2013

Estimated reading: 48 minutes 86 views

UPSC Algebra

\(cos\left(\theta +\phi \right)=cos\:\theta \:cos\:\phi -sin\:\theta \:sin\:\phi \)

1(a) Show that the set of matrices \(S=\left\{\left(\begin{array}{cc}a & -b \\ b & a\end{array}\right) \quad a, b \in \mathbb{R}\right\}\) is a field under the usual binary operations of matrix addition and matrix multiplication. What are the additive and multiplicative identities and what is the inverse of \(\left(\begin{array}{lr}1 & -1 \\ 1 & 1\end{array}\right)\)? Consider the map \(f: \mathbb{C} \rightarrow S\) defined by \(f(a+i b)=\left(\begin{array}{cc}a & -b \\ b & a\end{array}\right)\). Show that \(f\) is an isomorphism.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set S S SSS is a field under the usual binary operations of matrix addition and matrix multiplication, we need to establish the following properties:
  1. Closure under Addition and Multiplication
  2. Associativity of Addition and Multiplication
  3. Commutativity of Addition and Multiplication
  4. Existence of Additive and Multiplicative Identity
  5. Existence of Additive and Multiplicative Inverse
  6. Distributive Law
  7. Closure under Addition and Multiplication
Addition
Let A = ( a 1 b 1 b 1 a 1 ) A = a 1 b 1 b 1 a 1 A=([a_(1),-b_(1)],[b_(1),a_(1)])A = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right)A=(a1b1b1a1) and B = ( a 2 b 2 b 2 a 2 ) B = a 2 b 2 b 2 a 2 B=([a_(2),-b_(2)],[b_(2),a_(2)])B = \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)B=(a2b2b2a2) be two matrices in S S SSS.
Let’s substitute the values into the formula for matrix addition:
A + B = ( a 1 b 1 b 1 a 1 ) + ( a 2 b 2 b 2 a 2 ) A + B = a 1 b 1 b 1 a 1 + a 2 b 2 b 2 a 2 A+B=([a_(1),-b_(1)],[b_(1),a_(1)])+([a_(2),-b_(2)],[b_(2),a_(2)])A + B = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) + \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)A+B=(a1b1b1a1)+(a2b2b2a2)
After simplifying, we get:
A + B = ( a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 ) A + B = a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 A+B=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])A + B = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)A+B=(a1+a2(b1+b2)b1+b2a1+a2)
Since a 1 , a 2 , b 1 , b 2 a 1 , a 2 , b 1 , b 2 a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2a1,a2,b1,b2 are real numbers, a 1 + a 2 a 1 + a 2 a_(1)+a_(2)a_1 + a_2a1+a2 and b 1 + b 2 b 1 + b 2 b_(1)+b_(2)b_1 + b_2b1+b2 are also real numbers. Therefore, A + B A + B A+BA + BA+B is also in S S SSS.
Multiplication
Similarly, for multiplication, we have:
A × B = ( a 1 b 1 b 1 a 1 ) × ( a 2 b 2 b 2 a 2 ) A × B = a 1 b 1 b 1 a 1 × a 2 b 2 b 2 a 2 A xx B=([a_(1),-b_(1)],[b_(1),a_(1)])xx([a_(2),-b_(2)],[b_(2),a_(2)])A \times B = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) \times \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)A×B=(a1b1b1a1)×(a2b2b2a2)
After simplifying, we get:
A × B = ( a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 ) A × B = a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 A xx B=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])A \times B = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)A×B=(a1a2b1b2(a1b2+a2b1)a1b2+a2b1a1a2b1b2)
Since a 1 , a 2 , b 1 , b 2 a 1 , a 2 , b 1 , b 2 a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2a1,a2,b1,b2 are real numbers, a 1 a 2 b 1 b 2 a 1 a 2 b 1 b 2 a_(1)a_(2)-b_(1)b_(2)a_1a_2 – b_1b_2a1a2b1b2 and a 1 b 2 + a 2 b 1 a 1 b 2 + a 2 b 1 a_(1)b_(2)+a_(2)b_(1)a_1b_2 + a_2b_1a1b2+a2b1 are also real numbers. Therefore, A × B A × B A xx BA \times BA×B is also in S S SSS.
  1. Associativity of Addition and Multiplication
    Associativity of addition and multiplication follows directly from the associativity of addition and multiplication of matrices.
