# 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 $$\textrm{Ker}(\phi)=I$$.

• What does $K/I$ mean? – Brian Moehring Aug 21 at 21:49
• To answer my own question, since reuns already filled in the details I was trying to hint at, $I\neq 0$ is not an ideal of $K,$ so $K/I$ doesn't really make much sense. Based on what you wrote you are instead defining $K/I$ to be the image of the composition of canonical maps $$K \rightarrow R \rightarrow R/I$$ and then defining $\phi$ to be the composition, restricted to the image. As reuns mentioned, this definition gives $\ker(\phi) = I \cap K,$ not simply $I.$ – Brian Moehring Aug 21 at 22:56

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 $$R/I$$.

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 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 $$R/I$$. $$OE\Delta$$.