MHB Localization - D&F Proposition 38., part (2)

[Math Amateur]

I am reading Dummit and Foote,, Section 15.4: Localization.

I am working on Proposition 38 - see attachment page 709 (also see attachment page 708 for definitions of $$ ^eI $$ and $$ ^cJ $$.

I am having some trouble proving the second part of Section (2), which D&F leave largely to the reader.

Proposition 38, Section 15.4, page 709 reads as follows:

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(2) For any ideal I of R we have

$$ ^c{(^eI)} = \{ r \in R \ | \ dr \in I $$ for some $$ d \in D \} $$

Also $$ ^eI = D^{-1}R $$ if and only if $$ I \cap D \ne \emptyset $$

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I can follow the proof of the first part of the above. However, for the proof of the second part - viz.:

$$ ^eI = D^{-1}R $$ if and only if $$ I \cap D \ne \emptyset $$

D&F write "The second assertion of (2) then follows the definition of I' (where we have I' set equal to $$ \{ r \in R \ | \ dr \in I $$ for some $$ d \in D \} $$.

Can someone help me show (formally & rigorously) that $$ ^eI = D^{-1}R $$ if and only if $$ I \cap D \ne \emptyset $$ and thus help me to see how this follows easily from the definition of I'

Hope someone can help.

Peter

 

[Deveno]

One part of this is relatively easy:

Let $x \in I \cap D$. Then we have:

$r/d = (rx)/(dx) \in {}^eI$, which shows that $D^{-1}R = {}^eI$.

The reverse implication is a bit more complicated. Let's show first that $I'$ is indeed an ideal of $R$.

If $r,r' \in I'$ there exist $d,d' \in D$ such that: $rd,r'd' \in I$.

Hence $(r - r')(dd') = d'(rd) - d(r'd') \in I$, since $I$ is an ideal.

Since $dd' \in D,\ r - r' \in I'$, which shows $(I',+)$ is a subgroup of $(R,+)$.

If $r \in I'$ and $a$ is any element of $R$, we have for some $d \in D$:

$rd \in I$, so $a(rd) = (ar)d \in I$, showing $ar \in I'$.

Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$.

 

[Math Amateur]

One part of this is relatively easy:

Let $x \in I \cap D$. Then we have:

$r/d = (rx)/(dx) \in {}^eI$, which shows that $D^{-1}R = {}^eI$.

The reverse implication is a bit more complicated. Let's show first that $I'$ is indeed an ideal of $R$.

If $r,r' \in I'$ there exist $d,d' \in D$ such that: $rd,r'd' \in I$.

Hence $(r - r')(dd') = d'(rd) - d(r'd') \in I$, since $I$ is an ideal.

Since $dd' \in D,\ r - r' \in I'$, which shows $(I',+)$ is a subgroup of $(R,+)$.

If $r \in I'$ and $a$ is any element of $R$, we have for some $d \in D$:

$rd \in I$, so $a(rd) = (ar)d \in I$, showing $ar \in I'$.

Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$.

Appreciate the help, Deveno.

Working through your post now.

Peter

 

[Math Amateur]

One part of this is relatively easy:

Let $x \in I \cap D$. Then we have:

$r/d = (rx)/(dx) \in {}^eI$, which shows that $D^{-1}R = {}^eI$.

The reverse implication is a bit more complicated. Let's show first that $I'$ is indeed an ideal of $R$.

If $r,r' \in I'$ there exist $d,d' \in D$ such that: $rd,r'd' \in I$.

Hence $(r - r')(dd') = d'(rd) - d(r'd') \in I$, since $I$ is an ideal.

Since $dd' \in D,\ r - r' \in I'$, which shows $(I',+)$ is a subgroup of $(R,+)$.

If $r \in I'$ and $a$ is any element of $R$, we have for some $d \in D$:

$rd \in I$, so $a(rd) = (ar)d \in I$, showing $ar \in I'$.

Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$.

Hi Deveno,

You write:

"Let $x \in I \cap D$. Then we have:

$r/d = (rx)/(dx) \in {}^eI$, which shows that $D^{-1}R = {}^eI$."

I am having trouble following this argument. Can you help by being more explicit regarding the logic involved.

Peter

 

[Deveno]

Any element of $D^{-1}R$ is $r/d$ for some $r \in R$ and $d \in D$.

Since:

$d'(r(dx) - d(rx)) = 0$ for any $d' \in D$ it is clear that:

$r/d = (rx)/(dx)$ for any $x \in D$ (including the one we know exists in $I \cap D$).

But $rx \in I$ (since $I$ is an ideal, and $x \in I$), and $dx \in D$ (since $D$ is multiplicative, and $x \in D$).

