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I saw an exact sequence of ideals

$$0 \rightarrow I \cap J\rightarrow I \oplus J \rightarrow I + J \rightarrow 0$$In this sequence, maps are ring homomorpism? module homomorphism?

And can the above sequnce yield the exact sequnce?

$$0 \rightarrow R/I \cap J\rightarrow R/I \oplus R/J \rightarrow R/I + J \rightarrow 0$$

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2 Answers 2

up vote 6 down vote accepted

The maps are module homomorphisms; more precisely, they are:

$I \cap J \mapsto I \oplus J $ via $x \mapsto (x,x)$,

and

$I\oplus J \mapsto I + J $ via $(x,y) \mapsto x - y$.

If you embed this exact sequence as a subsequence of the obvious short exact sequence

$ 0 \to R \to R\oplus R \to R \to 0$

(with maps defined by the same formulas), then the quotient is the second short exact sequence that you ask about.

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I would just like to add that the short exact sequence of the quotients arises from the snake lemma. –  Manos Oct 6 at 16:25

Matt E. answers the question, but I just wanted to add a couple of things.

First note a proper ideals of $R$ does not inherit a ring structure from $R$ because it does not contain $1$. Also, there is no notion of exact sequences for rings, since the kernel of a ring homomorphism is not a ring (but an ideal).

The maps are $R$-module homomorphisms, so indeed the first and last term of the first sequence are ideals of $R$. The middle term $I \oplus J$ is just the direct sum of $I$ and $J$ as $R$-modules; it is not an ideal of $R$.

The second exact sequence comes from applying the snake lemma to the morphism of short exact sequences which Matt points to; its "cokernel" part is your sequence.

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There is a notion of exact sequence of non-unital rings, though! –  Qiaochu Yuan Aug 1 '12 at 14:05

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