Question

Question 7: If is a ring, show that $\mathit{R}\mathbf{/}\mathbf{\left(}{\mathbf{0}}_{\mathbf{R}}\mathbf{\right)}\mathbf{\cong }\mathit{R}$.

Short answer

Answer:

It is proved that $\mathit{R}\mathbf{/}\mathbf{\left(}{\mathbf{0}}_{\mathbf{R}}\mathbf{\right)}\mathbf{\cong }\mathit{R}$.

Step-by-step solution

 Step 1: First Isomorphism Theorem

Let$f:R\to S$ be a surjective homomorphism of rings with kernel . Then, the quotient ring R/K is isomorphic to S .

Proof the part

Let be a ring.

Consider the identity map=$\psi =R\to R$.

Since $\psi$is an identity map, so it is a surjective homomorphism with kernel (0_R).

Thus, by first Isomorphism theorem,$R/\left(0_{\mathrm{R}}\right)\cong R$ .

We can conclude that$R/\left(0_{\mathrm{R}}\right)\cong R$