We Have Numbers Of Free Samples

Question 1

1. Show that the ring Q[i], where Q is the field of rational numbers, and i2 = -1, is a (commutative) field. (Hint: give brief reasons for things like associativity, distributivity, inverses.)
2. Is the cardinality, |Q[i]|, of this field countable? (Hint: if so, this is equivalent to establishing a
3. bijection between Q[i] and a known countable set such as Z or Q.)
4. Is the cardinality of C, the complex numbers, countable? (Yes or no)
5. Explain why the fields Q[i] and C cannot be isomorphic?

Solution: –

a) Since Q is the field of rational numbers ,thus it has the property of the form of rational numbers such as the numbers in the form of a/b where a b are the integers where So ,lets say that we have two rational numbers as a=3/2+0i & b=4/3+0i so a/b =9/8 which is also rational as it satisfies the property of rational numbers .So ,now as per the commutative law:-

a+b =b+a

Which proves the commutative law as the number also belongs  Q .

b) Since in the above solution we have seen that for a particular number from Q on the operation process its commutative nature, and the result also exists in the same field. The cardinality of Q[i] is also proved as the order pairing of the numbers for the complex numbers are defined from the multiplicative identity.

c) Yes, the countability of the complex numbers exists and also possess continuum.

d) Since the complex numbers have the property of representing in the polar form where the numbers can be represented in the form of the sin as the real magnitude and the cosine as the virtual/complex feature, the representation of the rational numbers is not possible in doing so.

Question 2

1. Show how to construct (using 2×2 matrices over the ring Q[i], where i^2 = -1) a proper skew field H. (Hint: there is a similar exercise in the Hungerford book. Only give brief reasons why this is a skew field.)
2. What is the inverse of a general element of H.?

Solution: –

a) for the Construction of proper skew field H, we need following properties to be satisfied: –

Let’s say for a matrix defined as

So for the skew filed satisfaction: –

1. 1.The closure property, and thus, the abelian group must be satisfied
2. The above should satisfy distributive law. (associativity)
3. It must have unit element. (Identity)
4. Every non zero element of field has its multiplicative inverse. (Inversibility)
5. It shows commutativity for multiplicative inverse.

b) the inverse of the general element of H will be the conjugative inverse of that element of H.

So while constructing the proper skew field, the above properties needs to be verified, if it exists, then the field is a proper skew field.

Question 3

1. If p is prime, what is the factorization of x^p-x in Zp[x] into irreducible factors?
2. Use part (a) to find a polynomial fb in Zp[x] of degree p-1, where the evaluation fb(x) at x in Zp gives a function taking any element b in Zp to 1, and all the other elements of Zp{b} to zero.
3. Use part (b) to show that any function Zp to Zp is given by some polynomial of degree at most p-1 in Zp[x].
4. Show that the polynomial in part (c) is unique.
5. Calculate the polynomial function of degree at most 6 that takes each non-zero square in Z_7 to 1, takes non-squares to -1, and 0 to 0.

Solution: –

a) If p is prime then the criterion for the irreducible nature is as per the theorem: –

If F is a field and if f(x), 2 F[x] is a polynomial of degree 2 or 3, then f(x) is reducible over F[x] if and only if the polynomial f(x) has a root in F.

So based upon the theorem: –

At p=3 and x=1           Z₃[1]=1, Z₃[2]=2, Z₃[3]=1……………….

As we can see that the for finding the factorization of Zp[x]/x^p-x we need to find the divisor and for that we have to implement above formula as shown which gives the above result.

b) &c) for the function having degree p-1, say 2 from the above assumptions, the values are now as follows:

At p=2 and x=1           Z₂[1]=1, Z₂[2]=2, Z₂[3]=1/3……………….

Now, since the degree is reduced, the new polynomial is formed accordingly,

So fb(x) is xp^-x-1

On the evaluation of it, all the conditions are met.

d) As we know that in the polynomial, the nature of the degrees is in the sequential manner, but the clear property is maintained in which it is impossible to have same degree represented in two different variables. So, as per the property of polynomial, the degree is unique in nature as no other possession is identified in the same.

e) For this to be true, we take: –

f(x)=x²+x as the function, so for the values at most 6 we have:-

f(0)=0

f(1)=2

f(2)=6

f(3)=12

f(4)=20

f(5)=30

as there can be many examples for that, which dependent upon the condition of uniqueness of the polynomial.

Question 4

Consider a ring (R,+,.) with identity 1 that contains two non-identity elements a≠b that satisfy a² =a and b² =b. Suppose that every x in R satisfies the equation x³ = x.

Show that R is a commutative ring.

(Hint: consider the third powers of combinations of a and b such as a+b, a-b, or ab, etc. and look at the resulting equations involving a and b.)

Solution: – For proving the commutativity of the system, for the ring system

For the commutative rule to be proved: –

We have first identity as                             a+b=b+a

Which proves the commutative property for the x³=x.

Question 5

1. Assuming R is a (commutative) ring as in Question 4, show that the number of its elements, |R|, is finite.
2. Show how to construct two non-isomorphic rings R and R’ that satisfy the conditions of Question 4 and 5(a).

Solution: –

a)

b)

[Download not found]

Download