Answer to Question #135136 in Abstract Algebra for Ram

Question

Using Fundamental theorem of homomorphism to prove that the rings R^2 and R^2/R^4 are isomorphic.

Answer 1

$$Solution$$

Theorem 1 (The Fundamental Theorem of Ring Homomorphisms): Let $(R, +_1, *_1)$ and $(S, +_2, *_2)$ be homomorphic rings with ring homomorphism $\phi: R \to S.$ Then $R/ker(\phi) \cong \phi (R)$

.

Let $R_4 = ker(\phi)$ and let $\Phi :R_2/R_4 \to \phi(R_2)$ be defined for all $(a + R_4) \in R_2$ by $\phi (a+R_4)=\phi(a)$

Then $\psi$ is well defined, for if $a+R_4=b+R_4 \implies a=b+ r$ for some $r \in R_4$ and so:

$$\psi (a+R_4)=\psi ((b+r)+R_4)=\psi((b+R_4)+R_4)=\psi(b+R_4)$$

To show that $R_2/R_4$ is an isomorphism to $\phi (R_2)$, we need to show that $\psi$ is bijective.

Let $(a+R_4), (b+R_4) \in R_2/R_4$ and suppose that

$\psi(a+R_4)=\psi(b+R_4).$

Then $\phi(a)=\phi(b)$

So,

$\phi(a-b)=0$

So,

$a-b \in K,$

So, $a+R_4=b+R_4$

Hence $\psi$ is injective

Furthermore, $\forall a \in \phi(R_2)$ we have that $(a +I) \in R_2/R_4$ is such that $\psi (a + R_4)=a$.

So $\psi$ is surjective.

Hence $\psi$ is bijective and so $R_2/R_4$ is proven to be an isomorphism, that is

$$R_2/R_4\cong \phi(R_2)$$