So $r/d = (rx)/(dx)$ is an element of $D^{-1}R$ where the "numerator" is in $I$, and the "denominator" is in $D$, that is, an element of $D^{-1}I = {}^eI$.

This shows $D^{-1}R \subseteq D^{-1}I = {}^eI$.

Since ${}^eI$ is clearly a subset of $D^{-1}R$, we have equality.

 

[Math Amateur]

Any element of $D^{-1}R$ is $r/d$ for some $r \in R$ and $d \in D$.

Since:

$d'(r(dx) - d(rx)) = 0$ for any $d' \in D$ it is clear that:

$r/d = (rx)/(dx)$ for any $x \in D$ (including the one we know exists in $I \cap D$).

But $rx \in I$ (since $I$ is an ideal, and $x \in I$), and $dx \in D$ (since $D$ is multiplicative, and $x \in D$).

So $r/d = (rx)/(dx)$ is an element of $D^{-1}R$ where the "numerator" is in $I$, and the "denominator" is in $D$, that is, an element of $D^{-1}I = {}^eI$.

This shows $D^{-1}R \subseteq D^{-1}I = {}^eI$.

Since ${}^eI$ is clearly a subset of $D^{-1}R$, we have equality.

Thanks Deveno ... yes, followed most of that ... most helpful ... just one further issue ...

I may be reacting too quickly here ... but from my reading of Sharp and also D&F, the extension of an ideal I to $$ D^{-1}R $$ is defined as the ideal $$ \pi(I).D^{-1}R $$ generated by $$ \pi (I) $$ in $$ D^{-1}R $$

So $$ ^eI = \pi(I).D^{-1}R $$

Is this correct?

So I am concerned to see exactly how/why $$ ^eI $$ becomes $$ D^{-1}I $$? That is, why exactly can $$ \pi(I).D^{-1}R $$ be regarded as composed of elements r/d where $$ r \in R $$ and $$ d \in D $$?

Can you clarify?

Peter

 

[Math Amateur]

Thanks Deveno ... yes, followed most of that ... most helpful ... just one further issue ...

I may be reacting too quickly here ... but from my reading of Sharp and also D&F, the extension of an ideal I to $$ D^{-1}R $$ is defined as the ideal $$ \pi(I).D^{-1}R $$ generated by $$ \pi (I) $$ in $$ D^{-1}R $$

So $$ ^eI = \pi(I).D^{-1}R $$

Is this correct?

So I am concerned to see exactly how/why $$ ^eI $$ becomes $$ D^{-1}I $$? That is, why exactly can $$ \pi(I).D^{-1}R $$ be regarded as composed of elements r/d where $$ r \in R $$ and $$ d \in D $$?

Can you clarify?

Peter

Hi Deveno,

I have been reflecting on the above post, concerning my question as to why we would choose to regard:

$$ ^eI = \pi (I).D^{-1}R $$ ... ... ... (1)

as

$$ ^eI = D^{-1}I $$ ... ... ... (2)

So, for the above notation in (2) to make sense and be coherent, we need to show that the definition (1) above leads to elements being of the form b/d where $$ b \in I $$ and $$ d \in D $$.

Now, $$ \pi : \ R \to D^{-1}R $$ is of the form $$ \pi(r) = r/1 $$ for $$ r \in R $$.

Therefore elements of the ideal generated by $$ \pi(I) $$ in $$ D^{-1}R $$ are of the form:

$$ \frac{a_1}{1}\frac{r_1}{d_1} + \frac{a_2}{1}\frac{r_2}{d_2} + ... \ ... + \frac{a_n}{1}\frac{r_n}{d_n} $$ where $$ a_i \in I, r_i \in R $$ and $$ d_i \in D $$.

So then elements of the ideal generated by $$ \pi(I) $$ in $$ D^{-1}R $$ are of the form:

$$ \frac{a_1r_1}{d_1} + \frac{a_2r_2}{d_2} + ... \ ... + \frac{a_nr_n}{d_n} $$

But since $$ \pi (I).D^{-1}R $$ is an ideal and also I is an ideal, the above elements are of the form:

$$ \frac{b_1}{d_1} + \frac{b_2}{d_2} + ... \ ... + \frac{b_n}{d_n} = \frac{b}{d} $$ where $$ b_i \in I $$ and $$ d_i \in D $$

Thus the elements of $$ ^eI = \pi (I).D^{-1}R $$ are of the form $$ \frac{b}{d} $$ where $$ b \in I $$ and $$ d \in D $$ and so we can sensibly view $$ ^eI $$ as $$ D^{-1}I $$.

Can you confirm that the above reasoning is correct?

