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# Second isomorphic Thm, quotients Watch

1. 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
2. (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...
3. (Original post by Mark85)
Only if S and I are disjoint but then that is what the theorem says anyway...
Oh yeh lol. Thanks

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