How do you show that $\mathbb{Z}[x]/(x^2-29) \cong \{a+b\sqrt{29}|a,b \in \mathbb{Z}\}$? $\mathbb{Z}[x]/(x^2-29) \cong \{a+b\sqrt{29}|a,b \in \mathbb{Z}\}$.
I understand what $\mathbb{Z}[x]/(x^2-29)$ means: that every polynomial that can be factored by $(x^2-29)$ is equal to zero. However, I can't see how you show that isomorphism.
 A: Hint: Consider the homomorphism $\varphi: \mathbb{Z}[x] \to \mathbb{Z}[\sqrt{29}]$ given by $\varphi(f) = f(\sqrt{29})$. Show $\varphi$ is surjective and has kernel $(x^2-29)$, and use the first isomorphism theorem. Finally, show that $\mathbb{Z}[\sqrt{29}] \simeq \{a+b\sqrt{29} | a,b \in \mathbb{Z}\}$.
A: Note: $\mathbb{Z}[x]/(x^2-29)$ means more than just "every polynomial that is divisible by $x^2-29$ is equal to $0$". This is a quotient ring: the elements are equivalence classes of polynomials; two polynomials are in the same class if and only if their difference is a multiple of $x^2-29$. You have a ring structure by operating on representatives. Etc. 
By the universal property of polynomial rings, if you have an embedding $\mathbb{Z}\hookrightarrow R$ into a ring $R$, and you specify an $a$ in $R$, then you obtain a ring homomorphism $\varphi\colon\mathbb{Z}[x]\to R$ that maps $x$ to $a$. (In fact, given a ring homomorphism $\varphi\colon \mathbb{Z}\to \mathbb{R}$ and an element $a\in\mathbb{R}$ that commutes with $\varphi(b)$ for each $b\in\mathbb{Z}$, the homomorphism $\varphi$ extends to all of $\mathbb{Z}[x]$ mapping $x$ to $b$). 
Here, consider the natural embedding of $\mathbb{Z}\hookrightarrow \mathbb{R}$, and let $a=\sqrt{29}$. Then you automatically get a homomorphism
$$\varphi\colon \mathbb{Z}[x]\to \mathbb{R}$$
that acts as the identity on the integers, and maps $x$ to $\sqrt{29}$.
By the First Isomorphism Theorem, the image of this homomorphism is isomorphic to the quotient 
$$\frac{\mathbb{Z}[x]}{\mathrm{ker}(\varphi)}.$$
It is now easy to verify that the image contains $\{a+b\sqrt{29}\mid a,b\in\mathbb{Z}\}$; and using the fact that $\sqrt{29}^n$ is either an integer or an integer times $\sqrt{29}$, the converse inclusion also holds. So the image is $\{a+b\sqrt{29}\mid a,b\in\mathbb{Z}\}$.
Now just verify that the kernel of the map is exactly the ideal generated by $x^2-29$. It is easy to see that any multiple of $x^2-29$ is mapped to $0$. Now assume that $p(x)$ is mapped to zero. Dividing by $x^2-29$, we can write
$$p(x) = q(x)(x^2-29) + r(x)$$
where $q(x),r(x)\in\mathbb{Z}[x]$, and $r(x)=a+bx$ for some integers $a$ and $b$. Therefore, the image of $p(x)$ is equal to $a+b\sqrt{29}$, which is equal to $0$. If $b\neq 0$, then we would conclude that $\sqrt{29}$ is rational, which is impossible. Thus, $b=0$, hence $a=0$. That is, $p(x)$ is a multiple of $x^2-29$.
Thus, the First Isomorphism Theorem shows that
$$\frac{\mathbb{Z}[x]}{(x^2-29)} \cong \{a+b\sqrt{29}\mid a,b\in\mathbb{Z}\},$$
where the latter has the ring structure inherited from $\mathbb{R}$.
