Answer to Question #233064 in Abstract Algebra for 123

Question

Does { a + b √ 2 | a, b ∈ Z } form a ring (with the usual operations in R )?

Is it commutative? Does it have a unity? Be sure to justify your answers.

Answer 1

$\text{The set $A = \{ a + b\sqrt2\}$ is a subring of $R$ since for any arbitrary elements: }\\ a + b\sqrt2, c+ d\sqrt2\\ a + b\sqrt2 - c+ d\sqrt2 = (a-c) + (b-d)\sqrt2 \in A \\ \text{and } (a + b\sqrt2) (c+ d\sqrt2) = (ac + 2bd)(bc + ad)\sqrt2 \in A \\ \text{Hence since $A$ form a subring, thus it suffices to say that $A$ form a ring.} \\ \text{$A$ is commutative since:} (a + b\sqrt2) (c+ d\sqrt2) = (c+ d\sqrt2)(a + b\sqrt2) = (ac + 2bd)(bc + ad)\sqrt2 \in A \\ \text{It has a unity since $1 \in A$ and } 1\cdot(a + b\sqrt2) = (a + b\sqrt2) \cdot 1= (a + b\sqrt2)$