Why does the statement “p is prime if it is divisible by only itself and 1” define only one prime number? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Do we have negative prime numbers?In arbitrary commutative rings, what is the accepted definition of “associates”?Let $x$ be greater than $1$. Prove $x$ is prime if and only if for every integer $y$, either $gcd(x,y)=1$ or $xmid y$.Are zero and one relatively prime?How do I show that :$sigma(p^m)$ is divisible by $4$ if $m=4k+1$ , and $k$ is an integer number?Prove that there exist $n$ consecutive positive integers with the $i$th integer divisible by the $i$th primeRecall that an integer is said to be square-free if it is not divisible by the square of any prime. Prove that for any positive integer $n$…Is 2 both a prime and a highly composite number?Numbers which are divisible by the sum of their prime factors.Proving there exists a prime $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.What should we define as a prime number, as some definitions don’t go into enough detail, and some seem over the top.How many ways can we define prime number?
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Why does the statement “p is prime if it is divisible by only itself and 1” define only one prime number? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Do we have negative prime numbers?In arbitrary commutative rings, what is the accepted definition of “associates”?Let $x$ be greater than $1$. Prove $x$ is prime if and only if for every integer $y$, either $gcd(x,y)=1$ or $xmid y$.Are zero and one relatively prime?How do I show that :$sigma(p^m)$ is divisible by $4$ if $m=4k+1$ , and $k$ is an integer number?Prove that there exist $n$ consecutive positive integers with the $i$th integer divisible by the $i$th primeRecall that an integer is said to be square-free if it is not divisible by the square of any prime. Prove that for any positive integer $n$…Is 2 both a prime and a highly composite number?Numbers which are divisible by the sum of their prime factors.Proving there exists a prime $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.What should we define as a prime number, as some definitions don’t go into enough detail, and some seem over the top.How many ways can we define prime number?
$begingroup$
I'm having a bit of trouble understanding why this incorrect definition of primality only defines one prime number.
"p is prime if it is divisible by only itself and 1."
My understood definition of a prime number is a positive integer that has exactly 2 positive integer factors: 1 and itself. Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Sorry if my question is related to others, I am quite new to number theory and could not find any other resources that would help me with this specific problem.
elementary-number-theory prime-numbers divisibility
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
$endgroup$
closed as unclear what you're asking by Peter, Sil, B. Goddard, egreg, José Carlos Santos Mar 25 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 15 more comments
$begingroup$
I'm having a bit of trouble understanding why this incorrect definition of primality only defines one prime number.
"p is prime if it is divisible by only itself and 1."
My understood definition of a prime number is a positive integer that has exactly 2 positive integer factors: 1 and itself. Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Sorry if my question is related to others, I am quite new to number theory and could not find any other resources that would help me with this specific problem.
elementary-number-theory prime-numbers divisibility
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
$endgroup$
closed as unclear what you're asking by Peter, Sil, B. Goddard, egreg, José Carlos Santos Mar 25 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This related question may be helpful.
$endgroup$
– Brian
Mar 24 at 17:28
2
$begingroup$
What do you mean “defines only one prime number”??
$endgroup$
– symplectomorphic
Mar 24 at 17:39
2
$begingroup$
In my opinion your professor is wrong.
$endgroup$
– amsmath
Mar 24 at 17:43
1
$begingroup$
That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
$endgroup$
– B.Swan
Mar 24 at 17:50
2
$begingroup$
You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
$endgroup$
– symplectomorphic
Mar 24 at 17:54
|
show 15 more comments
$begingroup$
I'm having a bit of trouble understanding why this incorrect definition of primality only defines one prime number.
"p is prime if it is divisible by only itself and 1."
My understood definition of a prime number is a positive integer that has exactly 2 positive integer factors: 1 and itself. Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Sorry if my question is related to others, I am quite new to number theory and could not find any other resources that would help me with this specific problem.
elementary-number-theory prime-numbers divisibility
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
$endgroup$
I'm having a bit of trouble understanding why this incorrect definition of primality only defines one prime number.
"p is prime if it is divisible by only itself and 1."
My understood definition of a prime number is a positive integer that has exactly 2 positive integer factors: 1 and itself. Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Sorry if my question is related to others, I am quite new to number theory and could not find any other resources that would help me with this specific problem.
elementary-number-theory prime-numbers divisibility
elementary-number-theory prime-numbers divisibility
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
edited Mar 24 at 18:20
Bill Dubuque
214k29198660
214k29198660
asked Mar 24 at 17:26
PeterPeter
184
asked Mar 24 at 17:26
PeterPeter
184
asked Mar 24 at 17:26
PeterPeter
184
184
closed as unclear what you're asking by Peter, Sil, B. Goddard, egreg, José Carlos Santos Mar 25 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Peter, Sil, B. Goddard, egreg, José Carlos Santos Mar 25 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This related question may be helpful.
$endgroup$
– Brian
Mar 24 at 17:28
2
$begingroup$
What do you mean “defines only one prime number”??
$endgroup$
– symplectomorphic
Mar 24 at 17:39
2
$begingroup$
In my opinion your professor is wrong.
