# Examples of fields in $\Bbb{R}$

I'm having trouble understanding how to determine fields

I understand there are axioms it must satisfy to be considered a field like associativity, commutativity, distributivity, identity, and inverses but I'm having trouble actually applying it to the problems

Could someone help me to solve this:

Show that $$\{a+b\sqrt{3} \mid a,b \in \Bbb{Q}\}$$ is a field (a subfield of the field $\mathbb{R}$), but $$\{a+b\sqrt{3} \mid a,b \in \Bbb{Z}\}$$ is not a field.

• In both of these, the only tricky part of the proofs will be inverses - you must show (by construction, probably) an inverse for a nonzero element in the first, and prove in the second that there is no such inverse. All the other properties are easy to show computationally.
– user208649
Jan 21, 2016 at 2:10
• Welcome to Math SE. Please, show your efforts about the question. Jan 21, 2016 at 2:11
• When testing whether a subset of a given field is a subfield, you don't have to text every field axiom: some of them come for free from the known field. This question gives a smaller list of conditions to test whether a subset is a subfield. Jan 21, 2016 at 2:15

A strategy for solving this problem would be to check all the properties that need to be satisfied by a field. These are closure under addition and multiplication, associativity, commutativity, existence of additive and multiplicative identity elements, existence of additive and multiplicative inverses, and the distributivity of multiplication over addition.

All of the above, with the exception of multiplicative inverses, are satisfied by both of your examples, which make both of them rings. However, they differ in that the second example does not have a multiplicative inverse whereas the first one does.

To see this, we require that $\forall a, b \in \mathbb{Q}, \exists c, d \in \mathbb{Q}$ for which $$(a + b \sqrt{3})(c + d \sqrt{3}) = 1$$

We can multiply these out to give:

$$ac + 3 bd + (ad + bc)\sqrt{3} = 1$$

Since $\sqrt{3} \in \mathbb{R} \setminus \mathbb{Q}$ we have that:

$$ac + 3bd = 1$$ $$ad + bc = 0$$

Solving the above simultaneous system of equations yields (apart from the trivial solution where $a = b = 0$) that, if $a^2 \not= 3b^2$ (since here they'd both have to be zero):

$$c = \frac{a}{a^2 - 3b^2}$$ $$d = -\frac{b}{a^2 - 3b^2}$$

Now, since $a, b \in \mathbb{Q}$, the above shows that $c, d \in \mathbb{Q}$. However, this does not hold for the case where $a, b \in \mathbb{Z}$ since dividing two integers doesn't always give you an integer. Thus, the second case is not a field but a ring.