MHB Localization - Dummit and Foote, EXAMPLE 2, Ch. 15, Section 15.4, page 708

[Math Amateur]

I am reading Dummit and Foote, Example 2 of Section 15.4, page 708.

Rewriting the assertions of the example as exercise style questions, EXAMPLE 2, reads as follows:

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Let R be any commutative ring with 1 and let f be any element of R. Let D be the multiplicative set $$ \{ f^n \ | \ n \ge 0\} $$ of non-negative powers of f in R.

Define $$ R_f = D^{-1}R $$.

(a) Show that $$ R_f = 0 $$ if and only if f is nilpotent.

(b) Show that if f is not nilpotent, then f becomes a unit in R

(c) Show that $$ R_f \cong R[x]/(xf -1)$$

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I can manage (a) and (b) I think - my attempts follow - but need help to get started on (c)

Attempt at (a)

$$ R_f = \{ r/d \ | \ r \in R, d \in D \} $$

where r/d = s/e if and only if x(er - ds) = 0 for some $$ x \in D $$

$$ f $$ nilpotent $$ \Longrightarrow f^m = 0 $$ for some $$ m \in Z^+ $$

$$ \Longrightarrow \ 0 \in D $$

$$ \Longrightarrow \ R_f = D^{-1}R = \{ 0 \} $$

Attempt at (b)

$$ R_f = \{ \frac{r}{f^m} \ | \ r \in R, f^m \in R_f = D^{-1}R $$

Now $$ f \in R_f \Longrightarrow \ \frac{f^1}{f^0} \in R_f \Longrightarrow \ \frac{f^1}{1} $$

But we have $$ \frac{1}{f^1} \in R_f $$

So f is a unit in $$ R_f $$ since $$ \frac{f^1}{1} \times \frac{1}{f^1} = \frac{f^1}{f^1} = 1 $$ where $$ \frac{f^1}{1} \in R_f $$ and $$ \frac{1}{f^1} \in R_f $$

Can someone please confirm that arguments (a) and (b) above are OK?


Attempt at (c)

PROBLEM ... ... I cannot get started on the third assertion above ie (c) above. Can someone please help me get started on (c) ... ...

Peter

 

[Deveno]

Your argument for (a) is only "one way":

$f$ nilpotent $\implies R_f = 0$.

You also need to show that:

$R_f = 0 \implies f$ is nilpotent (which isn't hard: recall that

$D^{-1}R = 0 \iff 0_R \in D$).

For part (b) it is sometimes handy to draw a distinction between $R$ and its image in $R_f$. Also, your description of $R_f$ is a bit off, it should be:

$R_f = \{r/f^m : r \in R, f^m \in D\}$

Just because $f$ is not nilpotent doesn't mean we have a monomorphism $R \to R_f$ it is still possible we have some $g \in R$ with $fg = 0$ (we only know that $g$ is NOT a power of $f$).

Also, remember that $r/1$ is an *equivalence class*, so for example to show that:

$(f/1)(1/f) = f/f = 1/1 = 1_{R_f}$ we need to find some power of $f$ such that:

$f^m(1f - f1) = 0$ (and, of course ,any $m$ will do).

In other words, $f/f$ does not EQUAL 1, but it is EQUIVALENT to its image in $R_f$.

For (c) it may be more helpful to write the ideal generated as:

$(fx - 1)$ which makes it clear $f$ is a coefficient of $x$ in $R[x]$.

To prove these two rings are isomorphic, we need an surjective homomorphism:

$\phi: R[x] \to R_f$.

The next step is to show that $\text{ker}(\phi) = (fx - 1)$.

My suggestion, try:

$\phi(p(x)) = p(1/f)$.

Is this a ring homomorphism? Is it surjective? What is its kernel?

 

[Math Amateur]

Your argument for (a) is only "one way":

$f$ nilpotent $\implies R_f = 0$.

You also need to show that:

$R_f = 0 \implies f$ is nilpotent (which isn't hard: recall that

$D^{-1}R = 0 \iff 0_R \in D$).

For part (b) it is sometimes handy to draw a distinction between $R$ and its image in $R_f$. Also, your description of $R_f$ is a bit off, it should be:

$R_f = \{r/f^m : r \in R, f^m \in D\}$

Just because $f$ is not nilpotent doesn't mean we have a monomorphism $R \to R_f$ it is still possible we have some $g \in R$ with $fg = 0$ (we only know that $g$ is NOT a power of $f$).

Also, remember that $r/1$ is an *equivalence class*, so for example to show that:

$(f/1)(1/f) = f/f = 1/1 = 1_{R_f}$ we need to find some power of $f$ such that:

$f^m(1f - f1) = 0$ (and, of course ,any $m$ will do).

In other words, $f/f$ does not EQUAL 1, but it is EQUIVALENT to its image in $R_f$.

For (c) it may be more helpful to write the ideal generated as:

$(fx - 1)$ which makes it clear $f$ is a coefficient of $x$ in $R[x]$.

To prove these two rings are isomorphic, we need an surjective homomorphism:

$\phi: R[x] \to R_f$.

The next step is to show that $\text{ker}(\phi) = (fx - 1)$.

My suggestion, try:

$\phi(p(x)) = p(1/f)$.

Is this a ring homomorphism? Is it surjective? What is its kernel?

Thanks for the help Deveno.

I have now read and reflected on your advice regarding parts (a) and (b) - most helpful ... but just one clarification re part (b) ,,,

I see your point regarding equivalence classes,but do not fully understand the point:

"Just because $f$ is not nilpotent doesn't mean we have a monomorphism $R \to R_f$ it is still possible we have some $g \in R$ with $fg = 0$ (we only know that $g$ is NOT a power of $f$)."

Can you explain your point further and point out the exact implications/consequences of this point for the proof of (b).

By the way, is a monomorphism just an injective homomorphism?

Peter

 

[Deveno]

Yes, a monomorphism (in the case of rings) is an injective homomorphism.

The point is, if $R \to R_f$ is not injective, we cannot regard $R$ as a subring of $R_f$, so it is preferrable to speak of $[r] \in R_f$ rather than $r$.

 

[Math Amateur]

Your argument for (a) is only "one way":

$f$ nilpotent $\implies R_f = 0$.

You also need to show that:

$R_f = 0 \implies f$ is nilpotent (which isn't hard: recall that

$D^{-1}R = 0 \iff 0_R \in D$).

For part (b) it is sometimes handy to draw a distinction between $R$ and its image in $R_f$. Also, your description of $R_f$ is a bit off, it should be:

$R_f = \{r/f^m : r \in R, f^m \in D\}$

Just because $f$ is not nilpotent doesn't mean we have a monomorphism $R \to R_f$ it is still possible we have some $g \in R$ with $fg = 0$ (we only know that $g$ is NOT a power of $f$).

Also, remember that $r/1$ is an *equivalence class*, so for example to show that:

$(f/1)(1/f) = f/f = 1/1 = 1_{R_f}$ we need to find some power of $f$ such that:

$f^m(1f - f1) = 0$ (and, of course ,any $m$ will do).

In other words, $f/f$ does not EQUAL 1, but it is EQUIVALENT to its image in $R_f$.

For (c) it may be more helpful to write the ideal generated as:

$(fx - 1)$ which makes it clear $f$ is a coefficient of $x$ in $R[x]$.

To prove these two rings are isomorphic, we need an surjective homomorphism:

$\phi: R[x] \to R_f$.

The next step is to show that $\text{ker}(\phi) = (fx - 1)$.

My suggestion, try:

$\phi(p(x)) = p(1/f)$.

Is this a ring homomorphism? Is it surjective? What is its kernel?

Thanks Deveno.

So we need to use the First Isomorphism Theorem for Rings which states (Watson: Topics in Commutative Ring Theory, page 84)

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Let $$ \phi : \ R \to S $$ be an onto (surjective) homomorphism from a ring S.