  2. Commutativity of Addition and Multiplication
    Commutativity of addition and multiplication also follows directly from the commutativity of addition and multiplication of matrices.
  3. Existence of Additive and Multiplicative Identity
    The additive identity is ( 0 0 0 0 ) 0 0 0 0 ([0,0],[0,0])\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)(0000) and the multiplicative identity is ( 1 0 0 1 ) 1 0 0 1 ([1,0],[0,1])\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)(1001).
  4. Existence of Additive and Multiplicative Inverse
Additive Inverse
For each A = ( a b b a ) A = a b b a A=([a,-b],[b,a])A = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)A=(abba), the additive inverse is A = ( a b b a ) A = a b b a -A=([-a,b],[-b,-a])-A = \left(\begin{array}{cc} -a & b \\ -b & -a \end{array}\right)A=(abba).
Multiplicative Inverse
For each A = ( a b b a ) A = a b b a A=([a,-b],[b,a])A = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)A=(abba), the multiplicative inverse A 1 A 1 A^(-1)A^{-1}A1 is given by:
A 1 = 1 a 2 + b 2 ( a b b a ) A 1 = 1 a 2 + b 2 a b b a A^(-1)=(1)/(a^(2)+b^(2))([a,b],[-b,a])A^{-1} = \frac{1}{a^2 + b^2} \left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)A1=1a2+b2(abba)
Let’s find the multiplicative inverse of ( 1 1 1 1 ) 1 1 1 1 ([1,-1],[1,1])\left(\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right)(1111).
First, we calculate a 2 + b 2 a 2 + b 2 a^(2)+b^(2)a^2 + b^2a2+b2 for a = 1 a = 1 a=1a = 1a=1 and b = 1 b = 1 b=1b = 1b=1:
a 2 + b 2 = 1 2 + 1 2 a 2 + b 2 = 1 2 + 1 2 a^(2)+b^(2)=1^(2)+1^(2)a^2 + b^2 = 1^2 + 1^2a2+b2=12+12
After calculating, we get a 2 + b 2 = 2 a 2 + b 2 = 2 a^(2)+b^(2)=2a^2 + b^2 = 2a2+b2=2.
Now, we can find A 1 A 1 A^(-1)A^{-1}A1:
A 1 = 1 2 ( 1 1 1 1 ) A 1 = 1 2 1 1 1 1 A^(-1)=(1)/(2)([1,1],[-1,1])A^{-1} = \frac{1}{2} \left(\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right)A1=12(1111)
  1. Distributive Law
    The distributive law A × ( B + C ) = ( A × B ) + ( A × C ) A × ( B + C ) = ( A × B ) + ( A × C ) A xx(B+C)=(A xx B)+(A xx C)A \times (B + C) = (A \times B) + (A \times C)A×(B+C)=(A×B)+(A×C) also holds, as it does for matrices.
Isomorphism f : C S f : C S f:Crarr Sf: \mathbb{C} \rightarrow Sf:CS
The map f ( a + i b ) = ( a b b a ) f ( a + i b ) = a b b a f(a+ib)=([a,-b],[b,a])f(a + ib) = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)f(a+ib)=(abba) is an isomorphism if it preserves addition and multiplication, and is bijective.
