Let $R = \{m+n\sqrt{2} \mid m,n\in\mathbb Z\}$ and let $I = \{m+n\sqrt{2} \mid n\in\mathbb Z, m\in 2\mathbb Z\}$
$(a)$ show that $I$ is an ideal of $R$.
$(b)$ Find the well known commutative ring to which $R/I$ is isomorphic. Hint: How many congruence classes does $I$ determine?
To show that $I$ is an ideal of $R$, I need to show that $I$ is closed under addition and subtraction under multiplication by a member of $R$, which is fairly simple.
For part $(b)$ I believe that $R/I$ is isomorphic to $\mathbb{Z}_2$ but I have no idea how construct the isomorphism.
Any help is appreciated!
Your instincts are correct. To me it makes things easier to write them out as quotients:
$R=\mathbb Z[x]/(x^2-2)$ is your original ring, where $x$ is "$\sqrt 2$" in the quotient.
As for $I$, it is (if you examine it) the ideal generated by $2$ and $x$ in $R$. So you are looking at the quotient
$(\mathbb Z[x]/(x^2-2))/(2 + (x^2-2), x+(x^2-2))\cong \mathbb Z[x]/(2,x)\cong \mathbb Z/(2)$
via the third isomorphism theorem.
Of course, the homomorphism to use the first isomorphism theorem is the obvious one: $m+n\sqrt{2}\mapsto m\pmod 2$.