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.

  • $\begingroup$ What does $K/I$ mean? $\endgroup$ – Brian Moehring Aug 21 at 21:49
  • $\begingroup$ 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 $I.$ $\endgroup$ – 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 $R/I$. $OE\Delta$.


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