$endgroup$
– amsmath
Mar 24 at 17:43
1
$begingroup$
That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
$endgroup$
– B.Swan
Mar 24 at 17:50
2
$begingroup$
You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
$endgroup$
– symplectomorphic
Mar 24 at 17:54
|
show 15 more comments
1
$begingroup$
This related question may be helpful.
$endgroup$
– Brian
Mar 24 at 17:28
2
$begingroup$
What do you mean “defines only one prime number”??
$endgroup$
– symplectomorphic
Mar 24 at 17:39
2
$begingroup$
In my opinion your professor is wrong.
$endgroup$
– amsmath
Mar 24 at 17:43
1
$begingroup$
That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
$endgroup$
– B.Swan
Mar 24 at 17:50
2
$begingroup$
You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
$endgroup$
– symplectomorphic
Mar 24 at 17:54
1
1
$begingroup$
This related question may be helpful.
$endgroup$
– Brian
Mar 24 at 17:28
$begingroup$
This related question may be helpful.
$endgroup$
– Brian
Mar 24 at 17:28
2
2
$begingroup$
What do you mean “defines only one prime number”??
$endgroup$
– symplectomorphic
Mar 24 at 17:39
$begingroup$
What do you mean “defines only one prime number”??
$endgroup$
– symplectomorphic
Mar 24 at 17:39
2
2
$begingroup$
In my opinion your professor is wrong.
$endgroup$
– amsmath
Mar 24 at 17:43
$begingroup$
In my opinion your professor is wrong.
$endgroup$
– amsmath
Mar 24 at 17:43
1
1
$begingroup$
That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
$endgroup$
– B.Swan
Mar 24 at 17:50
$begingroup$
That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
$endgroup$
– B.Swan
Mar 24 at 17:50
2
2
$begingroup$
You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
$endgroup$
– symplectomorphic
Mar 24 at 17:54
$begingroup$
You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
$endgroup$
– symplectomorphic
Mar 24 at 17:54
|
show 15 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:
A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.
The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.
Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
$begingroup$
Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Actually is worse. There would be no prime numbers as for any $x in mathbb R$ then $x = n*frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $frac 32, frac 34, frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)
So.......
The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.
This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.
Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:
A natural number is prime if it has exactly two natural divisors; itself and $1$.
We can extend it extend this to all integers by saying a prime in any integer other than $pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($pm 1$) integers.
With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.
answered Mar 24 at 18:55
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:
A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).
Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.
answered Mar 24 at 17:58
FlowersFlowers
683410
$endgroup$
2
$begingroup$
Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
$endgroup$
– Bill Dubuque
Mar 24 at 18:10
2
$begingroup$
It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
$begingroup$
@amsmath I think I would have appreciated this explanation a few years ago.
$endgroup$
– Flowers
Mar 25 at 8:25
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:
A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.
The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.
Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
$begingroup$
Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:
A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.
The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.
Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
$begingroup$
Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:
A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.
The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.
Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
$endgroup$
Usually, in beginning number theory, we work only in the natural numbers (usually including $0$) so there are no negative numbers in this context. Assuming that, the definition of "divisor" would include the fact that a divisor is a natural number, than therefore positive. Then the definition:
A (natural) number $p$ greater than $1$ is prime iff it is divisible only by itself and $1$.
The "greater than $1$" part is important, because we don't want $1$ to be prime. (Although, historically it was considered a prime.) The implicit positiveness of the potential divisors is also important.
Later in the course, you can extend these definitions of "prime" and "divisor" to include the negative integers and then you'll have to adjust your definitions a bit.
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
answered Mar 24 at 17:40
B. GoddardB. Goddard
20.2k21543
20.2k21543
add a comment |
add a comment |
$begingroup$
Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Actually is worse. There would be no prime numbers as for any $x in mathbb R$ then $x = n*frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $frac 32, frac 34, frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)
So.......
The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.
This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.
Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:
A natural number is prime if it has exactly two natural divisors; itself and $1$.
We can extend it extend this to all integers by saying a prime in any integer other than $pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($pm 1$) integers.
With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.
answered Mar 24 at 18:55
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Actually is worse. There would be no prime numbers as for any $x in mathbb R$ then $x = n*frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $frac 32, frac 34, frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)
So.......
The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.
This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.
Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:
A natural number is prime if it has exactly two natural divisors; itself and $1$.
We can extend it extend this to all integers by saying a prime in any integer other than $pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($pm 1$) integers.
With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.
answered Mar 24 at 18:55
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Actually is worse. There would be no prime numbers as for any $x in mathbb R$ then $x = n*frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $frac 32, frac 34, frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)
So.......
The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.
This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.
Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:
A natural number is prime if it has exactly two natural divisors; itself and $1$.
We can extend it extend this to all integers by saying a prime in any integer other than $pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($pm 1$) integers.
With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.
answered Mar 24 at 18:55
fleabloodfleablood
1
$endgroup$
Is the problem with the definition related to the fact that it doesn't specify sign, so then -1 would be the only "prime" number?
