0
$\begingroup$

I am having a very hard understanding quotient rings. I never really understood quotient groups in abstract 1 but did well enough to get by. Now I am in abstract 2 and we hit quotient rings. I really want to understand quotient rings this time around.

We don't use a specific textbook in my class just handouts...but these rings may pr may not be commutative but with unity. From my understanding quotient rings are with $R$ being a ring and $I$ an ideal of $R$ then $R/I$ is a ring with the additive operation such that $(a+I)+(b+I)= (a+b)+I$. Likewise multiplication is defined as $(a+I)(b+I)= (ab+I)$. (Where $a$ and $b$ are elements of $R$)

I am having trouble understanding how to represent the quotient ring $R/I$. Is every element (which is a coset) in the form of an $r+I$?

Can someone help me go through what should be trivial (but not to me )of why $R/R = 0$ (zero ring) and $R/0 = R$.

I think that $R/R$ is in the form of $(r+R)$ but I don't see what to do to get zero.

$R/0$ is going to be in the form of $r+0=r$. Do I now say for all $r$ in $R$ ?

I understand these proofs should be very trivial and easy but I feel like going through in detail of why exactly this happens will help me solidify what I know about quotient rings so I can actually do my homework (which are more involved)

$\endgroup$
2
  • $\begingroup$ To show that $R/R=0$, what you need to do is show that every element is zero. Since the zero element is $0+R$, you need to show that, for all $r\in R$, $r+R = 0+R$. Likewise, to show that $R/0 \cong R$, you need to write down a map between the two rings and show it is an isomorphism. (Hint: There is an obvious function $R\to R/0$.) $\endgroup$
    – Andrew Dudzik
    Commented Apr 22, 2016 at 4:57
  • $\begingroup$ Thanks Slade. The function that makes sense is the inclusion map. Thanks so much guys I think this really helped clarify things. I'll buckle down and figure this all out now! $\endgroup$
    – Rey
    Commented Apr 22, 2016 at 5:08

1 Answer 1

Reset to default
0
$\begingroup$

Yes, every element of $R/I$ is of the form $r+I$.

Recall that $x+I$ and $y+I$ represent the same element if $x-y \in I$. As $r- 0 \in R$ for all $r \in R$, we have any coset $r+R$ is the same as $0+R$. Hence $R/R = 0$.

Likewise every $r \in R$ defines a distinct coset $r+0$, hence $R/0 \cong R$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .