MHB Proof of Exact Sequence Let $R$ & A Closed Set

[Fermat1]

Let $R$ be a commutative ring and $0\to L\to M\to N\to 0$ be a sequence of $R$ modules. Let $A$ be a multiplicativity closed subset of $R$ so that we can consider the corresponding localisation sequence: $0\to A^{-1}L\to A^{-1}M\to A^{-1}N\to 0$. Suppose that the localisation sequence is exact whenever $A=R\setminus m$ for some maximal ideal $m$ of $R$. Prove that $0\to L\to M\to N\to 0$ is exact.

Thanks

 

[Deveno]

Re: exact sequence proof

What have you tried?

Start with showing that if we have the sequence:

$0 \to L \stackrel{u}{\to}M\stackrel{v}{\to}N\to 0$

that the map:

$A^{-1}u:A^{-1}L \to A^{-1}M$ given by $A^{-1}u(x/a) = u(x)/a$ (for $x \in L$) is injective,

and the map:

$A^{-1}v:A^{-1}M \to A^{-1}N$ given by $A^{-1}(y/a) = v(y)/a$ (for $y \in M$) is surjective,

(this is implied by the statement of the problem: the maximality of the ideal $\mathfrak{m}$ is just to guarantee that this ideal is prime, so that $R - \mathfrak{m}$ is multiplicatively closed).

and argue that this implies $u$ is injective, and $v$ surjective.

The inclusion $\text{im u} \subseteq \text{ker }v$ should be straight-forward, what about the other inclusion?

(You may find it helpful to recall that $A^{-1}u$ is injective if and only if $A$ has no zero-divisors).

 

[Fermat1]

Re: exact sequence proof

What have you tried?

Start with showing that if we have the sequence:

$0 \to L \stackrel{u}{\to}M\stackrel{v}{\to}N\to 0$

that the map:

$A^{-1}u:A^{-1}L \to A^{-1}M$ given by $A^{-1}u(x/a) = u(x)/a$ (for $x \in L$) is injective,

and the map:

$A^{-1}v:A^{-1}M \to A^{-1}N$ given by $A^{-1}(y/a) = v(y)/a$ (for $y \in M$) is surjective,

(this is implied by the statement of the problem: the maximality of the ideal $\mathfrak{m}$ is just to guarantee that this ideal is prime, so that $R - \mathfrak{m}$ is multiplicatively closed).

and argue that this implies $u$ is injective, and $v$ surjective.

The inclusion $\text{im u} \subseteq \text{ker }v$ should be straight-forward, what about the other inclusion?

(You may find it helpful to recall that $A^{-1}u$ is injective if and only if $A$ has no zero-divisors).

Why is the map $A^{-1}u$ automatically injective? I get it when we are assuming the original sequence to be exact, but that's not what we're doing.

 

[Deveno]

Re: exact sequence proof

Is it not that the corresponding localization sequence is given to be exact?

 

[Fermat1]

Re: exact sequence proof

Is it not that the corresponding localization sequence is given to be exact?

No. This is partly the reason the question is throwing me. The hypothesis is that the localisation sequence is exact whenever $A=R\m$ for some maximal ideal $m$

 

[Deveno]

Re: exact sequence proof

We start with a sequence:

$0 \to L \stackrel{u}{\to}M\stackrel{v}{\to}N\to 0$

We know nothing about the nature of $u,v$ except they are $R$-linear maps.

Take ANY maximal ideal $\mathfrak{m}$ of $R$. Let $A = R - \mathfrak{m}$.

Form the corresponding sequence:

$0 \to A^{-1}L \stackrel{A^{-1}u}{\to}A^{-1}M\stackrel{A^{-1}v}{\to}A^{-1}N\to 0$

We are given that THIS sequence is exact.

In particular, this means that $A^{-1}u$ is injective, and $A^{-1}v$ is surjective (from exactness).

We know this is true because the problem SAYS so (we are free to pick any maximal ideal).

We aren't deriving exactness from some "other" exact sequence, we are using the exactness of the localized sequence to show the original sequence must have been exact originally.

 

[Fermat1]

Re: exact sequence proof

We start with a sequence:

$0 \to L \stackrel{u}{\to}M\stackrel{v}{\to}N\to 0$

We know nothing about the nature of $u,v$ except they are $R$-linear maps.

Take ANY maximal ideal $\mathfrak{m}$ of $R$. Let $A = R - \mathfrak{m}$.

Form the corresponding sequence:

$0 \to A^{-1}L \stackrel{A^{-1}u}{\to}A^{-1}M\stackrel{A^{-1}v}{\to}A^{-1}N\to 0$

We are given that THIS sequence is exact.

In particular, this means that $A^{-1}u$ is injective, and $A^{-1}v$ is surjective (from exactness).

We know this is true because the problem SAYS so (we are free to pick any maximal ideal).

We aren't deriving exactness from some "other" exact sequence, we are using the exactness of the localized sequence to show the original sequence must have been exact originally.