Preservation of Addition
f ( ( a 1 + i b 1 ) + ( a 2 + i b 2 ) ) = f ( a 1 + a 2 + i ( b 1 + b 2 ) ) = ( a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 ) f ( ( a 1 + i b 1 ) + ( a 2 + i b 2 ) ) = f ( a 1 + a 2 + i ( b 1 + b 2 ) ) = a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 f((a_(1)+ib_(1))+(a_(2)+ib_(2)))=f(a_(1)+a_(2)+i(b_(1)+b_(2)))=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])f((a_1 + ib_1) + (a_2 + ib_2)) = f(a_1 + a_2 + i(b_1 + b_2)) = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)f((a1+ib1)+(a2+ib2))=f(a1+a2+i(b1+b2))=(a1+a2(b1+b2)b1+b2a1+a2)
f ( a 1 + i b 1 ) + f ( a 2 + i b 2 ) = ( a 1 b 1 b 1 a 1 ) + ( a 2 b 2 b 2 a 2 ) = ( a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 ) f ( a 1 + i b 1 ) + f ( a 2 + i b 2 ) = a 1 b 1 b 1 a 1 + a 2 b 2 b 2 a 2 = a 1 + a 2 ( b 1 + b 2 ) b 1 + b 2 a 1 + a 2 f(a_(1)+ib_(1))+f(a_(2)+ib_(2))=([a_(1),-b_(1)],[b_(1),a_(1)])+([a_(2),-b_(2)],[b_(2),a_(2)])=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])f(a_1 + ib_1) + f(a_2 + ib_2) = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) + \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right) = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)f(a1+ib1)+f(a2+ib2)=(a1b1b1a1)+(a2b2b2a2)=(a1+a2(b1+b2)b1+b2a1+a2)
Preservation of Multiplication
f ( ( a 1 + i b 1 ) × ( a 2 + i b 2 ) ) = f ( a 1 a 2 b 1 b 2 + i ( a 1 b 2 + a 2 b 1 ) ) = ( a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 ) f ( ( a 1 + i b 1 ) × ( a 2 + i b 2 ) ) = f ( a 1 a 2 b 1 b 2 + i ( a 1 b 2 + a 2 b 1 ) ) = a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 f((a_(1)+ib_(1))xx(a_(2)+ib_(2)))=f(a_(1)a_(2)-b_(1)b_(2)+i(a_(1)b_(2)+a_(2)b_(1)))=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])f((a_1 + ib_1) \times (a_2 + ib_2)) = f(a_1a_2 – b_1b_2 + i(a_1b_2 + a_2b_1)) = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)f((a1+ib1)×(a2+ib2))=f(a1a2b1b2+i(a1b2+a2b1))=(a1a2b1b2(a1b2+a2b1)a1b2+a2b1a1a2b1b2)
f ( a 1 + i b 1 ) × f ( a 2 + i b 2 ) = ( a 1 b 1 b 1 a 1 ) × ( a 2 b 2 b 2 a 2 ) = ( a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 ) f ( a 1 + i b 1 ) × f ( a 2 + i b 2 ) = a 1 b 1 b 1 a 1 × a 2 b 2 b 2 a 2 = a 1 a 2 b 1 b 2 ( a 1 b 2 + a 2 b 1 ) a 1 b 2 + a 2 b 1 a 1 a 2 b 1 b 2 f(a_(1)+ib_(1))xx f(a_(2)+ib_(2))=([a_(1),-b_(1)],[b_(1),a_(1)])xx([a_(2),-b_(2)],[b_(2),a_(2)])=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])f(a_1 + ib_1) \times f(a_2 + ib_2) = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) \times \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right) = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)f(a1+ib1)×f(a2+ib2)=(a1b1b1a1)×(a2b2b2a2)=(a1a2b1b2(a1b2+a2b1)a1b2+a2b1a1a2b1b2)
Bijectiveness
The map f f fff is clearly bijective, as each complex number maps to a unique matrix in S S SSS, and each matrix in S S SSS maps to a unique complex number.
Therefore, f f fff is an isomorphism.
In summary, S S SSS is a field under the usual binary operations of matrix addition and matrix multiplication, and the map f : C S f : C S f:Crarr Sf: \mathbb{C} \rightarrow Sf:CS is an isomorphism.
Verified Answer
5/5
\(cos^2\left(\frac{\theta }{2}\right)=\frac{1+cos\:\theta }{2}\)
\(2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

2(a) What are the orders of the following permutations in \(\mathrm{S}_{10}\)? \(\left(\begin{array}{cccccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9\end{array}\right)\) and \((12345)(67)\).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To find the order of a permutation, we need to find the smallest positive integer n n nnn such that applying the permutation n n nnn times returns us to the original arrangement of elements.
  1. For the permutation ( 1 2 3 4 5 6 7 8 9 10 1 8 7 3 10 5 4 2 6 9 ) 1 2 3 4 5 6 7 8 9 10 1 8 7 3 10 5 4 2 6 9 ([1,2,3,4,5,6,7,8,9,10],[1,8,7,3,10,5,4,2,6,9])\left(\begin{array}{cccccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9\end{array}\right)(1234567891018731054269)
First, let’s write this permutation in disjoint cycle notation. The cycles are:
  • ( 2 8 ) ( 2 8 ) (28)(2 \, 8)(28)
  • ( 3 7 4 ) ( 3 7 4 ) (374)(3 \, 7 \, 4)(374)
  • ( 5 10 9 6 ) ( 5 10 9 6 ) (51096)(5 \, 10 \, 9 \, 6)(51096)
The lengths of these cycles are 2 , 3 , 2 , 3 , 2,3,2, 3,2,3, and 4 4 444 respectively.
