Let $R$ be a ring and $x,y \in R$ s.t. $ \langle x \rangle , \langle y \rangle $ are ideals of $R$, then prove that $(R/ \langle x \rangle )/ \langle y \rangle \approx R/ \langle x,y \rangle $

Not sure if the conditions listed are enough to prove this but here's my attempt:

Attempt: Consider a mapping $\Psi : R/ \langle x \rangle \rightarrow R/ \langle x,y \rangle $ Such that

$ \Psi(r + \langle x \rangle) = r + \langle x,y \rangle $

This is clearly a ring homomorphism as :

(i) $\Psi[ (r_1 + \langle x \rangle ) + (r_2 + \langle x \rangle )] = \Psi[ (r_1 + \langle x \rangle )] + \Psi[(r_2 + \langle x \rangle )] $

(ii) $\Psi[ (r_1 + \langle x \rangle )\cdot (r_2 + \langle x \rangle )] = \Psi[ (r_1 + \langle x \rangle )]\cdot \Psi[(r_2 + \langle x \rangle )]$

To find the kernel of this homomorphism : We need a particular $r + \langle x \rangle$ such that $\Psi(r + \langle x \rangle) \in \langle x,y \rangle$

$i.e ~~r + \langle x,y \rangle \in \langle x,y \rangle \implies r \in \langle x,y \rangle \implies r+\langle x \rangle=\langle x,y \rangle \implies $ Kernel of $\Psi=\langle x,y \rangle$

Where could I have gone wrong?

Thank you for your help..


The points (i) and (ii) do not verify that $\Psi$ is a ring homomorphism. You forgot to show that it is well defined, preserves the one element, and is compatible with subtraction. But nevertheless is not necessary here to check any of these properties, since the existence of $\Psi$ is given by the fundamental theorem on homomorphisms.

Regarding your question: In the ring $R/ \langle x \rangle$ we have $\langle x,y \rangle = \langle y \rangle$, since $x = 0$. Also note that the statement you want to show is a special case of the third isomorphism theorem, from which it easily follows.

  • $\begingroup$ The third isomorphism theorem says that $(R/B)/(A/B) \approx R/A$ where $A$ and $B$ are ideals of $R$ with $B \subseteq A$. If $A = \langle x,y \rangle , B = \langle x \rangle $, then $(R/B)/(A/B) \approx R/A \implies (R/\langle x \rangle)~/~(\langle x,y \rangle)/\langle x \rangle \approx R/\langle x,y \rangle $ $\endgroup$ – MathMan Aug 4 '14 at 9:55
  • $\begingroup$ @VHP: Right. And since $\langle x,y \rangle / \langle x \rangle$ is a principal ideal generated by $y$ this is exactly what you want. $\endgroup$ – Dune Aug 4 '14 at 9:57
  • $\begingroup$ uhm, So : $(\langle x,y \rangle)/\langle x \rangle = \langle y \rangle$ ; $(\langle x,y \rangle)/\langle x \rangle = ax+by + \langle x \rangle$ . Not sure why this is $= \langle y \rangle$ $\endgroup$ – MathMan Aug 4 '14 at 10:00
  • $\begingroup$ @VHP: As I said, $x$ equals the zero element of $R/\langle x \rangle$. You can always discard $0$ from a generating set of an ideal. $\endgroup$ – Dune Aug 4 '14 at 10:03
  • $\begingroup$ @VHP: I don't understand your second equation. Are you comparing an ideal to one of its elements? $\endgroup$ – Dune Aug 4 '14 at 10:05

IMO, the fact there are some technical details left out from the homomorphism verification is secondary to a more important problem: keeping ideals of $R$ and $R/\langle x\rangle$ separate, especially when using $\langle\rangle$ to denote ideals generated by elements in both of these rings.

If you overcome this first problem I might check you on the issue of the homomorphism too, but that would be a simpler issue :) Actually, your line of reasoning contains all the seeds of a reasonable answer (modulo the way you've expressed them.)

When you write "$(R/\langle x\rangle)/\langle y\rangle$" where $\langle x\rangle$ and $\langle y\rangle$ are ideals of $R$," we immediately need to point out that anything written under "$(R/\langle x\rangle)/-$" is a subset of $R/\langle x\rangle$, not of $R$. So the only way this is meaningful is if by $\langle y\rangle$, you meant the ideal generated in $R/\langle x\rangle$ rather than in $R$. (In the original problem, you may as well just have said $y\in R$ rather than $\langle y \rangle \lhd R$.)

For clarity, you'd probably rather write $\langle y +\langle x\rangle \rangle$ rather than $\langle y\rangle$ to sidestep this issue. We will see, after all, that you are really interested in the ideal of $R/\langle x\rangle$ generated by $y+\langle x\rangle$.

With this in mind, let's look at your last line:

$r + \langle x,y \rangle \in \langle x,y \rangle \implies r \in \langle x,y \rangle \implies r+\langle x \rangle=\langle x,y \rangle \implies \ker\Psi=\langle x,y \rangle$

The first half is fine, but the last half uses some fuzzy bookkeeping with cosets that leads to a confused conclusion.

Let's start from $r\in \langle x,y\rangle$ and try something naive. We know that $r=\sum a_ixb_i +\sum c_iyd_i$ since it's in $\langle x ,y\rangle$. But then $r+\langle x\rangle$ would be $$r+\langle x\rangle=\sum a_ixb_i +\sum c_iyd_i +\langle x\rangle=\sum c_iyd_i +\langle x\rangle\in \langle y +\langle x\rangle\rangle$$.

So we should conclude that $\langle y +\langle x\rangle\rangle$ is the kernel of $\Psi$ rather than $\langle x,y \rangle$. Notice that this is a subset of $R/\langle x\rangle$, as compared to $\langle y\rangle$ which you initially emphasized to be an ideal of $R$.)

What's that look like in your original expression? As you were angling for, the first isomorphism theorem says:

$(R/\langle x\rangle)/\langle y +\langle x\rangle\rangle\cong R/\langle x,y\rangle$

So, you were headed for the right idea, but the slightly ambiguous use of notation was making you unsure.

  • $\begingroup$ Ohh, I get this :-) . Thank you for your answer. :-) $\endgroup$ – MathMan Aug 4 '14 at 13:15

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.