Show that $\mathbb{Q} (\sqrt{3}) = \{ a + b\sqrt{3} : a,b\in\mathbb{Q}\}$ is a field.

Question

I need to show that $\mathbb{Q} (\sqrt{3}) = \{ a + b\sqrt{3} : a,b\in\mathbb{Q}\}$ is a field. Do I start with ring axioms? I am so confused.

Answer 1

You need to verify all the field axioms. Many of them, identities, commutativity, associativity, and distributivity follow from the properties of real numbers. You need to show that sums and products of the elements in the set are again in the set. You need to show that the inverses of elements of your set are also in the set. This is trivial for additive inverses, for multiplicative it takes a few lines of algebra.

Answer 2

One way to do your question is to verify the field axioms, showing that $(\mathbb{Q}[\sqrt3],+), (\mathbb{Q}[\sqrt3],\times)$ are abelian groups and that $\times$ is distributive over $+$.

However, indeed, a more general fact is true: as long as we have $\alpha$ to be algebraic over $\mathbb{Q}$, that is $\alpha$ is a root of some polynomial with coefficients in $\mathbb{Q}$, the set $\mathbb{Q}[\alpha]$ consisting of all polynomial expressions in $\alpha$ with coefficients in $\mathbb{Q}$ is a field. You will probably encounter such result in a course about field theory, or during the first few weeks of a course in Galois theory.

To let you have a favour of what you might see in the future, here is a proof that $\mathbb{Q}(\sqrt3)$ is a field using facts in ring theories without verifying through axioms. This proof can be generalized to the above general fact easily when you are mature enough to understand what is going on.

Claim: $\mathbb{Q}[\sqrt3]$ is a field.

Proof: Consider the evaluation map $\phi:\mathbb{Q}[x]\to\mathbb{C}$ by $f(x)\mapsto f(\sqrt3)$ for all $f\in\mathbb{Q}$. Note that $range(\phi)=\mathbb{Q}[\sqrt3]$ and $ker(\phi)=(x^2-3)$, which is the ideal generated by $x^2-3$. Hence, by the first isomorphism theorem of rings $\mathbb{Q}[x]/(x^2-3)\cong \mathbb{Q}[\sqrt3]$. It now suffices to show that $\mathbb{Q}[x]/(x^2-3)$ is a field.

Note that any quotient ring is a field if and only if the ideal one takes quotient of is maximal. We can then show $(x^2 -3)$ is a maximal ideal in turn. Since $\mathbb{Q}$ is a field, its polynomial ring $\mathbb{Q}[x]$ is a principal ideal domain (PID). In PID, every ideal generated by an irreducible element is maximal. Note that $x^2-3$ is an irreducible element in $\mathbb{Q}[x]$ as it cannot be factorized into linear polynomials with coefficients in $\mathbb{Q}$. Hence, $(x^2-3)$ is a maximal ideal.

By what's said previously, we have shown that $\mathbb{Q}[\sqrt3]$ is a field.