The order of a permutation in disjoint cycle notation is the least common multiple (LCM) of the lengths of its disjoint cycles.
Let’s substitute the values into the formula for LCM:
LCM ( 2 , 3 , 4 ) LCM ( 2 , 3 , 4 ) “LCM”(2,3,4)\text{LCM}(2, 3, 4)LCM(2,3,4)
After calculating, we get LCM ( 2 , 3 , 4 ) = 12 LCM ( 2 , 3 , 4 ) = 12 “LCM”(2,3,4)=12\text{LCM}(2, 3, 4) = 12LCM(2,3,4)=12.
So, the order of the first permutation is 12 12 121212.
  1. For the permutation ( 12345 ) ( 67 ) ( 12345 ) ( 67 ) (12345)(67)(12345)(67)(12345)(67)
This permutation is already in disjoint cycle notation. The lengths of these cycles are 5 5 555 and 2 2 222.
Let’s substitute the values into the formula for LCM:
LCM ( 5 , 2 ) LCM ( 5 , 2 ) “LCM”(5,2)\text{LCM}(5, 2)LCM(5,2)
After calculating, we get LCM ( 5 , 2 ) = 10 LCM ( 5 , 2 ) = 10 “LCM”(5,2)=10\text{LCM}(5, 2) = 10LCM(5,2)=10.
So, the order of the second permutation is 10 10 101010.
In summary, the order of the first permutation is 12 12 121212 and the order of the second permutation is 10 10 101010.
Verified Answer
5/5
\(a=b\:cos\:C+c\:cos\:B\)
\(b=c\:cos\:A+a\:cos\:C\)

3(a) Let \(J=\{a+b i \mid a, b \in \mathbb{Z}\}\) be the ring of Gaussian integers (subring of \(\mathbb{C}\)). Which of the following is \(J\): Euclidean domain, principal ideal domain, unique factorization domain? Justify your answer.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
The ring J = { a + b i a , b Z } J = { a + b i a , b Z } J={a+bi∣a,b inZ}J = \{ a + bi \mid a, b \in \mathbb{Z} \}J={a+bia,bZ} of Gaussian integers is a subring of C C C\mathbb{C}C. We will examine whether J J JJJ is a Euclidean Domain, a Principal Ideal Domain (PID), or a Unique Factorization Domain (UFD).
  1. Euclidean Domain
A Euclidean Domain is an integral domain equipped with a Euclidean function f f fff that satisfies the division algorithm property: for any two elements a , b a , b a,ba, ba,b in the domain with b 0 b 0 b!=0b \neq 0b0, there exist q , r q , r q,rq, rq,r in the domain such that a = b q + r a = b q + r a=bq+ra = bq + ra=bq+r and either r = 0 r = 0 r=0r = 0r=0 or f ( r ) < f ( b ) f ( r ) < f ( b ) f(r) < f(b)f(r) < f(b)f(r)<f(b).
For Gaussian integers, we can define the Euclidean function f ( a + b i ) = a 2 + b 2 f ( a + b i ) = a 2 + b 2 f(a+bi)=a^(2)+b^(2)f(a + bi) = a^2 + b^2f(a+bi)=a2+b2. Using this function, we can show that J J JJJ satisfies the division algorithm property. Therefore, J J JJJ is a Euclidean Domain.
  1. Principal Ideal Domain (PID)
A Principal Ideal Domain is an integral domain in which every ideal is generated by a single element. Since J J JJJ is a Euclidean Domain, it is also a PID. This is because in a Euclidean Domain, any ideal I I III can be generated by its element of smallest Euclidean value, making it a principal ideal.
  1. Unique Factorization Domain (UFD)
A Unique Factorization Domain is an integral domain in which every non-zero non-unit element can be uniquely factored into irreducible elements, up to ordering and units.
Since J J JJJ is a PID, it is also a UFD. This is a general property: every PID is a UFD.
In summary, the ring J J JJJ of Gaussian integers is a Euclidean Domain, a Principal Ideal Domain, and a Unique Factorization Domain.
