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Is this ring a PID? What are its invertible and irreducible elements?

A subring of the field of fractions of a PID is a PID as well.Show that the ring of all rational numbers, which when written in simplest form has an odd denominator, is a principal ideal domain.Can the product of two non invertible elements in a ring be invertible?Integral domain whose irreducible elements are not primeIrreducible elements and AssociatesAre there categorifications of prime or irreducible elements (of a ring, say)?Showing that right quasi regular elements are invertiblecollection of all non invertible elements in the ring is the maximal idealFind gcd and invertible elements of a ring.Finding irreducible elements in quotient ringLet $D$ be a PID and $a$ and $b$ be nonzero elements of $D$. Prove that there exist elements $s$ and $t$ in $D$ such that $gcd(a, b) = as + bt$.Proving that sum of two elements is not invertible in a ring

0

$begingroup$

$R$ = $left{ {left(dfrac{a}{3^{n}}right)|; n in N_{0}, a ∈ mathbb{Z} }right}$. Is $R$ a PID? What are the invertible elements of $R$? What are the irreducible elements of $R$?

My thoughts: I think this is a PID, but I don't know how to begin proving this. I think the invertible elements are the elements such that $a=3^{m}$ for some $m ∈ N_{0}$?

ring-theory

share|cite|improve this question

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Every ring between a PID $A$ and its quotient field is a PID, see here.
  $endgroup$
  – Dietrich Burde
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that link was very helpful. Do you have any thoughts about the irreducible elements in R?
  $endgroup$
  – hdhdhdhd
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, what are your thoughts about the irreducible elements? You only thought about invertible elements.
  $endgroup$
  – Dietrich Burde
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have considered both, but haven’t been able to come up with any irreducible elements. Sorry if this is a basic question but I’m struggling with fields of fractions - any point in the right direction would be appreciated
  $endgroup$
  – hdhdhdhd
  15 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

$R$ = $left{ {left(dfrac{a}{3^{n}}right)|; n in N_{0}, a ∈ mathbb{Z} }right}$. Is $R$ a PID? What are the invertible elements of $R$? What are the irreducible elements of $R$?

My thoughts: I think this is a PID, but I don't know how to begin proving this. I think the invertible elements are the elements such that $a=3^{m}$ for some $m ∈ N_{0}$?

ring-theory

share|cite|improve this question

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Every ring between a PID $A$ and its quotient field is a PID, see here.
  $endgroup$
  – Dietrich Burde
  17 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that link was very helpful. Do you have any thoughts about the irreducible elements in R?
  $endgroup$
  – hdhdhdhd
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, what are your thoughts about the irreducible elements? You only thought about invertible elements.
  $endgroup$
  – Dietrich Burde
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have considered both, but haven’t been able to come up with any irreducible elements. Sorry if this is a basic question but I’m struggling with fields of fractions - any point in the right direction would be appreciated
  $endgroup$
  – hdhdhdhd
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$R$ = $left{ {left(dfrac{a}{3^{n}}right)|; n in N_{0}, a ∈ mathbb{Z} }right}$. Is $R$ a PID? What are the invertible elements of $R$? What are the irreducible elements of $R$?

My thoughts: I think this is a PID, but I don't know how to begin proving this. I think the invertible elements are the elements such that $a=3^{m}$ for some $m ∈ N_{0}$?

ring-theory

share|cite|improve this question

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 17 hours ago

Yadati Kiran

1,8521620

edited 17 hours ago

Yadati Kiran

1,8521620

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

New contributor

hdhdhdhd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 17 hours ago

hdhdhdhdhdhdhdhd

61

1 Answer
1

active

oldest

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0

$begingroup$

To show that $R$ is a PID, you consider the "numerator" ideal of each ideal, which will be principal.

For example, let $I subset R$ be an ideal.

Let us see that $J = {a : frac a1 in I}$ is an ideal of $mathbb Z$. Therefore, as $mathbb Z$ is a PID we get that $J = langle rrangle$ for some $r in mathbb Z$.

We claim that $langle frac r1 rangle = I$. Why is this true? Certainly it is a subset.

Indeed, fix $i in I$, then $i$ is of the form $frac {c}{3^l}$ for some $c$. Now, $frac c1 = frac c{3^l} times frac{3^l}{1} in I$, so $c in J$ so $r | c$. Then $i = frac{r}{1} times frac{c/r}{3^l}$, where the latter belongs in $R$. Hence, $I$ is generated by $frac r1$.