Ok I 'get' the question now. I want to prove $u$ is injective. Suppose $x$ non-zero is in Ker$u$. The consider Ann($x$). This is a proper ideal hence contianed in a maximal ideal $m$. Let $A=R-m$. We have that $A^{-1}u(\frac{x}{1})=0$. So injectivity of $A^{-1}u$ implies that there exists $c$ in $A$ with $cx$=0. Then $c$ is in the annihalator of $x$ contary to $A$ being disjoint with the annihalator

 

[Deveno]

Re: exact sequence proof

Ok I 'get' the question now. I want to prove $u$ is injective. Suppose $x$ non-zero is in Ker$u$. The consider Ann($x$). This is a proper ideal hence contianed in a maximal ideal $m$. Let $A=R-m$. We have that $A^{-1}u(\frac{x}{1})=0$. So injectivity of $A^{-1}u$ implies that there exists $c$ in $A$ with $cx$=0. Then $c$ is in the annihalator of $x$ contary to $A$ being disjoint with the annihalator

Yes, that's one way to do it (and a clever use of the equivalence in $A^{-1}M$).

Now you need to show $v$ is surjective, and that $\text{ker }v = \text{im }u$. Again, you are free to use whatever ideals prove useful, here, to create the set $A$.

 

[Fermat1]

Re: exact sequence proof

Yes, that's one way to do it (and a clever use of the equivalence in $A^{-1}M$).

Now you need to show $v$ is surjective, and that $\text{ker }v = \text{im }u$. Again, you are free to use whatever ideals prove useful, here, to create the set $A$.

I can use the annihalator trick to show that Im(u) is a subset of ker(v). I don't see what ideal to choose to get the reverse inclusion

 

[ThePerfectHacker]

Re: exact sequence proof

I am confused. We do not need to assume that the localized sequence is exact, it is a true regardless of the multiplicative set we localize the modules at.

 

[Deveno]

Re: exact sequence proof

I am confused. We do not need to assume that the localized sequence is exact, it is a true regardless of the multiplicative set we localize the modules at.

Well, not quite...only localizations at maximal ideals are assumed to be exact. But yeah, pick one...anyone you like, the exact nature of the (prime) ideal chosen isn't that important.

As I understand it, given any prime ideal $\mathfrak{p}$, the functor: $M \to M_{\mathfrak{p}}$ (from $R-\mathbf{Mod} \to R-\mathbf{Mod}$) is exact (both ways). It seems to me the same should be true regardless of what the multiplicative set is, as well.

 

[Fermat1]

Re: exact sequence proof

Well, not quite...only localizations at maximal ideals are assumed to be exact. But yeah, pick one...anyone you like, the exact nature of the (prime) ideal chosen isn't that important.

As I understand it, given any prime ideal $\mathfrak{p}$, the functor: $M \to M_{\mathfrak{p}}$ (from $R-\mathbf{Mod} \to R-\mathbf{Mod}$) is exact (both ways). It seems to me the same should be true regardless of what the multiplicative set is, as well.

did you miss my post?

 

[ThePerfectHacker]

Let $R$ be a commutative ring and $0\to L\to M\to N\to 0$ be a sequence of $R$ modules. Let $A$ be a multiplicativity closed subset of $R$ so that we can consider the corresponding localisation sequence: $0\to A^{-1}L\to A^{-1}M\to A^{-1}N\to 0$.

This corresponding localization sequence is exact regardless of $A$, here is a proof.

Suppose $0 \to L \to M \to N\to 0$ is exact sequence of $R$-modules, then,
$ 0 \to A^{-1} L \to A^{-1} M \to A^{-1} N \to 0 $ is exact sequence of $A^{-1}R$-modules.

Let $f:L\to M$ and $g:M\to N$. The induced maps are $f^*:A^{-1}L \to A^{-1}M$ given by $f(l/a) = f(l)/a$ and $g^*:A^{-1} M \to A^{-1} N$ given by $g(m/a) = g(m)/a$.

To show that $ 0 \to A^{-1} L \to A^{-1} M \to A^{-1} N \to 0 $ is exact we have to show that $f^*$ is injective, $g^*$ is surjective, and $\text{im} f^* = \ker g^*$. We verify all of these three properties in order:

(i) $f^*$ injective: Suppose that $f^*(l/a) = 0$ so $f(l)/a = 0/a$ and it follows that $a'f(l) = 0$ for some $a' \in A$ from definition of fractions with denominators in $A$. But $f:L\to M$ is a map of $R$-modules, hence $a'f(l) = 0 \implies f(a'l) = 0$, as $f$ is injective we get $a'l = 0$. Thus, $l/a = a'l/aa' = 0/aa' = 0$.

(ii) $g^*$ is surjective: Take any $n/a \in A^{-1} N$, as $g:M\to N$ is surjective choose $m\in M$ such that $g(m) = n$. We then have that $g^*(m/a) = g(m)/a = n/a$.

(iii) $\text{im} f^* \subseteq \ker g^*$: Let $m/a \in \text{im} f^*$ and write $m/a = f^*(l/a')$ for some $l\in L$, and $a'\in A$, so $m/a = f(l)/a'$. Thus, $g^*(m/a) = g^*(f(l)/a) = g(f(l))/a = 0$ as the original sequence was exact and so $g(f(l)) = 0$.

(iv) $\ker g^* \subseteq \text{im} f^*$: Let $m/a \in A^{-1} M$ be such that $g^*(m/a) = 0$, so that, $g(m)/a = 0$, hence, $a'g(m) = 0$ for some $a'\in A$. Now as $g:M\to N$ is a map of modules we can write $g(a'm) = 0$. By the exactness of the original sequence we get $a'm = f(l)$ for some $l\in L$. Thus, $f(l/aa') = f(l)/aa' = a'm/aa' = m/a$.