Show that $A=\{x+y\sqrt{2}:x,y \in \mathbb{Z}\}$ is a commutative ring with unity, find the zero element, the unity and the negative of $a$ Show that $A=\{x+y\sqrt{2}:x,y \in \mathbb{Z}\}$ is a commutative ring with unity, find the zero element, the unity and the negative of an arbitrary $a$.
First thing first, I need to show it is a ring:
and I am having a hard time showing commutativity and possibly a bit confuse. 
I said let $x,y \in  \mathbb{Z}$.
Do need to show that
$x+y=y+x$
or do i need to show $A+B=B+A$
where A is defined as $A=\{x+y\sqrt{2}:x,y \in \mathbb{Z}\}$ and B is defined as $B=\{v+w\sqrt{2}:v,w \in \mathbb{Z}\}$.
The reason I am confuse is because I thought it was $x+y=y+x$ but then I get:
$$ x+y\sqrt{2}=y+x\sqrt{2}$$
which is not commutative.
 A: Note that your sets $A$ and $B$ here are the same, so defining $B$ is redundant.
Now, to show that the addition operation $\oplus$ of the ring $A$ is commutative, by definition we need to show that for any $a, b \in A$ we have $a \oplus b = b \oplus a$.
By definition, we can write
$a = p + q \sqrt{2}$ and $b = r + s \sqrt{2}$, and recall that $\oplus$ is defined by
$$(p + q \sqrt{2}) \oplus (r + s \sqrt{2}) := (p + r) + (q + s) \sqrt{2}.$$ Can you now use this definition to show commutativity?
A: Your ring is defined as $A=\{x+y\sqrt{2}:x,y\in \mathbb{Z}\}$. You want to show that any two elements commute. Two general elements can be written as $\alpha=x+y\sqrt{2}$ and $\beta=v+w\sqrt{2}$. To show commutativity, which for your question likely refers to the multiplication operation $\cdot$ in your ring, you want to show that $\alpha\cdot\beta = \beta \cdot \alpha$. 
A: For the commutativity part, let $a, b \in A \Rightarrow a = x+y\sqrt{2}, b = m + n\sqrt{2} \Rightarrow a+b = (x+y\sqrt{2})+(m+n\sqrt{2}) = (m+n\sqrt{2})+(x+y\sqrt{2}) = b+a$.
You said the "negative of a", I would think you meant "the negative of $A$", and it is: $-A = \{-a: a \in A\} = \{ -(x+y\sqrt{2}): x, y \in \mathbb{Z}\} = A$. 
For the zero element, it is $0$, and the unity, that is the multiplicative identity, it is $1$.
A: Hint $\ $ Apply the subring test to deduce that it is a subring of $\,\Bbb R,\,$ hence commutative.
