- Thread starter
- #1

- Jun 22, 2012

- 2,918

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:

-------------------------------------------------------------------------------------

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

Define \(\displaystyle R_f = D^{-1}R \).

(a) Show that \(\displaystyle R_f = 0 \) 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 \(\displaystyle R_f \cong R[x]/(xf -1)\)

------------------------------------------------------------------------------------

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

\(\displaystyle R_f = \{ r/d \ | \ r \in R, d \in D \} \)

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

\(\displaystyle f \) nilpotent \(\displaystyle \Longrightarrow f^m = 0 \) for some \(\displaystyle m \in Z^+ \)

\(\displaystyle \Longrightarrow \ 0 \in D \)

\(\displaystyle \Longrightarrow \ R_f = D^{-1}R = \{ 0 \} \)

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

Now \(\displaystyle f \in R_f \Longrightarrow \ \frac{f^1}{f^0} \in R_f \Longrightarrow \ \frac{f^1}{1} \)

But we have \(\displaystyle \frac{1}{f^1} \in R_f \)

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

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

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

Peter

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

-------------------------------------------------------------------------------------

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

Define \(\displaystyle R_f = D^{-1}R \).

(a) Show that \(\displaystyle R_f = 0 \) 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 \(\displaystyle R_f \cong R[x]/(xf -1)\)

------------------------------------------------------------------------------------

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

**Attempt at (a)**\(\displaystyle R_f = \{ r/d \ | \ r \in R, d \in D \} \)

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

\(\displaystyle f \) nilpotent \(\displaystyle \Longrightarrow f^m = 0 \) for some \(\displaystyle m \in Z^+ \)

\(\displaystyle \Longrightarrow \ 0 \in D \)

\(\displaystyle \Longrightarrow \ R_f = D^{-1}R = \{ 0 \} \)

**Attempt at (b)**\(\displaystyle R_f = \{ \frac{r}{f^m} \ | \ r \in R, f^m \in R_f = D^{-1}R \)

Now \(\displaystyle f \in R_f \Longrightarrow \ \frac{f^1}{f^0} \in R_f \Longrightarrow \ \frac{f^1}{1} \)

But we have \(\displaystyle \frac{1}{f^1} \in R_f \)

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

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

Attempt at (c)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

Last edited: