Is $\{a+b\sqrt{3}\mid a, b \in \Bbb Z\}$ Ring?

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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 $\text{unit in }R = \{ {a\over a^2-3b^2}-{b\over a^2-3b^2}\sqrt3\mid {a\over a^2-3b^2}, {b\over a^2-3b^2} \in \Bbb Z\}?$

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9
On

A ring does not need to have multiplicative inverses. A field does.

0
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The axioms you have to check are (if $R$ is to be a "ring with unit"):

1) $R$ is a group with respect to the addition

2) $R$ is a monoid with respect to multiplication

0
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Yes. The set $R$ is a ring with respect to the usual operations. The easiest way of proving that is to show that it is a subring of a known ring. Here I recommend that you prove that $R$ is a subring of the ring of real numbers $(\Bbb{R},+,\cdot,0,1)$. This is a simple application of the so called subring criterion that I invite you to verify yourself.

The question about units is more difficult. It turns out that there are infinitely many of those. Therefore a list is not expected. Judging from the rest of the question I expect that you are A) to exhibit infinitely many units with the aid of powers of a non-trivial unit you are to find, and possibly B) to show that $u=a+b\sqrt3$ is a unit of $R$ if and only if $v=a-b\sqrt3$ is its inverse. We can say a lot more about the group of units of $R$, but that question is met only (AFAICT) in the context of , and my judgement tells me that you are not expected to know that material. At least not yet.