Verified Answer
5/5
\(2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)\)
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

3(b) Let \(\mathrm{R}^{\mathrm{C}}=\) ring of all real-valued continuous functions on \([0,1]\), under the operations \[ \begin{aligned} &(f+g) x=f(x)+g(x) \\ &(f g) x=f(x) g(x) \end{aligned} \] Is \(M=\left\{f \in R^C \mid f\left(\frac{1}{2}\right)=0\right\}\) a maximal ideal of \(R\)? Justify your answer.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
An ideal M M MMM in a ring R R RRR is said to be a maximal ideal if M R M R M!=RM \neq RMR and there is no ideal N N NNN such that M N R M N R M⊊N⊊RM \subsetneq N \subsetneq RMNR.
Step 1: Verify M M MMM is an Ideal
First, let’s check if M M MMM is an ideal of R C R C R^(C)R^CRC. M M MMM is the set of all functions f f fff such that f ( 1 2 ) = 0 f 1 2 = 0 f((1)/(2))=0f\left(\frac{1}{2}\right) = 0f(12)=0.
  1. Closed under addition: If f , g M f , g M f,g in Mf, g \in Mf,gM, then f ( 1 2 ) = 0 f 1 2 = 0 f((1)/(2))=0f\left(\frac{1}{2}\right) = 0f(12)=0 and g ( 1 2 ) = 0 g 1 2 = 0 g((1)/(2))=0g\left(\frac{1}{2}\right) = 0g(12)=0. Therefore, ( f + g ) ( 1 2 ) = f ( 1 2 ) + g ( 1 2 ) = 0 + 0 = 0 ( f + g ) 1 2 = f 1 2 + g 1 2 = 0 + 0 = 0 (f+g)((1)/(2))=f((1)/(2))+g((1)/(2))=0+0=0(f + g)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) + g\left(\frac{1}{2}\right) = 0 + 0 = 0(f+g)(12)=f(12)+g(12)=0+0=0. So, f + g M f + g M f+g in Mf + g \in Mf+gM.
  2. Closed under multiplication by an element in R C R C R^(C)R^CRC: If f M f M f in Mf \in MfM and g R C g R C g inR^(C)g \in R^CgRC, then f ( 1 2 ) = 0 f 1 2 = 0 f((1)/(2))=0f\left(\frac{1}{2}\right) = 0f(12)=0. Therefore, ( f g ) ( 1 2 ) = f ( 1 2 ) g ( 1 2 ) = 0 ( f g ) 1 2 = f 1 2 g 1 2 = 0 (f*g)((1)/(2))=f((1)/(2))*g((1)/(2))=0(f \cdot g)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right) = 0(fg)(12)=f(12)g(12)=0. So, f g M f g M f*g in Mf \cdot g \in MfgM.
Since M M MMM is closed under addition and multiplication by an element in R C R C R^(C)R^CRC, M M MMM is an ideal of R C R C R^(C)R^CRC.
Step 2: Check for Maximal Ideal
To show that M M MMM is a maximal ideal, we need to show that there is no ideal N N NNN such that M N R C M N R C M⊊N⊊R^(C)M \subsetneq N \subsetneq R^CMNRC.
Suppose there exists an ideal N N NNN such that M N R C M N R C M⊊N⊊R^(C)M \subsetneq N \subsetneq R^CMNRC. Then there must be a function g N g N g in Ng \in NgN but g M g M g!in Mg \notin MgM. This means g ( 1 2 ) 0 g 1 2 0 g((1)/(2))!=0g\left(\frac{1}{2}\right) \neq 0g(12)0.
Consider the function h ( x ) = 1 g ( x ) h ( x ) = 1 g ( x ) h(x)=1-g(x)h(x) = 1 – g(x)h(x)=1g(x) for x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in [0, 1]x[0,1]. Since N N NNN is an ideal and g N g N g in Ng \in NgN, h N h N h in Nh \in NhN as well. Now, h ( 1 2 ) = 1 g ( 1 2 ) 0 h 1 2 = 1 g 1 2 0 h((1)/(2))=1-g((1)/(2))!=0h\left(\frac{1}{2}\right) = 1 – g\left(\frac{1}{2}\right) \neq 0h(12)=1g(12)0 and h ( x ) 0 h ( x ) 0 h(x)!=0h(x) \neq 0