Question

Prove that Z[√2] is isomorphic to

Matrix H = a 2b

b a

Where a,b∈Z as rings.

Answer 1

$Let \\Z[ √ 2] =\{ a + b √ 2|a, b ∈ Z\} and$

$H=\{ \begin{bmatrix} a&2b\\ b&a \end{bmatrix} | a,b \isin Z \}$

show that Z[√2] and Hare isomorphic as rings

solution.

define a function

$f: H\to Z[\sqrt{2}] by \\ f\begin{pmatrix}\begin{bmatrix}a&2b\\b&a\end{bmatrix}\end{pmatrix}=a+b\sqrt{2}\\ then, for A=\begin{bmatrix}a&2b\\b&a\end{bmatrix}, C=\begin{bmatrix}c&2d\\d&c\end{bmatrix}\isin H, we have\\ f\begin{pmatrix}\begin{bmatrix}a&2b\\b&a\end{bmatrix}+\begin{bmatrix}c&2d\\d&c\end{bmatrix}\end{pmatrix}=\\ f\begin{pmatrix}\begin{bmatrix}a+c&2b+2d\\b+d&a+c\end{bmatrix}\end{pmatrix}\\= (a+b)+(b+d)\sqrt{2}\\ =(a+b\sqrt{2})+(c+d\sqrt{2})\\ =f\begin{pmatrix}\begin{bmatrix}a&2b\\b&a\end{bmatrix}\end{pmatrix}+ f\begin{pmatrix}\begin{bmatrix}c&2d\\d&c\end{bmatrix}\end{pmatrix}\\ and\\ f\begin{pmatrix}\begin{bmatrix}a&2b\\b&a\end{bmatrix}.\begin{bmatrix}c&2d\\d&c\end{bmatrix}\end{pmatrix}=\\ f\begin{pmatrix}\begin{bmatrix}ac+2bd&2ad+2bc\\ad+bc&ac+2bd\end{bmatrix}\end{pmatrix}=(ac+2bd)+(ad+bc)\sqrt{2}\\ =(a+b\sqrt{2})(c+d\sqrt{2})\\ =f\begin{pmatrix}\begin{bmatrix}a&2b\\b&a\end{bmatrix}\end{pmatrix}.f\begin{pmatrix}\begin{bmatrix}c&2d\\d&c\end{bmatrix}\end{pmatrix} \\$

Hence, f is a ring homomorphism. Now, f is surjective since, given any x ∈ Z[√2], we have that x = a + b√2 for some a, b ∈ Z, so A ∈ H and f(A)=a+b√2=x.

finally to see that f is injective

suppose that A,C ∈ H and f(A)=f(C)

this mean that a+b√2=c+d√2.

if b not equal to d (so that d-b is not equal to zero). then this would imply that √2 =(a-c)/(d-b) is rational, a contradiction. hence we must have that b=d.

hence we have

a + b√2 = c + d√2 = c + b√2.

Subtracting b√2 from both sides of the above equation shows that a = c. Thus,

A=C

, so it follows that f is injective. Hence, f is a ring isomorphism.