Then $$ S \ \cong \ R/ker (\phi) $$

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So, as you say, we need a surjective homomorphism:

$$ \phi : \ R[x] \to R_f $$ where, as established previously:

$$ R_f = \{ r/f^m \ : \ r \in R, \ f^m \in D \} $$

Then, of course, we have, by the First Isomorphism Theorem that:

$$ R_f \ \cong \ R[x]/ker (\phi) $$

As you say we have to show that $$ \phi $$ is a homomorphism with kernel (fx -1), and that $$ \phi $$ is surjective.

I can demonstrate the first two points, but am having trouble showing that $$ \phi $$ is surjective.

Can you help?

Note that I can see that the elements of $$ \phi (p(x)) $$ belong to $$ R_f $$.

For example, if

$$ h(x) = 3x^2 + 5x + 2 $$

then $$ h(1/f) = 3/f^2 + 5/f + 2 = (3 +5f + 2f^2)/f^2 = r/f^2 $$ where $$ r \in R$$ and so clearly $$ r/f^2 \in R_f $$ ... ...

But we need to show that taking any element $$ r/f^m \in R_f $$ we can find a polynomial h(x) in R[x] such that:

$$ \phi (h(x)) = h(1/f) = r/f^m $$

Can you help?

Peter

 

[Math Amateur]

Yes, a monomorphism (in the case of rings) is an injective homomorphism.

The point is, if $R \to R_f$ is not injective, we cannot regard $R$ as a subring of $R_f$, so it is preferable to speak of $[r] \in R_f$ rather than $r$.

Thanks for the help Deveno.

After mentioning the isomorphism $$ R_f \cong R[x]/(xf -1) $$, D&F go on in Example 2 page 708 to say the following:

"Note also that $$ R_f $$ and $$ R_{f^n} $$ are naturally isomorphic for any $$ n \ge 1 $$ since both $$ f $$ and $$ f^n $$ are units in both rings."

I have several issues/problems with the above statement.

1. What is the exact nature of $$ R_{f^n} $$

[I suspect that for $$ R_{f^n} = D^{-1}R $$ we have $$ D =\{ ({f^n})^m \ | \ n \in \mathbb{z} $$ and $$ m \ge o \} = \{ f^n, \ f^{2n}, \ f^{3n}, ... \} $$ and $$ R_{f^n} = D^{-1}R $$ , but I am unsure if this is correct.]
2. What is meant by "naturally isomorphic"?

[I suspect there may be some link to the natural map $$ R \to R/I$$?]
3. Why exactly does it follow from $$ f $$ and $$ f^n $$ being units in both rings that $$ R_f $$ and $$ R_{f^n} $$ are naturally isomorphic for any $$ n \ge 1 $$
(Note: I feel that I am not fully appreciating the significance of an element of $$ D^{-1}R $$ being a unit.)I would appreciate help/clarification of these issues.

Peter

 

[Deveno]

1) Isomorphic to $R_f$.

Your suspicion is correct, the isomorphism is:

$r/f^m \mapsto r/f^{mn}$

This is an isomorphism *because* $f$ is non-nilpotent, and is because:

$\Bbb N \to n\Bbb N$ given by $k \mapsto kn$ is an isomorphism of monoids (under addition in $\Bbb N$). This is a consequence of the familiar distributive law for natural numbers.

The important part is that you verify the isomorphism is "well-defined" that is:

$r/f^m \sim r'/f^{m'} \implies r/f^{mn} \sim r'/f^{m'n}$

The isomorphism is called "natural" because you define it the same way no matter what $n$ is (in the most naive way possible). In a similar way the isomorphism:

$F[\alpha_1] \cong F[\alpha_2]$ where $\alpha_1,\alpha_2$ are both roots of the same minimal polynomial is also "natural".

Again, we can use $\Bbb Z_6$ as an example with $D = \{1,2,4\}$, we get the same ring as a result as if we had used $D = \{1,4\}$. You may wish to verify this for yourself.

In answer to your previous post:

what is $\phi(rx^n)$?