---

Now, what are the invertible elements? Let $r = frac c{3^l}$ be invertible, then it basically means that (noting $c neq 0$) $frac {3^l}{c} in R$. Now, a simple look at the structure of $R$ will convince you that $c$ must be plus or minus a power of $3$, and that after cancellation, $r = pm 3^{-l}$ for some $l in mathbb Z$.

---

The irreducibles? If $r = frac c{3^l}$ is irreducible, then we remove the factors of $3$ i.e. write $r = d times 3^{g}$ where $g in mathbb Z, d in mathbb Z$ and $3 nmid d$. Now, $r$ is irreducible if and only if $d$ is(in $R$) : but can you proceed now?

---

Actually, the ring $R$ is an example of a more common construction. Given an integral domain (can be defined for commutative rings, but I'll skip that) $R$ and a multiplicatively closed set $S subset R$ we define the localization of $R$ at $S$, denotes by $S^{-1}R$, to be the set ${frac rs : r in R , s in S}$ , quotiented by the equivalence relation $frac rs sim frac xy iff sx = ry$. With addition and subtraction defined as for usual fractions, this becomes a ring.

Furthermore, the localization of a PID is a PID, and the localization "inverts" $S$, so that elements of $S$ become invertible. So the invertible elements are of the form $frac rs$ for $r,s in S$.

It turns out there is a correspondence between the prime ideals of the localization, and that of the original ring, via the numerator ideal, and the embedding $r to frac r1$. Since we have PIDs here, every prime ideal is generated by an irreducible element, so we can use the mapping to easily find the relation between the irreducibles of $R$ and its localization.

share|cite|improve this answer

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

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0

$begingroup$

To show that $R$ is a PID, you consider the "numerator" ideal of each ideal, which will be principal.

For example, let $I subset R$ be an ideal.

Let us see that $J = {a : frac a1 in I}$ is an ideal of $mathbb Z$. Therefore, as $mathbb Z$ is a PID we get that $J = langle rrangle$ for some $r in mathbb Z$.

We claim that $langle frac r1 rangle = I$. Why is this true? Certainly it is a subset.

Indeed, fix $i in I$, then $i$ is of the form $frac {c}{3^l}$ for some $c$. Now, $frac c1 = frac c{3^l} times frac{3^l}{1} in I$, so $c in J$ so $r | c$. Then $i = frac{r}{1} times frac{c/r}{3^l}$, where the latter belongs in $R$. Hence, $I$ is generated by $frac r1$.

---

Now, what are the invertible elements? Let $r = frac c{3^l}$ be invertible, then it basically means that (noting $c neq 0$) $frac {3^l}{c} in R$. Now, a simple look at the structure of $R$ will convince you that $c$ must be plus or minus a power of $3$, and that after cancellation, $r = pm 3^{-l}$ for some $l in mathbb Z$.

---

The irreducibles? If $r = frac c{3^l}$ is irreducible, then we remove the factors of $3$ i.e. write $r = d times 3^{g}$ where $g in mathbb Z, d in mathbb Z$ and $3 nmid d$. Now, $r$ is irreducible if and only if $d$ is(in $R$) : but can you proceed now?

---

Actually, the ring $R$ is an example of a more common construction. Given an integral domain (can be defined for commutative rings, but I'll skip that) $R$ and a multiplicatively closed set $S subset R$ we define the localization of $R$ at $S$, denotes by $S^{-1}R$, to be the set ${frac rs : r in R , s in S}$ , quotiented by the equivalence relation $frac rs sim frac xy iff sx = ry$. With addition and subtraction defined as for usual fractions, this becomes a ring.

Furthermore, the localization of a PID is a PID, and the localization "inverts" $S$, so that elements of $S$ become invertible. So the invertible elements are of the form $frac rs$ for $r,s in S$.

It turns out there is a correspondence between the prime ideals of the localization, and that of the original ring, via the numerator ideal, and the embedding $r to frac r1$. Since we have PIDs here, every prime ideal is generated by an irreducible element, so we can use the mapping to easily find the relation between the irreducibles of $R$ and its localization.

share|cite|improve this answer

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

0

$begingroup$

To show that $R$ is a PID, you consider the "numerator" ideal of each ideal, which will be principal.

For example, let $I subset R$ be an ideal.

Let us see that $J = {a : frac a1 in I}$ is an ideal of $mathbb Z$. Therefore, as $mathbb Z$ is a PID we get that $J = langle rrangle$ for some $r in mathbb Z$.

We claim that $langle frac r1 rangle = I$. Why is this true? Certainly it is a subset.