Actually is worse. There would be no prime numbers as for any $x in mathbb R$ then $x = n*frac xn$ for any integer $n$ so $x$ has infinitely many divisors and can't be prime. $3$ is not prime because $frac 32, frac 34, frac 35$, etc all divide $3$. (The definition of $a$ divides $b$ is that there exists an integer $k$ so that $b = ka$. There is nothing in the definition that says $a$ and $b$ are integers)
So.......
The definition of "Having only $1$ and itself as divisor" assumes that we are only considering natural numbers (i.e. positive integers) and even then we are considering $1$ and "itself" to be different.
This definition is not even considering negative numbers to be considered. Nor is it considering that that rational non integers can be divisors.
Which is okay. We just have to be specific. The following would be a perfectly acceptable definition:
A natural number is prime if it has exactly two natural divisors; itself and $1$.
We can extend it extend this to all integers by saying a prime in any integer other than $pm 1$ with only two positive integer divisors; $1$ and the absolute value of itself. ... or a prime number is an integer which can not be expressed as a product of two or more non-unit ($pm 1$) integers.
With more abstract algebraic concepts we can define primes more precisely as Flowers' answer and various comment links explain. These also allow for primes in systems other than just the integers.
answered Mar 24 at 18:55
fleabloodfleablood
1
answered Mar 24 at 18:55
fleabloodfleablood
1
answered Mar 24 at 18:55
fleabloodfleablood
1
answered Mar 24 at 18:55
fleabloodfleablood
1
1
add a comment |
add a comment |
$begingroup$
The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:
A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).
Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.
answered Mar 24 at 17:58
FlowersFlowers
683410
$endgroup$
2
$begingroup$
Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
$endgroup$
– Bill Dubuque
Mar 24 at 18:10
2
$begingroup$
It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
$begingroup$
@amsmath I think I would have appreciated this explanation a few years ago.
$endgroup$
– Flowers
Mar 25 at 8:25
add a comment |
$begingroup$
The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:
A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).
Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.
answered Mar 24 at 17:58
FlowersFlowers
683410
$endgroup$
2
$begingroup$
Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
$endgroup$
– Bill Dubuque
Mar 24 at 18:10
2
$begingroup$
It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
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@amsmath I think I would have appreciated this explanation a few years ago.
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– Flowers
Mar 25 at 8:25
add a comment |
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The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:
A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).
Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.
answered Mar 24 at 17:58
FlowersFlowers
683410
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The precise way to define prime and irreducible elements in a commutative ring are given here and here. For the integers, the definitions are equivalent, so you can use either. Rephrasing the second link for integers:
A unit in a ring is an element that has a multiplicative inverse. In the integers, the only units are $1$ and $-1$. So a nonzero non-unit element of the integers is irreducible (prime) if it cannot be written as the product of two integers which are both non-units (not 1 or -1).
Note that this implies that negative primes exist. Unique factorization is defined up to multiplication by units.
answered Mar 24 at 17:58
FlowersFlowers
683410
edited Mar 24 at 18:30
answered Mar 24 at 17:58
FlowersFlowers
683410
answered Mar 24 at 17:58
FlowersFlowers
683410
answered Mar 24 at 17:58
FlowersFlowers
683410
683410
2
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Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
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– Bill Dubuque
Mar 24 at 18:10
2
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It's always a good choice confusing newbies with very general theory. So, well done.
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– amsmath
Mar 25 at 0:12
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@amsmath I think I would have appreciated this explanation a few years ago.
$endgroup$
– Flowers
Mar 25 at 8:25
add a comment |
2
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Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
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– Bill Dubuque
Mar 24 at 18:10
2
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It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
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@amsmath I think I would have appreciated this explanation a few years ago.
$endgroup$
– Flowers
Mar 25 at 8:25
2
2
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Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
$endgroup$
– Bill Dubuque
Mar 24 at 18:10
$begingroup$
Beware that the 2nd link only defines "irreducible" for integral domains. In more general rings there may be zero divisors, which forces common notions of "associate" to bifurcate into a few inequivalent notions. This variation percolates into related notions, e.g. "irreducib;e" - see this answer
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– Bill Dubuque
Mar 24 at 18:10
2
2
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It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
$begingroup$
It's always a good choice confusing newbies with very general theory. So, well done.
$endgroup$
– amsmath
Mar 25 at 0:12
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@amsmath I think I would have appreciated this explanation a few years ago.
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– Flowers
Mar 25 at 8:25
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@amsmath I think I would have appreciated this explanation a few years ago.
$endgroup$
– Flowers
Mar 25 at 8:25
add a comment |
1
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This related question may be helpful.
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– Brian
Mar 24 at 17:28
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What do you mean “defines only one prime number”??
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– symplectomorphic
Mar 24 at 17:39
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In my opinion your professor is wrong.
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– amsmath
Mar 24 at 17:43
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That definition would be "wrong" because $1$ is divisible by $1$ and itself, but $1$ is not prime.
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– B.Swan
Mar 24 at 17:50
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You should ask your professor what he or she meant. You are most likely interpreting what he or she said incorrectly.
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– symplectomorphic
Mar 24 at 17:54