product of two elements that factor in a unique way factors in a unique way?
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I asked myself a question that I can't really answer. If I have a ring R and two elements $x$ and y that factor in a unique way into irreductible elements $x=x_1 ldots x_m$ and $y=y_1ldots y_m$ does this imply that $xy$ factors uniquely as $xy=x_1ldots x_n y_1 ldots y_m$ or does the fact that $x$ and $y$ factor in a unique way is not sufficient (i.e. we need R to be a UFD)?
abstract-algebra ring-theory unique-factorization-domains
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
add a comment |
up vote
0
down vote
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I asked myself a question that I can't really answer. If I have a ring R and two elements $x$ and y that factor in a unique way into irreductible elements $x=x_1 ldots x_m$ and $y=y_1ldots y_m$ does this imply that $xy$ factors uniquely as $xy=x_1ldots x_n y_1 ldots y_m$ or does the fact that $x$ and $y$ factor in a unique way is not sufficient (i.e. we need R to be a UFD)?
abstract-algebra ring-theory unique-factorization-domains
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I asked myself a question that I can't really answer. If I have a ring R and two elements $x$ and y that factor in a unique way into irreductible elements $x=x_1 ldots x_m$ and $y=y_1ldots y_m$ does this imply that $xy$ factors uniquely as $xy=x_1ldots x_n y_1 ldots y_m$ or does the fact that $x$ and $y$ factor in a unique way is not sufficient (i.e. we need R to be a UFD)?
abstract-algebra ring-theory unique-factorization-domains
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
I asked myself a question that I can't really answer. If I have a ring R and two elements $x$ and y that factor in a unique way into irreductible elements $x=x_1 ldots x_m$ and $y=y_1ldots y_m$ does this imply that $xy$ factors uniquely as $xy=x_1ldots x_n y_1 ldots y_m$ or does the fact that $x$ and $y$ factor in a unique way is not sufficient (i.e. we need R to be a UFD)?
abstract-algebra ring-theory unique-factorization-domains
abstract-algebra ring-theory unique-factorization-domains
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
edited Nov 15 at 16:29
José Carlos Santos
141k19111207
141k19111207
asked Nov 15 at 16:14
roi_saumon
31117
asked Nov 15 at 16:14
roi_saumon
31117
asked Nov 15 at 16:14
roi_saumon
31117
31117
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
This is not enough. Indeed, there are examples when $x$ and $y$ are irreducible (then the decomposition is obviously unique), and there is another decomposition $xy = pq$.
Take for example $x=2$ and $y=3$ in $mathbb Z[i sqrt{5}]$, and the well-known example:
$$2 cdot 3 = (1+i sqrt{5}) cdot (1-i sqrt{5}).$$
answered Nov 15 at 16:23
J. Darné
3365
add a comment |
up vote
2
down vote
In $mathbb{Z}[sqrt5,i]$, all elements $2$, $3$, $1+sqrt5,i$ and $1-sqrt5,i$ are irreducible. However,$$2times3=left(1+sqrt5,iright)timesleft(1-sqrt5,iright).$$
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This is not enough. Indeed, there are examples when $x$ and $y$ are irreducible (then the decomposition is obviously unique), and there is another decomposition $xy = pq$.
Take for example $x=2$ and $y=3$ in $mathbb Z[i sqrt{5}]$, and the well-known example:
$$2 cdot 3 = (1+i sqrt{5}) cdot (1-i sqrt{5}).$$
answered Nov 15 at 16:23
J. Darné
3365
add a comment |
up vote
1
down vote
accepted
This is not enough. Indeed, there are examples when $x$ and $y$ are irreducible (then the decomposition is obviously unique), and there is another decomposition $xy = pq$.
Take for example $x=2$ and $y=3$ in $mathbb Z[i sqrt{5}]$, and the well-known example:
$$2 cdot 3 = (1+i sqrt{5}) cdot (1-i sqrt{5}).$$
answered Nov 15 at 16:23
J. Darné
3365
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This is not enough. Indeed, there are examples when $x$ and $y$ are irreducible (then the decomposition is obviously unique), and there is another decomposition $xy = pq$.
Take for example $x=2$ and $y=3$ in $mathbb Z[i sqrt{5}]$, and the well-known example:
$$2 cdot 3 = (1+i sqrt{5}) cdot (1-i sqrt{5}).$$
answered Nov 15 at 16:23
J. Darné
3365
This is not enough. Indeed, there are examples when $x$ and $y$ are irreducible (then the decomposition is obviously unique), and there is another decomposition $xy = pq$.
Take for example $x=2$ and $y=3$ in $mathbb Z[i sqrt{5}]$, and the well-known example:
$$2 cdot 3 = (1+i sqrt{5}) cdot (1-i sqrt{5}).$$
answered Nov 15 at 16:23
J. Darné
3365
answered Nov 15 at 16:23
J. Darné
3365
answered Nov 15 at 16:23
J. Darné
3365
answered Nov 15 at 16:23
J. Darné
3365
3365
add a comment |
add a comment |
up vote
2
down vote
In $mathbb{Z}[sqrt5,i]$, all elements $2$, $3$, $1+sqrt5,i$ and $1-sqrt5,i$ are irreducible. However,$$2times3=left(1+sqrt5,iright)timesleft(1-sqrt5,iright).$$
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
add a comment |
up vote
2
down vote
In $mathbb{Z}[sqrt5,i]$, all elements $2$, $3$, $1+sqrt5,i$ and $1-sqrt5,i$ are irreducible. However,$$2times3=left(1+sqrt5,iright)timesleft(1-sqrt5,iright).$$
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
add a comment |
up vote
2
down vote
up vote
2
down vote
In $mathbb{Z}[sqrt5,i]$, all elements $2$, $3$, $1+sqrt5,i$ and $1-sqrt5,i$ are irreducible. However,$$2times3=left(1+sqrt5,iright)timesleft(1-sqrt5,iright).$$
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
In $mathbb{Z}[sqrt5,i]$, all elements $2$, $3$, $1+sqrt5,i$ and $1-sqrt5,i$ are irreducible. However,$$2times3=left(1+sqrt5,iright)timesleft(1-sqrt5,iright).$$
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
answered Nov 15 at 16:22
José Carlos Santos
141k19111207
141k19111207
add a comment |
add a comment |
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