Show R = {a + b√2} and R₂ are subrings and find an isomorphism

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3. Let [tex]R = a+b \sqrt{2}[/tex] , a,b is integer and let [tex]R_{2}[/tex] consist of all 2 x 2
matrices of the form [tex][\begin{array}{cc} a & 2b \\ b & a \\ \end{array} }][/tex]

Show that R is a subring of [tex]Z(integer)[/tex] and [tex]R_{2}[/tex] is a subring of [tex]M_{2} (Z)[/tex]. Also. Prove that the mapping from R to [tex]R_{2}[/tex] is a isomorphism.
 
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I am lost totally in this question. I know i need to do this.
need to show it is closed under addition, multiplication containing identity to prove it is a subring and show that it is surjective and injective. to show that it is isomorphism.

Can you help me out in this question?
 
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that the part i am struck with. [tex]a+b\sqrt{2} + c+d\sqrt{2}[/tex] ? but [tex]\sqrt{2} not an integer[/tex]
 
dreamer.ande said:
that the part i am struck with. [tex]a+b\sqrt{2} + c+d\sqrt{2}[/tex] ? but [tex]\sqrt{2} not an integer[/tex]

You will want to write [tex]a+b\sqrt{2}+c+d\sqrt{2}[/tex] in the form [tex]e+f\sqrt{2}[/tex], since by definition, elements in R have this form.

We are not claiming that [tex]\sqrt{2}[/tex] is an integer!
 
how many axiom do i need to show to prove that it is a subring of a ring?
There are five axiom to show?
How to show that first axiom : containment. R belong to Z?

Can you show me the proper step of proving in such question as I always have problem in such question?
 
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dreamer.ande said:
how many axiom do i need to show to prove that it is a subring of a ring?

Normally all of them. But luckily you have theorems that limit the amount of work you need to do.

A certain theorem tells you that you only need to show that R is closed under addition and multiplication, that 0 and 1 are in R, and that -a is in R for every a.

So you only need to show 5 things. However, if you did not see such a theorem, then you will need to show all the axioms!
 
Could you show me how to do it for the first part and I try out the second part?
Sorry, I am new to this thing. Very Confusing for me.