Peter

 

[Deveno]

Yes..."loosely speaking", we want numerators in $I$, denominators in $D$.

The contraction does the opposite: we want the numerators of the elements in $J$, more precisely the pre-image under the ring homomorphism:

$\pi: R \to \pi(R) \subseteq D^{-1}R$

of the set $J \cap \pi(R)$.

Suppose we take $R = \Bbb Z$ with $D = \{2^n\}$. let's look at the extension of $I = (3)$. Note that $I \cap D = \emptyset$.

Well the direct images under $\pi$ of the elements of $(3)$ are rational numbers of the form:

$\dfrac{3k}{1}$. Multiplying by one of the "inverses" of $D$, we get fractions of the form:

$\dfrac{3k}{2^m}$

Let's add two of these together. We'll assume $n = m + m'$:

$\dfrac{3k}{2^m} + \dfrac{3k'}{2^n} = \dfrac{3k 2^{m'}}{2^n} + \dfrac{3k'}{2^n}$

$= \dfrac{3k 2^{m'} + 3k'}{2^n} = \dfrac{3k''}{2^n}$

where $k'' = k 2^{m'} + k' \in \Bbb Z$.

If we multiply a "typical" element by any element of $D^{-1}R$, we get:

$\dfrac{a}{2^{m'}}\dfrac{3k}{2^m} = \dfrac{3(ak)}{2^{m+m'}}$

On the other hand, consider the ideal $J$ in $D^{-1}R$ generated by $\dfrac{3}{1}$.

Clearly this ideal contains $\dfrac{1}{2^n}\dfrac{3}{1} = \dfrac{3}{2^n}$

for any $n \in \Bbb N$ (from the multiplicative property of an ideal), and repeated sums of such elements gives us any element:

$\dfrac{3k}{2^n}$.

Since $(3)$ is a prime ideal of $\Bbb Z$, we know that $\left(\dfrac{3}{1}\right)$ is a likewise a prime ideal of $D^{-1}R$, but this can also be shown directly:

If: $\dfrac{ab}{2^{m+m'}} = \dfrac{3k}{2^n}$

one of $a,b$ must be a multiple of $3$ (use "cross-multiplication" (since we have an integral domain for $R$) to get a relation between integers).

It turns out that $D^{-1}R$ preserves "almost all" the properties of $\Bbb Z = R$ except that factors of 2 "disappear" into the denominators (in the ring of dyadic fractions, 2 is no longer prime, as it becomes a unit...we have created (loosely speaking) "a ring out of odd numbers" (something that is not possible in the integers alone)).

Personally, I conjecture that this construction is the origin of the phrase: "what are the odds?"...a game of chance might typically have an outcome space of $2^m$ events, in which case the chance that a certain desired outcome will occur is the ratio of two odd numbers: if you have a 1 in 4 chance of winning, the odds are 3 to 1.

 

[Math Amateur]

One part of this is relatively easy:

Let $x \in I \cap D$. Then we have:

$r/d = (rx)/(dx) \in {}^eI$, which shows that $D^{-1}R = {}^eI$.

The reverse implication is a bit more complicated. Let's show first that $I'$ is indeed an ideal of $R$.

If $r,r' \in I'$ there exist $d,d' \in D$ such that: $rd,r'd' \in I$.

Hence $(r - r')(dd') = d'(rd) - d(r'd') \in I$, since $I$ is an ideal.

Since $dd' \in D,\ r - r' \in I'$, which shows $(I',+)$ is a subgroup of $(R,+)$.

If $r \in I'$ and $a$ is any element of $R$, we have for some $d \in D$:

$rd \in I$, so $a(rd) = (ar)d \in I$, showing $ar \in I'$.

Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$.

Hi Deveno,

I was just working carefully through the above post again to ensure that I understood it all ... but found I did not completely follow one of the steps in your logic ...

You write:

"Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$."Can you clarify exactly why $$ {}^c({}^eI) = I' $$.Peter

 

[Math Amateur]

Hi Deveno,

I was just working carefully through the above post again to ensure that I understood it all ... but found I did not completely follow one of the steps in your logic ...

You write:

"Now if ${}^eI = D^{-1}R$, for any $d \in D$ we know that $d/1 \in {}^eI$.

But this means that $d \in {}^c({}^eI) = I'$ so for some other element

$d' \in D$, we have $dd' \in I$. Since $dd' \in D$, we have $dd' \in I \cap D$."Can you clarify exactly why $$ {}^c({}^eI) = I' $$.Peter

Oh! ... how foolish of me ... $$ {}^c({}^eI) = I' $$ is the result of the first part of the proposition we are proving ... forgive me ... I am studying too myopically :-)

Peter