Answer to Question #145988 in Abstract Algebra for Sourav Mondal

Question

Let R = Z + √2 Z and S = the ring of 2 x 2
matrices of the form [
a 2b
b. a
. ] Where a,b belongs to Z

Show that f : R --> S defined by
f (a + √2 b) = [
a. 2b
b. a
] is an isomorphism of rings.

Answer 1

Surjectivity is obvious since S is the set of matrices of that special form.

Injectivity: $f(a+\sqrt{2}b)=f(c+\sqrt{2}d)$. Hence, equating the columns of the matrices we get, $a=c,b=d.$ Hence injective.

Homomorphism: $f[(a+\sqrt{2}b)+(c+\sqrt{2}d)]=f[a+c+\sqrt{2}(b+d)]$$=\begin{bmatrix} a+c & 2b+2d \\ b+d & a+c \end{bmatrix}.$ $=\begin{bmatrix} a & 2b \\ b & a \end{bmatrix} +\begin{bmatrix} c & 2d \\ d & c \end{bmatrix} = f(a+\sqrt{2}b)+f(c+\sqrt{2}d).$ Also $f[(a+\sqrt{2}b)(c+\sqrt{2}d)]=f[ac+2bd+\sqrt{2}(bc+ad)]$ $=\begin{bmatrix} ac+2bd & 2bc+2ad \\ bc+ad & ac+2bd \end{bmatrix}.$ $=\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}\begin{bmatrix} c & 2d \\ d & c \end{bmatrix}=$ $f(a+\sqrt{2}b)f(c+\sqrt{2}d).$ Also $f(1)=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ . Hence homomorphism