacing each coefficient a a aaa of f f fff by q ( a ) q ( a ) q(a)q(a)q(a), we obtain f q f q f^(q)f^qfq.
Therefore, there exists a field F F F^(‘)F^{\prime}F and an embedding q : F F q : F F q:F rarrF^(‘)q: F \rightarrow F^{\prime}q:FF such that the polynomial f q F [ x ] f q F [ x ] f^(q)inF^(‘)[x]f^q \in F^{\prime}[x]fqF[x] has a root in F F F^(‘)F^{\prime}F, where f q f q f^(q)f^qfq is obtained by replacing each coefficient a a aaa of f f fff by q ( a ) q ( a ) q(a)q(a)q(a).
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 Q Q Q\mathbb{Q}Q, consists of all rational numbers under the operation of addition. To show that there are infinitely many subgroups of Q Q Q\mathbb{Q}Q, we can construct an infinite family of subgroups.
Construction of Subgroups:
Consider the set n Q n Q nQn\mathbb{Q}nQ, where n n nnn is any positive integer. This set consists of all rational numbers that can be written in the form n q n q nqnqnq, where q q qqq is any rational number. Formally, n Q = { n q q Q } n Q = { n q q Q } nQ={nq∣q inQ}n\mathbb{Q} = \{nq \mid q \in \mathbb{Q}\}nQ={nqqQ}.
Properties:
  1. Identity Element: 0 0 000 is in n Q n Q nQn\mathbb{Q}nQ because 0 = n × 0 0 = n × 0 0=n xx00 = n \times 00=n×0.
  2. Closure under Addition: If a , b n Q a , b n Q a,b in nQa, b \in n\mathbb{Q}a,bnQ, then a + b = n q 1 + n q 2 = n ( q 1 + q 2 ) a + b = n q 1 + n q 2 = n ( q 1 + q 2 ) a+b=nq_(1)+nq_(2)=n(q_(1)+q_(2))a + b = nq_1 + nq_2 = n(q_1 + q_2)a+b=nq1+nq2=n(q1+q2) is also in n Q n Q nQn\mathbb{Q}nQ.
  3. Closure under Inverse: If a n Q a n Q a in nQa \in n\mathbb{Q}anQ, then a = n q = n ( q ) a = n q = n ( q ) -a=-nq=n(-q)-a = -nq = n(-q)a=nq=n(q) is also in n Q n Q nQn\mathbb{Q}nQ.
Therefore, n Q n Q nQn\mathbb{Q}nQ is a subgroup of Q Q Q\mathbb{Q}Q for each positive integer n n nnn.
Infinitude of Subgroups:
Since n n nnn can be any positive integer, and there are infinitely many positive integers, we have constructed an infinite family of subgroups n Q n Q nQn\mathbb{Q}nQ of Q Q Q\mathbb{Q}Q.
Conclusion:
Hence, we have shown that there are infinitely many subgroups of the additive group Q Q Q\mathbb{Q}Q of rational numbers.
Verified Answer
5/5
\(cos\left(\theta -\phi \right)=cos\:\theta \:cos\:\phi +sin\:\theta \:sin\:\phi \)

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