Let's consider a commutative ring $R(+,.)$ and let $I$ be an ideal of $R$. Now there may be several ways to prove that if $R/I$ is a field then $I$ is a maximum ideal. I thought of the following proof which already might have been discussed: Consider any $r_1 \in R , \notin I$. Then we can consider the ideal $I + r_1.R$ which is not equal to $I$. Now if $R/I$ is a field then for every such $r_1$ there exists another element $r_2 \in R, \notin I$ such that $r_1.r_2= 1+i$ for some $i \in I$. Since $r_2.R \subset R$, $I + r_1.R=R$. This proves that $I$ is a maximum ideal.
If someone can point out whether there are any loopholes/incorrectness in this proof, that will be very helpful. Thank you.
What you have given is one of the standard proofs of the fact that $\frac RI$ is a field implies that $I$ is maximal. For completeness, I wish to add another : the ring homomorphism theorems establish a one-one correspondence between the ideals of $R / I$ and the ideals of $R$ containing $I$, via $I \subset J \subset R \to \frac JI$, which is one-one and onto i.e. any ideal of $\frac RI$ is of the form $\frac JI$ for some ideal $I \subset J \subset R$.
Thus, if $\frac RI$ is a field, this means that there are only two ideals (the zero ideal, and the whole ring itself), which gives only two possibilities for $J$ i.e. $J=I,J=R$, giving that $I$ is maximal.
Conversely, if $I$ is maximal, this implies that $\frac RI$ has only two ideals, and is therefore a field(why?). Hence, the reverse direction is also completed in this manner.