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The image contains questions related to Algebra-4. Below, I will provide concise answers for each of the problems so that they can be easily written on paper.

### 1. **Prove that matrices form a ring. Does this hold for invertible matrices?**

   - **Answer**: Yes, matrices over a ring (like \( \mathbb{R} \) or \( \mathbb{Z} \)) form a ring because they satisfy the ring axioms: closure under addition and multiplication, associativity, distributivity, existence of an additive identity, and the presence of additive inverses. However, invertible matrices (non-singular matrices) do not form a ring because they are not closed under addition (the sum of two invertible matrices may not be invertible).

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### 2. **Is the homomorphism of rings preserved when cubing or squaring \( A \)?**

   - **Answer**: No, in general, a ring homomorphism does not necessarily preserve the squaring or cubing of elements. The property preserved by homomorphisms is that \( f(A^2) = f(A) \cdot f(A) \), but \( f(A^3) \) is not necessarily equal to \( f(A)^3 \).

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### 3. **Is the mapping \( g \rightarrow g^{-1} \) a homomorphism for invertible elements of a ring \( R \)?**

   - **Answer**: No, the map \( g \mapsto g^{-1} \) for invertible elements of a ring is generally not a homomorphism. A homomorphism requires that the product be preserved, i.e., \( f(ab) = f(a)f(b) \), but for \( f(g) = g^{-1} \), \( f(ab) = (ab)^{-1} \neq a^{-1}b^{-1} \) unless \( a \) and \( b \) commute (which happens in commutative rings).

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### 4. **Prove that \( f: A \rightarrow B \) is an isomorphism of rings if and only if \( \ker f = 0 \).**

   - **Answer**: The statement is incorrect. A homomorphism \( f: A \to B \) is injective (one-to-one) if and only if \( \ker(f) = \{0\} \), but for it to be an isomorphism, \( f \) must also be surjective (onto). Hence, the isomorphism condition is both \( \ker(f) = 0 \) and \( f \) being surjective.

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### 5. **Provide an example of an ideal where \( \ker f = 0 \).**

   - **Answer**: Consider the ring \( \mathbb{Z} \) and the identity map \( f: \mathbb{Z} \rightarrow \mathbb{Z} \), which is an isomorphism. Here, \( \ker(f) = 0 \) because the only element mapped to 0 is 0 itself. Hence, \( \mathbb{Z} \) is an example where the kernel is trivial.

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### 6. **Prove that every ideal in \( \mathbb{Z} \) is principal.**

   - **Answer**: Every ideal \( I \) in \( \mathbb{Z} \) is of the form \( I = (d) \), where \( d \) is the greatest common divisor (gcd) of all the elements in the ideal. Hence, \( I \) is generated by a single integer, making it a principal ideal.

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### 7. **Prove that the matrix ring \( M_n(K) \) over a field \( K \) has no non-trivial two-sided ideals.**

   - **Answer**: The matrix ring \( M_n(K) \), where \( K \) is a field, is simple, meaning it has no non-trivial two-sided ideals. This is because any two-sided ideal \( I \) in \( M_n(K) \) must be either \( \{0\} \) or the whole ring \( M_n(K) \). This result is a consequence of the fact that every matrix can be expressed in terms of elementary matrices, which generate the whole ring.

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Feel free to ask for more details or clarifications!

### Follow-up Questions:
1. What are the properties of ring homomorphisms in detail?
2. Why do invertible matrices fail to form a ring under addition?
3. Can you explain the kernel of a ring homomorphism with more examples?
4. What is the structure of ideals in commutative vs. non-commutative rings?
5. How can you prove that the matrix ring \( M_n(K) \) is simple?

### Tip:
Remember that for a ring homomorphism to be an isomorphism, it must be both injective and surjective.

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Explore detailed solutions for Algebra-4 questions covering topics like ring theory, homomorphisms, matrix theory, and ideals. Suitable for undergraduate students studying advanced algebra concepts.

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