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: $(b + c) ⋅ a = (b ⋅ a) + (c ⋅ a)$
  • 1
    $\begingroup$ What is the definition of "ring" ... what clauses of the definition are difficult for you to check? $\endgroup$ – GEdgar May 29 '14 at 14:42
  • $\begingroup$ All you have to do is prove the ring axioms hold. $\endgroup$ – Deepak May 29 '14 at 14:48

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.


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 $\,x\mapsto \sqrt 3.$


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.


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