I've got an question to prove
Let $R = \Bbb Z [\sqrt 3] = \{a+b\sqrt{3}\mid a, b \in \Bbb Z\}$. Prove that R forms a ring with the addition and multiplication considered R as a subset of the field of real numbers. Find all units in this ring.
While I am proceeding with a proof of $R$ to be ring, for its requirement to have multiplicative inverse, I've got ${a\over a^2-3b^2}-{b\over a^2-3b^2}\sqrt3$ where each coefficient not in $\Bbb Z$
IIs $\{a+b\sqrt{3}\mid a, b \in \Bbb Z\}$ really to be Ring?
Additionally, for unit in R, one must be represented in a form of ${a\over a^2-3b^2}-{b\over a^2-3b^2}\sqrt3$ and those coefficient must be in $\Bbb Z$.
Is it sufficient to say $\text{unit in }R = \{ {a\over a^2-3b^2}-{b\over a^2-3b^2}\sqrt3\mid {a\over a^2-3b^2}, {b\over a^2-3b^2} \in \Bbb Z\}?$
A ring does not need to have multiplicative inverses. A field does.