Why Is $I$ A Maximal Ideal If $R/I$ Is A Field?

Question

How can I prove that if $R/I$ is a field then $I$ is a maximal ideal? In my book it says that this is a corollary of the following theorem:

If $φ:\mathbb{F}→S$ is a non trivial homomorphism and $\mathbb{F}$ is a field then $φ$ is injective.

Can someone explain to me why this is true?

edit: any proof would be helpful, not just using the theorem above.

Answer 1

Here is a proof that uses the given theorem.

Consider any ideal $J \supseteq I$ of $R$. We wish to prove that $J$ must be either $R$ or $I$ (so that $I$ is a maximal ideal).

Consider the canonical homomorphism $\varphi: R/I \to R/J$, $\varphi(x + I) = x + J$, $\forall x \in R$.

If $\varphi$ is trivial, then $x + J = J$ for all elements $x \in R$, and this implies that $J = R$.

If $\varphi$ is non-trivial, then it is injective (by the given theorem), so that $\ker \varphi = \{I\}$ ($I$ being the zero of $R/I$). Thus, for any element $x \in R$, $\varphi(x + I) = x + J = J$ if and only if $x + I = I$ or equivalently, $x \in J$ if and only if $x \in I$. This proves that $I = J$.

Answer 2

If $I$ is a maximal ideal, let $p:R\rightarrow R/I$ the canonical projection, if $x$ is not an element of $I$, then the ideal generated by $I$ and $x$ is $R$ thus contains $1$, we deduce that $1=xy+i, i\in I$ and $p(xy)=p(x)p(y)=1$. This implies that $p(x)$ is invertible and henceforth $R/I$ is a field.

On the other hand, if $R/I$ is a field, consider $J$ an ideal which contains $I$ distinct of $I$, there exists $x\in J$ not in $I$. This implies that $p(x)$ is not trivial thus is invertible, there exists $y$ such that $p(x)p(y)=p(1)$ or equivalently $xy-1\in I\subset J$. Since $x\in J$, $xy\in J$ and henceforth $1\in J$. This implies that $J=R$ and $I$ is maximal.

You can characterize a field (I believe it is what is saying your book) by the property every non trivial morphism $f:F\rightarrow S$ is injective. If you have a morphism $f:R/I\rightarrow S$, it $f\circ p$ is a morphism whose kernel contains $I$ thus its kernel is $I$ or $R$ since $I$ is maximal, thus $f$ is trivial or injective.