Prove that all the non-zero elements in an integral domain have same additive order, which is the characteristic of \(R\) if char \(R >0\)and infinite if char \(R=0\) .

If \(a\) is a nilpotent element in a ring \(R\) with identity 1, then show that  \(1-a\) is a unit in \(R\). Further, show that if \(R\) is commutative, then \(1-ab\) is a unit for all  \(b \in R\) .

Let \(R\) be a non-zero commutative ring with identity. If \(R\) has no non-trivial ideals, prove that \(R\) is a field.

Show that there cannot be an integral domain with 6 elements.

Let \(R\) be a commutative ring with characteristic \(p\), where \(p\) is a prime. Show that \((a+b)^{p^n}=a^{p^n}+b^{p^n}\) and \((a-b)^{p^n}=a^{p^n}-b^{p^n}\) for all integers \(n \ge0\) . Also show that if \(R\) is an integral domain, then the mapping \(a \mapsto a^p\) is a monomorphism of \(R\).

Show that every endomorphism of a field is either trivial or a monomorphism. Hence show that every non-trivial endomorphism of a finite field is an automorphism.

Let \(I\) and \(J\) be non-zero ideals of an integral domain. Prove that \(I \cap J \neq \{0\}\) .

Show that the ring \(M_n(\mathbb{F})\) of \(n \times n\) matrices over a field \(\mathbb{F}\) is a simple ring. Hence deduce that simple rings may have one-sided proper ideals.

Let \(R\) be a commutative ring. Show that the set \(N=\{ a\in R |a^m =0\) for some positive integer m\(\}\) is an ideal of \(R\) which is contained in every prime ideal of \(R\). Further show that \(R/N\) is a ring with no non-zero nilpotent elements. \((N\) is called the nilradical of \(R)\).

Let \(R\) be a commutative ring and \(I\) an ideal of \(R\), and let \(\sqrt{I} =\{a \in R |a^m \in I \) for some positive integer \(m\}\) Show that

(i) \(\sqrt{I}\) is an ideal of \(R\) and \(I \subseteq \sqrt{I}\) ,

(ii) \(\sqrt{\sqrt{I}}=\sqrt{I}\) ,

(iii) If \(R\) is with identity and \(\sqrt{I} =R\), then \(I=R\) . What is \(\sqrt{I}\), when \(I=\{0\}\). (\(\sqrt{I}\) is called the radical of \(I\)).

If \(n > 1\) is a positive integer and \(n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\) is the prime factorization of \(n\), then show that  \(\sqrt{\langle n \rangle }=\langle p_1p_2 \cdots p_k \rangle\) .