Verified answer

1) Suppose that a + b√d = 0 for some a, b in Q.

==> a = -b√d

Since d is squarefree, we know that √d is irrational.

==> -b√d is irrational if b is nonzero.

This is a contradiction of a being rational unless b = 0.

So, b = 0, and thus a = 0.

Hence, {1, √d} is linearly independent over Q.

Spanning is straightforward to show, because √d^2 = d and so any polynomial in √d

can be simplified to that of the form a + b√d for some a, b in Q.

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2) We need to be a bit more specific about the subring/field of matrices.

We'll call it L:

L = {(a bd; b a): a, b in Q}; it's easy to check if needed that this is closed under addition, multiplication, as well as their inverses.)

So, T: K---> L as defined in your problem.

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First, we show that T is a ring homomorphism.

(Once we show that T is also a bijection, it follows that K is also a field, because Q(√d) is a field; so inverse correspond to inverses.)

Let a + b√d, and r + s√d be elements of Q(√d).

T(a + b√d) + T(r + s√d)

=.[a bd]....[r sd]

...[b a].+.[s r]

=.[(a+r) (b+s)d]

...[(b+s) (a+r)..]

= T((a+r) + (b+s)√d)

= T((a + b√d) + (r + s√d)).

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T(a + b√d) * T(r + s√d)

=.[a bd][r sd]

...[b a][s r]

=.[ar+bds...(as+br)d]

...[as+br.......ar+bds]

= T((ar+bds) + (as+br)√d)

= T((a + b√d) * (r + s√d)).

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Next, we show that T is onto.

Given

[a bd]

[b a] in L, note that a + b√d is in Q(√d), and T(a+b√d) = (a bd; b a) as required.

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Finally, we check that T is 1-1 by examining its kernel.

ker T = {a+b√d in Q(√d) : T(a+b√d) = 0 [matrix]}

.........= {a+b√d in Q(√d) : (a bd; b a) = (0 0; 0 0)}

.........= {a+b√d in Q(√d) : a = b = 0}, by equating like matrix entries

.........= {0}.

Hence, T is an isomorphism.

I hope this helps!

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