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).

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 $f(A)^3$.

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).

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.

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.

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.

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.

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.