Is $\mathbb R[x] / \langle x^2+1 \rangle = \mathbb R[x] / \langle x^2+2 \rangle$? I am starting to study rings. One of the first examples in my book about ring factors is
$$\mathbb{R}[x] / \langle x^2+1 \rangle = \{ ax + b +  \langle x^2 +1 \rangle  \mid a,b \in \mathbb{R} \}$$
I am wondering if 
$$\mathbb{R}[x] / \langle x^2+1 \rangle = \mathbb{R}[x] / \langle x^2+2 \rangle\text{ ?}$$
and more generally
$$\mathbb{R}[x] / \langle x^2+1 \rangle = \mathbb{R}[x] / \langle a x^2+b x + c \rangle \, \, \forall (a,b,c) \in \mathbb{R}^3 \setminus \{(0,0,0) \}?$$
How can i show that they are equal? (or that they are not?)
 A: You want your polynomial to have two conjugate complex roots. Does this give you an idea on the isomorphism?
A: For any $u,v\in\mathbb R$ (i.e. for any $z=u+iv\in\mathbb C$), there is a ring homomorphism $f\colon \mathbb R[X]\to \mathbb C$ that is the identity on $\mathbb R$ and that maps $X\mapsto u+iv$. (This is the universal peroperty of the polynomial ring).
If $v=0$, obviously the image is just $\mathbb R$, not all of $\mathbb C$.
If $v\ne 0$, then $f$ is onto; in fact $f(\frac bvX-\frac {ub}v+a)=a+bi$.
One verifies that $f(X^2)=u^2-v^2+2uvi=2uf(X)-(u^2+v^2)$ and hence $X^2-2uX+u^2+v^2$ is in the kernel of $f$. As $\mathbb R[X]$ is a principal ideal domain and no linear polynomial can be in the kernel, we conclude that the kernel is $\langle X^2-2uX+u^2+v^2\rangle$ and by the isomorphism theorems
$$ \mathbb R[X]/\langle X^2-2uX+u^2+v^2\rangle\cong \mathbb C\qquad\text{if }v\ne0.$$ 
The discriminant of $X^2-2uX+u^2+v^2$ is simply $(-2u)^2-4(u^2+v^2)=-4v^2<0$.
We note that for any quadratic polynomial $X^2+pX+q$ with discriminant $D=p^2-4q<0$ we can simply let $u=-\frac p2$ and $v=\frac12\sqrt{-D}$ and thus find that 
$$ \mathbb R[X]/\langle X^2+pX+q\rangle\cong \mathbb C\qquad\text{if }D=p^2-4q<0.$$ 

What is $\mathbb R[X]/\langle X^2+pX+q\rangle$ if the discriminant is nonnegative? If $D=0$, then $\epsilon:=X+\frac p2+\langle X^2+pX+q\rangle$ is a strange element: it is nonzero, but its square is $\epsilon^2=X^2+pX+\frac{p^2}4+\langle X^2+pX+q\rangle$, and that is just zero (because $q=\frac{p^2}4$). This ring is often just written $\mathbb R[\epsilon]$ with the mnemonic that $\epsilon$ is very small, but $\epsilon^2$ is so negligibly small that it really equals $0$.
$$ \mathbb R[X]/\langle X^2+pX+q\rangle\cong \mathbb R[\epsilon]\qquad\text{if }D=p^2-4q=0.$$ 

If $D>0$, $X^2+pX+q$ has two distinct real roots $a,b$. Can you see what the ring looks like now? As the previous example, it has zero divisors, so at least it is definitely not isomorphic to $\mathbb C$.
A: They are not equally set theoretically, but are isomorphic as fields. As long as $ax^2+bx+c$ is an irreducible polynomial over $\mathbb{R}[x]$, then the ideal $\langle ax^2+bx+c\rangle$ is maximal, thus the quotient $\mathbb{R}[x]/\langle ax^2+bx+c\rangle$ is a field. In fact the quotient is isomorphic to $\mathbb{C}$ whenever the irreducible polynomial is degree 2, and is isomorphic to $\mathbb{R}$ whenever the irreducible polynomial is degree 1. 
A: Once you get to the isomorphism theorem (which should be very soon), this is fairly easy to prove.  Unfortunately, this might not make a whole lot of sense right now, but it should be rather illuminating in a week or two.  The isomorphism theorem says that, given a homomorphism $\phi:R \rightarrow S$, then:
$$R/\ker(\phi) \cong Im(\phi)$$
Typically for polynomial rings, the evaluation homomorphism $ev_a$ does the trick, which basically evaluates all polynomials in the ring at $a$.  That is, $f(x) \mapsto f(a)$.  You should prove that this is indeed a homomorphism.
So for the first, we are going to look at the evaluation homomorphism $ev_i:\mathbb{R}[x] \rightarrow \mathbb{C}$.  Note that its kernel is $\langle x^2 + 1 \rangle$, since that is the minimal polynomial with $i$ as a root.  Further, this is surjective onto $\mathbb{C}$ since $(ax + b) \mapsto  (ai + b)$ for all $a, b \in \mathbb{R}$.  So by the isomorphism theorem, we get $\mathbb{R}[x]/\langle x^2 + 1 \rangle \cong \mathbb{C}$.
Now let's look at the second.  This time, we will consider the evaluation homomorphism $ev_{i\sqrt{2}}: \mathbb{R}[x] \rightarrow \mathbb{C}$.  Note that the kernel is $\langle x^2 + 2 \rangle$, since that is the minimal polynomial with $i\sqrt{2}$ as a root.  Again, this is surjective onto $\mathbb{C}$ since $(a\sqrt{2}x + b) \mapsto (ai + b)$ for all $a, b \in \mathbb{R}$.  So by the isomorphism theorem, we get $\mathbb{R}[x]/\langle x^2 + 2 \rangle \cong \mathbb{C}$.
Thus, both quotient rings are isomorphic to $\mathbb{C}$, and so we conclude $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{R}[x]/\langle x^2+2 \rangle$.
