If $G$ is a field and there is an isomorphism $f\colon H/I \to G$, then does $I$ have to be a principal ideal? I noticed that the ideal $I = \left(2, 1 + \sqrt{-7}\right)$ follows the definition of a non-principal ideal. I took two random elements from $\mathbb Z[-7]$, say $ a + b\sqrt{-7}$ and $ c + d\sqrt{-7}$ and multiplied each respectively with $2$ and $1+\sqrt{-7}$ and we find out that $$\sqrt{-7} \left(2b + c + d \right) + (2a + c - 7d) = 0.$$ 
Label $ x = 2b + c + d$ and $ y = 2a + c - 7d$ and $x-y$ is divisible by $2$. 
Doesn't that mean that $\dfrac{\mathbb Z[-7]}{(2,\, 1 + \sqrt{-7})}$ is isomorphic to $\mathbb{Z}_2$? But $\mathbb{Z}_2$ is a field. 
Also, I hope my question is clear enough. I don't have the "privilege" of going to a college and asking questions to an instructor. 
 A: Well I do not think you claim is true. Just take the ring polynomial ring in $2$ variables, say the variables are $x,y$. Take the ideal $I=\langle x,y\rangle$. The quotient by this ideal is field, but $I$ is not principal 
A: No, $I$ does not have to be a principal ideal. For example, $k[x,y]/(x,y)\cong k$, as noted by GGT.
To expand on this, for a (commutative, with $1$) ring $H$ and proper ideal $I$, we have necessary and sufficient conditions for when $H/I$ is a field. Namely, the quotient $H/I$ is a field if and only if $I$ is maximal with respect to inclusion; that is, if $J\subset H$ is an ideal so $I\subset J$, then $J=I$ or $J=R$, and there are plenty examples of maximal ideals not generated by a single element. This fact follows as if $I$ is maximal, then $H/I$ has no proper nontrivial ideals, so every element must be unit.
A: You do in fact have $\mathbb{Z}[\sqrt{-7}]/\langle2, 1+\sqrt{-7}\rangle \simeq \mathbb{F}_{2}$ (or $\mathbb{Z}_{2}$, essentially the same thing).
It is not true that if you mod a ring $R$ by an ideal $I$ and get a field, then $I$ must be principal; the correct statement is that if $R/I$ is a field, then $I$ is a maximal ideal (if and only if, in fact).  It is also true that $R/I$ is an integral domain if and only if $I$ is prime.  This shows that maximal ideals are prime ideals, but the converse does not necessarily hold.
Out of curiosity, what book are you working through?
A: What the "correspondence theorem" (also known as the "Fourth Isomorphism Theorem") says for rings, is that there is a bijection (actually a lattice-isomorphism) between the ideals of $R$ containing $\text{ker }f$ and the ideals of $S$, when $f: R \to S$ is a surjective ring homomorphism.
If $S = F$, a field, we know that $F$ has only two ideals, $\{0\}$, and $F$ itself. Therefore, $R$ has only two ideals containing $I = \text{ker }f$, $I$ (which maps to $\{0\}$) and $R$ (which maps to $F$). Hence $I$ must be a maximal ideal.
Of course, it goes without saying that such an ideal is principal if $R$ is a PID (or even better, a Euclidean domain). And many of the rings commonly used for examples are PID's (because these act "more like integers", from which our ring-intuitions spring).
What we cannot do, is work this backwards to deduce what kind of ring $R$ might be. All we know is $R$ possesses a maximal ideal. We cannot even say that $I$ is a prime ideal, since that requires further restrictions on $R$ (namely: that $R^2 = \{\sum_i a_ib_i: a_i,b_i \in R\}$ is equal to $R$). For an example of a ring where this actually happens, consider $R = 2\Bbb Z$, and $I = (4)$, in this ring, $(4)$ is a maximal ideal, but is not prime, since $2\cdot 2 = 4$, but $2 \not\in (4)$.
Rings like $\Bbb Z[\sqrt{-7}]$ tend to confound our intuition in a big way: not only is it not a PID (and therefore non-Euclidean), it isn't even a unique factorization domain:
$8 = 2^3 = (1 + \sqrt{-7})(1 - \sqrt{-7})$.
