Irreducible and prime elements
$begingroup$
In my commutative algebra lecture notes it says:
A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.
Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$
Thanks :)
abstract-algebra ring-theory
edited Dec 21 '14 at 12:22
user26857
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
$endgroup$
add a comment |
$begingroup$
In my commutative algebra lecture notes it says:
A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.
Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$
Thanks :)
abstract-algebra ring-theory
edited Dec 21 '14 at 12:22
user26857
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
$endgroup$
$begingroup$
You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
$endgroup$
– drhab
Dec 21 '14 at 11:58
$begingroup$
You are correct. This is the definition of irreducible element.
$endgroup$
– Crostul
Dec 21 '14 at 12:19
$begingroup$
Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
$endgroup$
– egreg
Dec 21 '14 at 14:12
add a comment |
$begingroup$
In my commutative algebra lecture notes it says:
A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.
Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$
Thanks :)
abstract-algebra ring-theory
edited Dec 21 '14 at 12:22
user26857
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
$endgroup$
In my commutative algebra lecture notes it says:
A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.
Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$
Thanks :)
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 21 '14 at 12:22
user26857
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
edited Dec 21 '14 at 12:22
user26857
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
edited Dec 21 '14 at 12:22
user26857
39.3k124183
edited Dec 21 '14 at 12:22
user26857
39.3k124183
edited Dec 21 '14 at 12:22
user26857
39.3k124183
39.3k124183
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
asked Dec 21 '14 at 11:48
MattBurrowsMattBurrows
30929
30929
$begingroup$
You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
$endgroup$
– drhab
Dec 21 '14 at 11:58
$begingroup$
You are correct. This is the definition of irreducible element.
$endgroup$
– Crostul
Dec 21 '14 at 12:19
$begingroup$
Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
$endgroup$
– egreg
Dec 21 '14 at 14:12
add a comment |
$begingroup$
You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
$endgroup$
– drhab
Dec 21 '14 at 11:58
$begingroup$
You are correct. This is the definition of irreducible element.
$endgroup$
– Crostul
Dec 21 '14 at 12:19
$begingroup$
Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
$endgroup$
– egreg
Dec 21 '14 at 14:12
$begingroup$
You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
$endgroup$
– drhab
Dec 21 '14 at 11:58
$begingroup$
You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
$endgroup$
– drhab
Dec 21 '14 at 11:58
$begingroup$
You are correct. This is the definition of irreducible element.
$endgroup$
– Crostul
Dec 21 '14 at 12:19
$begingroup$
You are correct. This is the definition of irreducible element.
$endgroup$
– Crostul
Dec 21 '14 at 12:19
$begingroup$
Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
$endgroup$
– egreg
Dec 21 '14 at 14:12
$begingroup$
Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
$endgroup$
– egreg
Dec 21 '14 at 14:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An element in an integral domain is irreductible if it is not a unit nor a product of non-units.
An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)
Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p mid ab$ implies that $p mid a$ or $p mid b$. Say $p mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.
The implication "irreducible implies prime" is false in general, for instance in the ring $A = mathbf{Z}[sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - sqrt{-5}) (2 + sqrt{-5})$ but does not divide either of the two elements $2 pm sqrt{-5}$.
The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 times 3 = (2 - sqrt{-5}) (2 + sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = mathbf{Z}[sqrt{-5}]$ is not a unique factorization domain.
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
$endgroup$
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
add a comment |
Your Answer
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1 Answer
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$begingroup$
An element in an integral domain is irreductible if it is not a unit nor a product of non-units.
An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)
Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p mid ab$ implies that $p mid a$ or $p mid b$. Say $p mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.
The implication "irreducible implies prime" is false in general, for instance in the ring $A = mathbf{Z}[sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - sqrt{-5}) (2 + sqrt{-5})$ but does not divide either of the two elements $2 pm sqrt{-5}$.
The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 times 3 = (2 - sqrt{-5}) (2 + sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = mathbf{Z}[sqrt{-5}]$ is not a unique factorization domain.
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
$endgroup$
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
add a comment |
$begingroup$
An element in an integral domain is irreductible if it is not a unit nor a product of non-units.
An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)
Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p mid ab$ implies that $p mid a$ or $p mid b$. Say $p mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.
The implication "irreducible implies prime" is false in general, for instance in the ring $A = mathbf{Z}[sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - sqrt{-5}) (2 + sqrt{-5})$ but does not divide either of the two elements $2 pm sqrt{-5}$.
The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 times 3 = (2 - sqrt{-5}) (2 + sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = mathbf{Z}[sqrt{-5}]$ is not a unique factorization domain.
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
$endgroup$
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
add a comment |
$begingroup$
An element in an integral domain is irreductible if it is not a unit nor a product of non-units.
An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)
Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p mid ab$ implies that $p mid a$ or $p mid b$. Say $p mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.
The implication "irreducible implies prime" is false in general, for instance in the ring $A = mathbf{Z}[sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - sqrt{-5}) (2 + sqrt{-5})$ but does not divide either of the two elements $2 pm sqrt{-5}$.
The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 times 3 = (2 - sqrt{-5}) (2 + sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = mathbf{Z}[sqrt{-5}]$ is not a unique factorization domain.
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
$endgroup$
An element in an integral domain is irreductible if it is not a unit nor a product of non-units.
An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)
Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p mid ab$ implies that $p mid a$ or $p mid b$. Say $p mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.
The implication "irreducible implies prime" is false in general, for instance in the ring $A = mathbf{Z}[sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - sqrt{-5}) (2 + sqrt{-5})$ but does not divide either of the two elements $2 pm sqrt{-5}$.
The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 times 3 = (2 - sqrt{-5}) (2 + sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = mathbf{Z}[sqrt{-5}]$ is not a unique factorization domain.
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
edited Mar 5 at 22:37
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
answered Dec 21 '14 at 13:19
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
9,25711638
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
add a comment |
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
The nicest definition of a prime ideal $mathfrak{p}$ of a commutative ring $A$ is to say that $A backslash mathfrak{p}$ ($A$ deprived from $mathfrak{p}$) is stable by finite products.
$endgroup$
– ujsgeyrr1f0d0d0r0h1h0j0j_juj
Dec 21 '14 at 13:34
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
Equivalently in a domain, a nonunit $pne 0,$ is irred if $,p = ab,Rightarrow,pmid a,$ or $,pmid b.,$ Being so similar to the definition of a prime, it makes obvious the implication $rm, color{#c00}{prime},Rightarrow$ irred, $ $ viz. $$p = ab,Rightarrow, pmid ab overset{large p rmcolor{#c00}{prime}}Rightarrow pmid a {rm or} pmid b,$$
$endgroup$
– Bill Dubuque
Dec 21 '14 at 16:47
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
$begingroup$
I think we don't even need commutativity for fact: R/P integral domain iff P prime ideal in R. And definition of prime ideal makes sense even in non-commuataive ring
$endgroup$
– Sushil
Feb 15 '15 at 15:01
add a comment |
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You are correct in observing that your lecture notes are not quite right here. It is indeed the definition of an irreducible element. What you read elsewhere is indeed the definition of prime element.
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– drhab
Dec 21 '14 at 11:58
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You are correct. This is the definition of irreducible element.
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– Crostul
Dec 21 '14 at 12:19
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Your lecture notes use an uncommon definition of “prime”. The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students.
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– egreg
Dec 21 '14 at 14:12