Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field

Question

Question:

Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the operations - addition and multiplication where addition is given by: $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$

My work:

Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.

Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$

The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$

The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$

The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$

The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.

If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.

Answer 1

We just need to simplify that expression so that it has the form: $$ \alpha + \beta\sqrt{5} $$ for some $\alpha,\beta \in \mathbb Q$. Indeed, we can do this via conjugates: \begin{align*} \frac{1}{a + b\sqrt 5} &= \frac{1}{a + b\sqrt 5} \cdot \frac{a - b\sqrt 5}{a - b\sqrt 5} \\ &= \frac{a - b\sqrt 5}{a^2 - 5b^2} \\ &= \underbrace{\frac{a}{a^2 - 5b^2}}_{\in ~ \mathbb Q} + \underbrace{\frac{-b}{a^2 - 5b^2}}_{\in ~ \mathbb Q}\sqrt 5 \end{align*} To finish this off, can you see why the denominator $a^2 - 5b^2$ is never zero if $a$ and $b$ are not both zero?

Answer 2

Every non-zero element of $\mathbb Q[√5]$ is a unit. And other proof you have already shown. That's why we can conclude it's a field.

Answer 3

What if there is a condition $a^2+b^2\neq 0?$ How to prove closure under multiplication?

Answer 4

You can use a similar trick as in complex numbers. Note that $(a+b\sqrt5)(a-b\sqrt5)=a^2-5b^2$, so $\frac{1}{a+b\sqrt5}=\frac{a-b\sqrt5}{a^2-5b^2}$.