Dtyjlui

I am trying to show surjectivity is a local property, that is, for a ring $R$ and $R$-modules $M$ and $N$,

$f: M to N$ is surjective if and only if $f_{mathscr{M}}:M_{mathscr{M}} to N_{mathscr{M}}$ is surjective for all maximal ideals $mathscr{M}$ of $R$, where $M_{mathscr{M}}$ is the localization at $mathscr{M}$.

One of the directions is easy but I am having trouble seeing how to prove that $f_{mathscr{M}}:M_{mathscr{M}} to N_{mathscr{M}}$ is surjective for all maximal ideals implies that $f: M to N$ is surjective.

Atiyah Macdonald says to reverse the arrows in the injectivity proof (that uses exact sequences) but I am not sure how to go about it. Any help is much appreciated.

Exactness is a local property. So looking at the exactness of $M to N to 0$ gives that surjectivity is a local property, just as looking at the exactness of $0 to M to N$ gives that injectivity is a local property.

Doing as Atiyah and Macdonald say, reversing the arrows in their injectivity proof, works with some care.

Claim: $fcolon M to N$ is surjective if and only if $f_m colon M_m to N_m$ is surjective for every maximal ideal $m$ of $R$.

Proof: (References are to Atiyah Macdonald and this proof is an adaptation of their proof of Proposition 3.9)

$(to)$ Suppose that $f$ is surjective. Then $M to N to 0$ is an exact sequence. Because localization preserves exactness (Proposition 3.3), $M_m to N_m to 0$ is exact for all maximal ideals $m$ of $R$ and therefore each $f_m colon M_m to N_m$ is surjective.

($leftarrow$) Suppose that $f_m colon M_m to N_m$ is surjective for every maximal ideal $m$ of $R$. Let $N' = N/text{Im}(f)$ be the cokernel of $f$. Then the sequence $M to N to N' to 0$ is exact and, again because localization preserves exactness, $M_m to N_m to N'_m to 0$ is exact. Because localization commutes with quotients (Corollary 3.4), $N'_m$ is isomorphic to the cokernel of $f_m$. Because $f_m$ is surjective, this cokernel is $0$, so $N'_m = 0$ for every maximal ideal $m$ of $R$. Therefore, because being the $0$-module is a local property (Proposition 3.8), $N' = 0$. So the first three terms of the exact sequence $M to N to N' to 0$ are really $M to N to 0$, which means that $f$ is surjective. $square$

The point is to not only just reverse the arrows in the injectivity proof, but also to look at the cokernel of $f$ instead of at the kernel.

You want to prove $f(M)=N$. Consider the quotient module $frac{N}{f(M)}$.

There is a result which says that, an $R$-module $M$ is zero module if $M_{mathfrak{m}}=0$ for every maximal ideal $mathfrak{m}$ of $R$.

Look at the quotient $left(frac{N}{f(M)}right)_{mathfrak{m}}$. An element in $left(frac{N}{f(M)}right)_{mathfrak{m}}$ will be of the for $frac{[n]}{a}$ where $ain Rsetminus mathfrak{m}$.

As $frac{n}{a}in N_{mathfrak{m}}$ there exists $b(mathfrak{m})in M_{mathfrak{m}}$ such that $f_{mathfrak{m}}(b(mathfrak{m}))=frac{n}{a}$. Show that this means $frac{[n]}{a}=0$.

Thus, $left(frac{N}{f(M)}right)_{mathfrak{m}}=0 $ for every maximal ideal $mathfrak{m}$. Thus, $frac{N}{f(M)}=0$ i.e., $f(M)=N$.

Hint for the following result

$M_{mathfrak{m}}=0$ for all maximal ideals $mathfrak{m}$ of $R$ implies $M=0$.

Let $ain M$ is such that $aneq 0$.

Consider the annihilator ideal $text{Ann}(a)$ of $a$. This will be in some maximal ideal $mathfrak{m}$.

Localize $M$ at this $mathfrak{m}$ and use that $M_{mathfrak{m}}=0$. You will end up with $a=0$. Thus, $M=0$