Question

I want an original answer.

Answer 1

The map
`\(f: \mathbb{Q}[√(2), √(3)] \to \mathbb{Q}[√(2), √(3)]\)` given by
`\(f(a+b√(2)+c√(3)+d√(6)) = a - b√(2) + c√(3) - d√(6)\)` is a ring isomorphism, preserving addition and multiplication, and is bijective.

To prove that the given map
`\( f: \mathbb{Q}[√(2), √(3)] \to \mathbb{Q}[√(2), √(3)] \)` defined by
`\( f(a + b√(2) + c√(3) + d√(6)) = a - b√(2) + c√(3) - d√(6) \)` is a ring isomorphism, we need to show two things:

1. f is a homomorphism: This means that for any elements x, y in the domain,
`\( f(x + y) = f(x) + f(y) \)` and
`\( f(xy) = f(x)f(y) \)`.

2. f is bijective: This means that f is both injective (one-to-one) and surjective (onto).

Let's go through each step:

Homomorphism:

Let
`\( x = a_1 + b_1√(2) + c_1√(3) + d_1√(6) \) and \( y = a_2 + b_2√(2) + c_2√(3) + d_2√(6) \)` be elements in the domain.

`\[ f(x + y) = f((a_1 + a_2) + (b_1 + b_2)√(2) + (c_1 + c_2)√(3) + (d_1 + d_2)√(6)) \]\[ = (a_1 + a_2) - (b_1 + b_2)√(2) + (c_1 + c_2)√(3) - (d_1 + d_2)√(6) \]\[ = (a_1 - b_1√(2) + c_1√(3) - d_1√(6)) + (a_2 - b_2√(2) + c_2√(3) - d_2√(6)) \]\[ = f(x) + f(y) \]`

This shows that f preserves addition.

Now let's consider multiplication:

`\[ f(xy) = f((a_1a_2 + 2b_1b_2 + 3c_1c_2 + 6d_1d_2) + (a_1b_2 + b_1a_2 + 2c_1d_2 + d_1c_2)√(2) + (a_1c_2 + c_1a_2 + b_1d_2 + d_1b_2)√(3) + (a_1d_2 + d_1a_2 + b_1c_2 + c_1b_2)√(6)) \]`

`\[ = (a_1a_2 + 2b_1b_2 + 3c_1c_2 + 6d_1d_2) - (a_1b_2 + b_1a_2 + 2c_1d_2 + d_1c_2)√(2) + (a_1c_2 + c_1a_2 + b_1d_2 + d_1b_2)√(3) - (a_1d_2 + d_1a_2 + b_1c_2 + c_1b_2)√(6) \]`

`\[ = (a_1 - b_1√(2) + c_1√(3) - d_1√(6))(a_2 - b_2√(2) + c_2√(3) - d_2√(6)) \]\[ = f(x)f(y) \]`

This shows that f preserves multiplication.

Bijectiveness:

Now, let's show that f is bijective.

- Injective (One-to-One):

Suppose f(x) = f(y) for some x, y in the domain. Then,

`\[ a_1 - b_1√(2) + c_1√(3) - d_1√(6) = a_2 - b_2√(2) + c_2√(3) - d_2√(6) \]`

This implies that
`\( a_1 = a_2 \), \( b_1 = b_2 \)`,
`\( c_1 = c_2 \)`, and
`\( d_1 = d_2 \)`. Therefore, x = y . and f is injective.

- Surjective (Onto):

For any element
`\( y = a - b√(2) + c√(3) - d√(6) \)` in the codomain, consider
`\( x = a + b√(2) + c√(3) + d√(6) \)` in the domain. Then, f(x) = y showing that f is surjective.

Since f is both injective and surjective, it is bijective.

Therefore, the map f is a ring isomorphism.