False statement

Let the ring $R$ and its ideals $I$ and $J$ s.t. $I\subset J$.

$R/J$ is a ideal (or subring) of the $R/I$ ?

Since $R/J$ is not a subset of the $R/I$ (actually $R/J$ is a isomorphic with some quotient ring of the $R/I$, the above statement is surely incorrect.

For example $Q[x]/\langle(x-1) \rangle $ is not a subset of the $Q[x]/\langle(x-1)(x-2) \rangle$

(When we taking the $I (= \langle (x-1)(x-2) \rangle) \subset J(= \langle (x-1) \rangle) $)

The other example $Z/\langle 2 \rangle $ is not a subset of the $Z /\langle 4 \rangle$ (When we taking the $I (= \langle 4 \rangle) \subset J(= \langle 2 \rangle) $)


Let the ring $R$ and its ideal $I$ and $J$ s.t. $I\subset J$

Let the $R_I$ be a subring or ideal of $R/I$

Then Does $R_I$ exist $s.t.$ isomorphic with the $R/J$ ?

(I.E. I want to know existence of $R/I$'s subring or ideal who is ismorphic with the $R/J$ )

It looks like a true when we considering the above two examples.(It's just my thoght.)

$Q \times \{0\}(\simeq Q[x]/\langle(x-1) \rangle \simeq Q )$ is a subring or ideal of the $Q[x]/\langle(x-1)(x-2) \rangle$ .

$\{ 1,3\}(\simeq Z/\langle 2 \rangle \simeq Z_2) $ is a subring or ideal of the $Z /\langle 4 \rangle$.

But I'm not sure this is right or not.

(Cause I couldn't find the other counterexmaples and don't know how to prove. :()

What do you think about that?

Any help would be appreciated.

Thanks for reading my post.

  • 1
    $\begingroup$ If I understood your notations correct, $R_I=\{ 1,3\}$ is not a subring of $Z /\langle 4 \rangle$, becuase $1+1=2\not\in R_I$. $\endgroup$ – Alex Ravsky Jul 27 at 4:16
  • 1
    $\begingroup$ Yes, you've already understood my notation correctly! Ah thanks for answering. $\endgroup$ – se-hyuck yang Jul 27 at 6:31

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