1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $R/I$.

$\endgroup$
0
$\begingroup$

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 $K$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.