Question #145988
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.
1
Expert's answer
2020-11-24T16:53:11-0500

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

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

Homomorphism: f[(a+2b)+(c+2d)]=f[a+c+2(b+d)]f[(a+\sqrt{2}b)+(c+\sqrt{2}d)]=f[a+c+\sqrt{2}(b+d)]=[a+c2b+2db+da+c].=\begin{bmatrix} a+c & 2b+2d \\ b+d & a+c \end{bmatrix}. =[a2bba]+[c2ddc]=f(a+2b)+f(c+2d).=\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+2b)(c+2d)]=f[ac+2bd+2(bc+ad)]f[(a+\sqrt{2}b)(c+\sqrt{2}d)]=f[ac+2bd+\sqrt{2}(bc+ad)] =[ac+2bd2bc+2adbc+adac+2bd].=\begin{bmatrix} ac+2bd & 2bc+2ad \\ bc+ad & ac+2bd \end{bmatrix}. =[a2bba][c2ddc]==\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}\begin{bmatrix} c & 2d \\ d & c \end{bmatrix}= f(a+2b)f(c+2d).f(a+\sqrt{2}b)f(c+\sqrt{2}d). Also f(1)=[1001]f(1)=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} . Hence homomorphism


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