Answer to Question #294228 in Abstract Algebra for Delightful

Question

P rove that all real of the form a + b 2 , a; b ∈ Z form s a ring.

Answer 1

We are to prove that the set $\displaystyle S=\{a+b\sqrt{2}:a,b\in\Z\}$ is a ring.

Let $p=a_1+b_1\sqrt{2},\ q=a_2+b_2\sqrt{2},\ r=a_3+b_3\sqrt{2}$ belong to $S$, then

1] $S$  is an abelian group under addition, meaning that:

$-(p+q)+r=p+(q+r)$

we have:

$(a_1+b_1\sqrt2+a_2+b_2\sqrt2)+a_3+b_3\sqrt2=a_1+b_1\sqrt2+(a_2+b_2\sqrt2+a_3+b_3\sqrt2)$

$--p+q=q+p$

we have:

$a_1+b_1\sqrt2+a_2+b_2\sqrt2=a_2+b_2\sqrt2+a_1+b_1\sqrt2$

$---$ There is an element $0=0+0\sqrt{2}$ in $S$ if $a_1,\ b_1=0$  such that $p + 0 = p$

$----$ For each p in S there exists −p in S such that p + (−p) = 0

we have:

$a_1+b_1\sqrt2+(-a_1-b_1\sqrt2)=0$

2] $S$  is a monoid under multiplication, meaning that:

$(p ⋅ q) ⋅ r = p ⋅ (q ⋅ r)$

we have:

$((a_1+b_1\sqrt2)(a_2+b_2\sqrt2))(a_3+b_3\sqrt2)=(a_1+b_1\sqrt2)((a_2+b_2\sqrt2)(a_3+b_3\sqrt2))$

There is an element $1=1+0\sqrt{2}$ in $S\ \text{if }a=1, b=0$  such that p ⋅ 1 = p and 1 ⋅ p = p

3] Multiplication is distributive with respect to addition, meaning that:

p ⋅ (q + r) = (p ⋅ q) + (p ⋅ r)

we have:

$(a_1+b_1\sqrt2)(a_2+b_2\sqrt2+a_3+b_3\sqrt2)=(a_1+b_1\sqrt2)(a_2+b_2\sqrt2)+$

$+(a_1+b_1\sqrt2)(a_3+b_3\sqrt2)$

(q + r) ⋅ p = (q ⋅ p) + (r ⋅ p)

we have:

$(a_2+b_2\sqrt2+a_3+b_3\sqrt2)(a_1+b_1\sqrt2)=(a_1+b_1\sqrt2)(a_2+b_2\sqrt2)+$

$+(a_1+b_1\sqrt2)(a_3+b_3\sqrt2)$