Indeed, fix $i in I$, then $i$ is of the form $frac {c}{3^l}$ for some $c$. Now, $frac c1 = frac c{3^l} times frac{3^l}{1} in I$, so $c in J$ so $r | c$. Then $i = frac{r}{1} times frac{c/r}{3^l}$, where the latter belongs in $R$. Hence, $I$ is generated by $frac r1$.

---

Now, what are the invertible elements? Let $r = frac c{3^l}$ be invertible, then it basically means that (noting $c neq 0$) $frac {3^l}{c} in R$. Now, a simple look at the structure of $R$ will convince you that $c$ must be plus or minus a power of $3$, and that after cancellation, $r = pm 3^{-l}$ for some $l in mathbb Z$.

---

The irreducibles? If $r = frac c{3^l}$ is irreducible, then we remove the factors of $3$ i.e. write $r = d times 3^{g}$ where $g in mathbb Z, d in mathbb Z$ and $3 nmid d$. Now, $r$ is irreducible if and only if $d$ is(in $R$) : but can you proceed now?

---

Actually, the ring $R$ is an example of a more common construction. Given an integral domain (can be defined for commutative rings, but I'll skip that) $R$ and a multiplicatively closed set $S subset R$ we define the localization of $R$ at $S$, denotes by $S^{-1}R$, to be the set ${frac rs : r in R , s in S}$ , quotiented by the equivalence relation $frac rs sim frac xy iff sx = ry$. With addition and subtraction defined as for usual fractions, this becomes a ring.

Furthermore, the localization of a PID is a PID, and the localization "inverts" $S$, so that elements of $S$ become invertible. So the invertible elements are of the form $frac rs$ for $r,s in S$.

It turns out there is a correspondence between the prime ideals of the localization, and that of the original ring, via the numerator ideal, and the embedding $r to frac r1$. Since we have PIDs here, every prime ideal is generated by an irreducible element, so we can use the mapping to easily find the relation between the irreducibles of $R$ and its localization.

share|cite|improve this answer

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

add a comment | 

$begingroup$

To show that $R$ is a PID, you consider the "numerator" ideal of each ideal, which will be principal.

For example, let $I subset R$ be an ideal.

Let us see that $J = {a : frac a1 in I}$ is an ideal of $mathbb Z$. Therefore, as $mathbb Z$ is a PID we get that $J = langle rrangle$ for some $r in mathbb Z$.

We claim that $langle frac r1 rangle = I$. Why is this true? Certainly it is a subset.

Indeed, fix $i in I$, then $i$ is of the form $frac {c}{3^l}$ for some $c$. Now, $frac c1 = frac c{3^l} times frac{3^l}{1} in I$, so $c in J$ so $r | c$. Then $i = frac{r}{1} times frac{c/r}{3^l}$, where the latter belongs in $R$. Hence, $I$ is generated by $frac r1$.

---

Now, what are the invertible elements? Let $r = frac c{3^l}$ be invertible, then it basically means that (noting $c neq 0$) $frac {3^l}{c} in R$. Now, a simple look at the structure of $R$ will convince you that $c$ must be plus or minus a power of $3$, and that after cancellation, $r = pm 3^{-l}$ for some $l in mathbb Z$.

---

The irreducibles? If $r = frac c{3^l}$ is irreducible, then we remove the factors of $3$ i.e. write $r = d times 3^{g}$ where $g in mathbb Z, d in mathbb Z$ and $3 nmid d$. Now, $r$ is irreducible if and only if $d$ is(in $R$) : but can you proceed now?

---

Actually, the ring $R$ is an example of a more common construction. Given an integral domain (can be defined for commutative rings, but I'll skip that) $R$ and a multiplicatively closed set $S subset R$ we define the localization of $R$ at $S$, denotes by $S^{-1}R$, to be the set ${frac rs : r in R , s in S}$ , quotiented by the equivalence relation $frac rs sim frac xy iff sx = ry$. With addition and subtraction defined as for usual fractions, this becomes a ring.

Furthermore, the localization of a PID is a PID, and the localization "inverts" $S$, so that elements of $S$ become invertible. So the invertible elements are of the form $frac rs$ for $r,s in S$.

It turns out there is a correspondence between the prime ideals of the localization, and that of the original ring, via the numerator ideal, and the embedding $r to frac r1$. Since we have PIDs here, every prime ideal is generated by an irreducible element, so we can use the mapping to easily find the relation between the irreducibles of $R$ and its localization.

share|cite|improve this answer

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

$endgroup$

share|cite|improve this answer

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477

answered 15 hours ago

астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

39.1k33477