Prove that $\{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$ is a ring (or not) How to prove that $(R, +, \cdot)$ is a ring (or not), where $R = \{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$?
Update. Is this proof correct?
$(R, +)$ is an abelian group:


*

*Closure: $a, b \in R \implies (a + b) \in R$.

*Associativity: $a, b, c \in R \implies (a + b) + c = a + (b + c)$.

*Identity: $0 \in R$.

*Inverse: $-(x + y \sqrt{3}) \in R$.

*Commutativity: $a, b \in R \implies a + b = b + a$.


$(R, \cdot)$ is a monoid:


*

*Closure: $(x_1 + y_1 \sqrt{3}) \cdot (x_2 + y_2 \sqrt{3}) = x + y \sqrt{3}$, where $x = x_1 x_2 + 3 y_1  y_2 \in \mathbb{Z}$ and $y = x_1 y_2 + x_2 y_1 \in \mathbb{Z}$.

*Identity: $1 \in R$.


Multiplication distributes over addition: 
$a, b, c \in R \implies$


*

*Left distributivity: $a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)$

*Right distributivity: $(b + c) ⋅ a = (b ⋅ a) + (c ⋅ a)$

 A: The set in question is a subset of the real numbers, which is a ring. The operations on the set are the same as those on the reals. Hence, you only have to prove that the set contains 0 and 1, and is closed under addition, additive inverses, and multiplication. The other properties hold because they hold in the reals. 
A: Hint $\ $ Employ the subring test to prove that $\,S\,$ is a subring of some known larger ring (e.g. $\,\Bbb R),\,$ i.e. show $\,1\in S\,$ and $\,S\,$ is closed under  subtraction and multiplication (by $\,\sqrt 3^2 = 3\in S)\,$ 
Alternatively, show that $\,S\,$ is the image of $\,\Bbb Z[x]\,$ in $\,\Bbb R\,$ under the evaluation hom $\,x\mapsto \sqrt 3.$
A: That $(R,+)$ is an abelian group is easy to verify.
Also, it has a multiplicative identity $1_{R} = 1 + 0\cdot \sqrt{3} = 1$, and $0_{R} = 0 +  0\cdot \sqrt{3} = 0$ is its additive identity.
For each element $x + y\cdot \sqrt{3}$ in $R$, its additive inverse is $-x - y\cdot \sqrt{3}$.
The distributive property holds as well, and also easily verified.
So: $(R,+,\cdot)$ is a ring.
