If $R$ is a ring, $K$ a field (and subring of $R$), and $I$ a proper ideal of $R$, $R/I$ contains a field isomorphic to $K$ Let $R$ a ring, $K$ subring of $R$ and $I$ a proper ideal of $R$. Now suppose $K$ is a field. I need to prove that $R/I$ contains some field isomorphic to $K$.
My idea is to take $K/I$ as that subfield of $R/I$. I tried to prove that the function
$$\phi:K\rightarrow K/I,\quad \phi(k)=k+I$$
is a isomorphism.
It was easy to see that $K/I$ is a field, $\phi$ is homomorphism and surjective, but I think that it is not injective, since $\textrm{Ker}(\phi)=I$. 
Am I right? Please, only hints, not the whole answer.
 A: For $K$ unital subring of $R$ then $\phi : R \to R/I$ restricts to an homomorphism $\phi|_K:K \to R/I$ whose kernel is $I \cap K$. Since $K$ is a field if $I \cap K$ is larger than $\{0\}$ then $1 \in I$ and $I = R$. Otherwise $\phi|_K$ is injective and its image $\{ a+I, a \in K\}$ is a copy of $K$ in $R/I$.
A: Any non-trivial ring homomorphism of a field is necessarily injective: If $k\in K$ is a non-zero element and $\psi:K\to S$ a non-trivial ring homomorphism, then
$$
1_S=\psi(1_K)=\psi(kk^{-1})=\psi(k)\psi(k^{-1})
$$
so $\psi(k)$ cannot be $0_S$.
So, assuming $K\not\subseteq I$ (which may follow from "$I$ is a proper ideal of $R$", depending on your definition of "subring"), we must have that $\phi(K)$ is isomorphic to $K$.
A: A slightly different take on it:
Note that with $R$ a unital ring, 
$K \subset R \tag 1$
a subfield, and 
$I \subsetneq R \tag 3$
a proper ideal, we must have
$K \cap I = \{ 0_R \}, \tag 4$
for if
$\exists 0 \ne k \in K \cap I, \tag 5$
then for any $r \in R$,
$r = (rk^{-1})k \in I \Longrightarrow I = R, \tag 6$
contradicting the assumption (3) that $I$ is proper.
These observations in turn imply that for
$k_1, k_2 \in K, \; k_1 \ne k_2, \tag 7$
their cosets satisfy
$k_1 + I \ne k_2 + I, \tag 8$
i.e., are distinct; indeed,
$k_1 + I = k_2 + I \Longleftrightarrow k_1 - k_2 \in I, \tag 9$
whence by virtue of (4),
$k_1 - k_2 \in K \cap I \Longrightarrow k_1 - k_2 = 0, \tag{10}$
contradicting (7).  It is now easy to see that the canonical projection map
$\pi: R \to R/I, \; \pi(r) = r + I, \; \forall r \in R, \tag{11}$
is injective when restricted to the subfield $K$, and thus that $\pi(K)$ is an isomorphic image of $K$ in $R/I$.  $OE\Delta$.
