# Field Show that F is a field under + and $\cdot$ and F = $\{a + b\sqrt{3}\} | a, b \in \mathbb Q\}$ [duplicate]

Show that F = $\{a + b\sqrt{3}\} | a, b \in \mathbb Q\}$ is a field under the operations + and $\cdot$, where + is given by $(a + b\sqrt{3}) +(c + d\sqrt{3}) = (a + c) + (b+d)\sqrt{3}$ and $\cdot$ is given by $(a + b\sqrt{3}) \cdot(c + d\sqrt{3}) = (ac + 3bd)+ (ad + bc)\sqrt{3}$

Do I just have to show it satisfies all the $A1 - A4, S1 - S4$ axioms?

For example:

$A1 = u + (v + w) =$

$$=(a + b\sqrt{3} + ((c + d\sqrt{3}) + (e + f\sqrt{3})) = (a + b\sqrt{3}) + ((c + e) + (d + f)\sqrt3{}))$$

$$= (a + c + e) + (b + d + f)\sqrt{3}$$

Also:

$$(u + v) + w = ((a + b\sqrt{3}) + (c + d\sqrt{3}) ) + (e + f\sqrt{3})$$

$$= (a + c + e) + (b + d + f)\sqrt{3}$$

Satisfies one of the eight of the axioms. Is this what I am suppose to do?

If so, I don't get how to check the scalars for example

S1: $r(u + v) = ru + rv$

$$= r((a + b\sqrt{3}) + (c + d\sqrt{3}))$$

$$= r((a+c) + (b+d)\sqrt{3})$$

Unsure what to do after this

• Yes, that is what you are supposed to do. Commented Sep 10, 2017 at 4:56
• Yes!!! You have to check all of them. Unless, of course, you come up with an alternative proof (e.g. using ring homomorphisms etc). By the way, you made some typo while defining $\cdot$. Commented Sep 10, 2017 at 4:59
• Sorry fixed. It was in the scalar part Commented Sep 10, 2017 at 5:05

Yes, you should check that it satisfies all of the field axioms.

Note that the A1−A4, S1−S4 axioms you quote are not the axioms for a field! They are the axioms for a vector space (over an already given field). You should locate the actual list of axioms for a field. For example, there should be an axiom that guarantees the existence of multiplicative inverses. And there should not be any distinction between vectors and scalars.

The axiom you linked to are those of a vector space, not those for a field.

It is true (but it is probably not your intention) that $F$ is a vector space with scalar field $\mathbb{Q}$:
$F$ is just the $\mathbb{Q}$-span of $\{1, \sqrt{3}\}$ in $\mathbb{R}$ seen as a vector space over $\mathbb{Q}$, so $F$ is a vector subspace of this vector space.

For any two fields $F_1 \subseteq F_2$, we can check quite easily that $F_2$ is a vector space with $F_1$ as scalars: all vector space axiom are field axioms in particular, and the equations thus hold in $F_2$. (multiplicative inverses are not part of vector spaces though)

The axioms for a field $(F, +, \cdot ,0, 1)$ just say that $(F,+,0)$ is an Abelian group, and so is $(F\setminus\{0\},\cdot, 1)$ while we have the distributive law $a(x+y) = ax+ay$ to link the operations $+$ and $\cdot$.

It's clear that $F \subseteq \mathbb{R}$ with the inherited operations from $\mathbb{R}$ (we don't really have to define multiplication with a formula, as the formula just follows from $\sqrt{3}^2 =3$ and the field axioms in the reals:

$$(a + b\sqrt{3})(c + d\sqrt{3}) = \text{ (by distributivity)} =a(c+d\sqrt{3}) + b\sqrt{3}(c+d\sqrt{3}) =\\ ac + ad\sqrt{3} + bc\sqrt{3} + bd\sqrt{3}^2 = ac + (ad+bc)\sqrt{3} + 3bd$$. So the only thing to check is that $F$ is closed under inverses and this is clear as

$$\frac{1}{a + b\sqrt{3}} = \frac{a-b\sqrt{3}}{(a-b\sqrt{3})(a+b\sqrt{3})} = \frac{a}{a^2 - 3b^2} + \frac{b}{a^2-3b^2}\sqrt{3}$$

and $-(a+b\sqrt{3}) = (-a) + (-b)\sqrt{3}$, so if a number $x$ is in $F$ so is $-x$ and $\frac{1}{x}$. All other field axioms are just the same equations that hold for $\mathbb{R}$. So $F$ is just a subfield of $\mathbb{R}$.