The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Second isomorphic Thm, quotients

Hi,

I just want some clarification. I am always overthinking, but would like to know the reasoning behind this:

R Ring, I ideal of R, S subring of R

(S+I)/I ~ S/(I n S)

Now I was just thinking about the first term.

S+I = {s + a: seS, aeI}

Now if u take the quotient ring of this (S+I)/I, then essentially you are turning every element a to be zero. So isnt (S+I)/I just S?

Thanks
Reply 1
Avatar for Mark85
Mark85
16
Original post by 2710
Hi,

I just want some clarification. I am always overthinking, but would like to know the reasoning behind this:

R Ring, I ideal of R, S subring of R

(S+I)/I ~ S/(I n S)

Now I was just thinking about the first term.

S+I = {s + a: seS, aeI}

Now if u take the quotient ring of this (S+I)/I, then essentially you are turning every element a to be zero. So isnt (S+I)/I just S?

Thanks


Only if S and I are disjoint but then that is what the theorem says anyway...
Reply 2
Avatar for 2710
2710
OP
18
Original post by Mark85
Only if S and I are disjoint but then that is what the theorem says anyway...


Oh yeh lol. Thanks

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you