# Image of ideal under the isomorphism given by the Chinese Remainder Theorem.

Suppose that $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ are maximal ideals of a ring $R$.

Then $\mathfrak{p}_i+\mathfrak{p}_j=R$ with $i\neq j$ and $\mathfrak{p}_i^a+\mathfrak{p}_j^b=R$ with $a,b$ positive integers.

By the Chinese remainder theorem $$R/\mathfrak{p}_1^{a_n}\cdots \mathfrak{p}_n^{a_n}\approx R/\mathfrak{p}_1^{a_1}\oplus \cdots \oplus R/\mathfrak{p}_n^{a_n}$$

with $a_i$ positive integers.

My question is:

Under this isomorphism the image of the ideal $\mathfrak{p}_1^{b_1}\cdots\mathfrak{p}_n^{b_n}/\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}$ is $\mathfrak{p}_1^{b_1}/\mathfrak{p}_1^{a_1}\oplus\cdots\oplus \mathfrak{p}_n^{b_n}/\mathfrak{p}_n^{a_n}$ for positive integers $b_i\leq a_i$?

It is clear that the image of the ideal $\mathfrak{p}_1^{b_1}\cdots\mathfrak{p}_n^{b_n}/\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}$ is a subset of $\mathfrak{p}_1^{b_1}/\mathfrak{p}_1^{a_1}\oplus\cdots\oplus \mathfrak{p}_n^{b_n}/\mathfrak{p}_n^{a_n}$.

But the other inclusion not is clear for my.

Thank you all.

Note: The isomorphism in the Chinese Remainder Theorem is:

$$R/\mathfrak{p}_1^{a_n}\cdots \mathfrak{p}_n^{a_n}\approx R/\mathfrak{p}_1^{a_1}\oplus \cdots \oplus R/\mathfrak{p}_n^{a_n}, \ x+\mathfrak{p}_1^{a_n}\cdots \mathfrak{p}_n^{a_n}\mapsto (x+\mathfrak{p}_1^{a_1},\ldots,x+\mathfrak{p}_n^{a_n}).$$

• The statement is trivial after localization at each maximal Ideal of $R$. – MooS Feb 22 '15 at 9:11
• @MooS; why it is trivial? – user126033 Feb 23 '15 at 22:14
• After localization the map is either the identity or the zero map between zero modules. In particular your two ideals identify. – MooS Feb 24 '15 at 5:50
• I don't understand; "the map is either the identity or the zero map between zero modules". – user126033 Feb 28 '15 at 16:48