Showing that set $\mathbb{S} = \{a + bi\mid a,b\in\mathbb{Q}\}$ is closed under addition and multiplication

Question

I need to prove that set $\mathbb{S} = \{a + bi\mid a,b\in\mathbb{Q}\}$ follows certain axioms and is indeed a field. The question is, can I prove that set is closed under addition and multiplication operations, for example using facts that $(\forall a \in \mathbb{S})(\exists \,{-a} \in \mathbb{S})$ and $(\forall a \in \mathbb{S})(\exists \frac{1}{a} \in \mathbb{S})$ we have $a + (-a) = 0 \in \mathbb{S}$ and $a\cdot a^{-1} = 1 \in \mathbb{S}$, which have to show that they are closed under addition and multiplication. For other cases, can I refer that other elements can be generated using $1$,$0$ and defined operations on the set?

I.e. can I state that because $(\forall a \in \mathbb{S})(\exists \frac{1}{a} \in \mathbb{S})$ we have $a\cdot a^{-1} = 1$ then set is closed under multiplication?

Answer 1

The set $\left\{ 2, \dfrac 1 2 \right\}$ is closed under multiplicative inversion but not under multiplication. So it is not true that every set closed under multiplicative inversion is closed under multiplication.

You have $(a+bi)(c+di) = (ac-bd) + i(ad+bc)$. That is the core of the proof of closure under multiplication.

Similarly $\dfrac 1 {a+bi} = \dfrac{a-bi}{a^2+b^2}$ is at the center of the proof that every complex number has a multiplicative inverse.

Answer 2

To show "closed under addition", you need to show that for $\alpha,\beta\in\Bbb S$, also $\alpha+\beta\in\Bbb S$. This is not related to the question wheter $-\alpha\in\Bbb S$ or $0\in\Bbb S$.

Similarly, to show "closed unde rmultiplication", you need to show that for $\alpha,\beta\in\Bbb S$, also $\alpha\cdot\beta\in\Bbb S$. This is not related to the question wheter $\alpha^{-1}\in\Bbb S$ for $\alpha\ne0$, or $1\in\Bbb S$.