Rings & Ideals MCQ Quiz in తెలుగు - Objective Question with Answer for Rings & Ideals - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Explanation:

Explanation:

Let A, B ∈ S then

A = \(\begin{pmatrix}a_1&0\\b_1&0\end{pmatrix}\), B = \(\begin{pmatrix}a_2&0\\b_2&0\end{pmatrix}\)

Now, A - B = \(\begin{pmatrix}a_1&0\\b_1&0\end{pmatrix}\) - \(\begin{pmatrix}a_2&0\\b_2&0\end{pmatrix}\) = \(\begin{pmatrix}a_1-a_2&0\\b_1-b_2&0\end{pmatrix}\) ∈ S

Let X = \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) be an arbitrary element of R and let A = \(\begin{pmatrix}p&0\\q&0\end{pmatrix}\)

XA = \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\)\(\begin{pmatrix}p&0\\q&0\end{pmatrix}\) = \(\begin{pmatrix}ap+bq&0\\cp+dq&0\end{pmatrix}\) ∈  S

∵ ap + bq, cp + dq are integers.

∴ A ∈ S, X ∈ R ⇒ XA ∈ S

Hence S is a left ideal of R

Also if A = \(\begin{pmatrix}1&0\\1&0\end{pmatrix}\) ∈ S and X = \(\begin{pmatrix}1&3\\0&1\end{pmatrix}\)   ∈ R then 

AX = \(\begin{pmatrix}1&0\\1&0\end{pmatrix}\)\(\begin{pmatrix}1&3\\0&1\end{pmatrix}\) = \(\begin{pmatrix}1&3\\1&3\end{pmatrix}\) ∉ S

Hence S is not the right ideal of R

(1) is correct

Explanation:

x2 - 5x + 6 = (x - 2)(x - 3) in \(\mathbb Z\)

x2 - 7x + 12 = (x - 3)(x - 4) in \(\mathbb Z\)

x2 - 9x + 20 = (x - 4)(x - 5) in \(\mathbb Z\)

All of these are not irreducible over \(\mathbb Z\)

(4) is correct

Explanation:

R = {(aij)2×2 : aij ∈ ℤ} is a ring with respect to matrix addition and matrix multiplication

aij is 2 × 2 matrix whose elements are integers

Let \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) be a units of R then a, b, c, d ∈ ℤ

\(\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\) = \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) which implies

a = ± 1, b = ± 1, c = ± 1, d = ± 1

So, each entries of the matrix has two possibilities.

So, 8 such matrix exist.

Hence number of units in R is 8.

(1) is true.

Concept:

Maximal ideal: An ideal m in a ring A is called maximal if m 6= A and the only ideal strictly containing m is A.

Prime ideal: An ideal P in a ring A is called prime if P ≠A and if for every pair x, y of elements in A \ P we have xy ∉ P.

Integral domain: A commutative ring with identity is said to be an integral domain if it has no zero divisors.

Explanation:

X = (0, 1) is the open unit interval and C(X, ℝ) is the ring of continuous functions from X to ℝ.

For any x ∈ (0, 1), I(x) = {f ∈ C(X, ℝ)∣ f(x) = 0}.

Then by theorem, I(x) is a maximal ideal and is primal ideal

Hence option (1) and option (2) are true.

 Let f(x) = \(\begin{cases}x-\frac12& 0 and g(x) = \(\begin{cases}0& 0

Then f(x) and g(x) are non-zero but their product is zero.

Hence option (4) is false

C(X, ℝ) is equal to I(x) if X is a compact set not for all x ∈ X

Hence option (3) is false.

Given:

𝒮1 = ℤ[𝑥] / 〈2, 𝑥3〉 ⁄ and 𝒮2 = ℤ2[𝑥] / 〈𝑥2〉

Concept:

Every non zero prime ideal of commutative ring with unity is maximal .

Calculation:

𝒮1 = ℤ[𝑥] / 〈2, 𝑥3〉 

= {ax² + bx + c + 〈2, 𝑥3〉 | a, b, c ∈ {0, 1} }

= { x² + x +1, x² + 1, x, 1, 0, x +1, x² + x , x² }

Now,

\(\rm (x)^2=0,\ (x^2)^3=0\)

\(\rm (x+1)(x^2+x+1)=x^3+2x+2x^2+1=1\)

\(\rm (x^2+1)(x^2+1)=x^4+2x^2+1=1\)

Hence All elements area either nilpotent or unit .

Now,

\(\rm x\cdot x^2 ∈\ <0>\) Then this is not prime ideal .

we know that Every non zero prime ideal of commutative ring with unity is maximal .

Hence every prime ideal of S₁ is maximal .

 and 𝒮2 = ℤ2[𝑥] / 〈𝑥2〉

= { ax + b + 〈𝑥2〉 | a, b ∈ {0, 1} }

= { x, x+1, 0 , 1 }

( x + 1 ) is an unit .

(x) is nilpotent .

Hence every element of S₂ is either nilpotent or unit.

〈𝑥〉 is only maximal ideal of S_{2 .}

Hence the options (1), (2) and (3) are correct.

Explanation:

Explanation:

U is an ideal of ring R and 1 ∈ U

So for r ∈ R, 1r ∈ U ⇒ r ∈ U

∴ r ∈ R ⇒ r ∈ U

So, R ⊆ U

Also, U ⊆ R always

Combiniting both we get U = R

(2) is correct

Explanation -

An element ' a ' in an Euclidean domain is a unit  iff d(a)=d(1)

Therefore option (2) is correct.

Concept:

(i) If R₁ & R₂ are two commutative ring with unity and ϕ∶ R₁→R₂ is a ring homomorphism then, If I is a prime ideal in R₂ then ϕ^-1 (I) is a prime ideal in R₁.

(ii) If R is finite commutative ring with unity then every ideal of R is maximal.

(iii) Let R be commutative ring with unity such that ∀ a ∈ R, ∃ n ∈ N such that aⁿ = a, then every prime ideal is maximal.

(iv) In a Principal ideal domain, every non-trivial prime ideal is a maximal ideal.

Explanation:

Here Q ⊆ S be a non-zero prime ideal so by result (i),  f-1(Q) is a non-zero prime ideal in R

Option (1) is true

Since f-1(Q) is a non-zero prime ideal in R so by result (iv), f-1(Q) is a maximal ideal in R if R is a PID

Option (2) is true

Since f-1(Q) is a non-zero prime ideal so f-1(Q) is ideal. So by result (ii), f-1(Q) is a maximal ideal in R if R is a finite commutative ring with unity.

Option (3) is true

By result (iii), here n = 5, so f-1(Q) is a maximal ideal in R if x5 = x for all x ∈ R

Option (4) is true