Non unique factorization of integer valued polynomials
$begingroup$
Is there a nice example of a polynomial with non unique factorization in the subring of
$mathbb Q[X,Y]$ of polynomials that defines functions $mathbb Z^2tomathbb Z$?
I don't think this subring is a UFD because of the possibility to multiply the unique factors in $mathbb Q[X,Y]$ in different ways to sometimes obtain different irreducible factors in the subring.
abstract-algebra polynomials ring-theory factoring
edited Dec 10 '18 at 5:21
PatrickR
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
$endgroup$
add a comment |
$begingroup$
Is there a nice example of a polynomial with non unique factorization in the subring of
$mathbb Q[X,Y]$ of polynomials that defines functions $mathbb Z^2tomathbb Z$?
I don't think this subring is a UFD because of the possibility to multiply the unique factors in $mathbb Q[X,Y]$ in different ways to sometimes obtain different irreducible factors in the subring.
abstract-algebra polynomials ring-theory factoring
edited Dec 10 '18 at 5:21
PatrickR
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
$endgroup$
add a comment |
$begingroup$
Is there a nice example of a polynomial with non unique factorization in the subring of
$mathbb Q[X,Y]$ of polynomials that defines functions $mathbb Z^2tomathbb Z$?
I don't think this subring is a UFD because of the possibility to multiply the unique factors in $mathbb Q[X,Y]$ in different ways to sometimes obtain different irreducible factors in the subring.
abstract-algebra polynomials ring-theory factoring
edited Dec 10 '18 at 5:21
PatrickR
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
$endgroup$
Is there a nice example of a polynomial with non unique factorization in the subring of
$mathbb Q[X,Y]$ of polynomials that defines functions $mathbb Z^2tomathbb Z$?
I don't think this subring is a UFD because of the possibility to multiply the unique factors in $mathbb Q[X,Y]$ in different ways to sometimes obtain different irreducible factors in the subring.
abstract-algebra polynomials ring-theory factoring
abstract-algebra polynomials ring-theory factoring
edited Dec 10 '18 at 5:21
PatrickR
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
edited Dec 10 '18 at 5:21
PatrickR
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
edited Dec 10 '18 at 5:21
PatrickR
715623
edited Dec 10 '18 at 5:21
PatrickR
715623
edited Dec 10 '18 at 5:21
PatrickR
715623
715623
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
asked Dec 7 '18 at 12:53
LehsLehs
7,00831662
7,00831662
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This can even be done with one variable:
$$
2cdot left(frac{x(x+1)}{2}right)=big(xbig)cdotbig(x+1big).
$$
If you prefer to avoid irreducibles that become units in $mathbb{Q}$:
$$
left(frac{x(x+1)}{2}right)cdotleft(frac{(x+2)(x+3)}{2}right) = left(frac{x(x+3)}{2}right)cdotleft(frac{(x+1)(x+2)}{2}right).
$$
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
$endgroup$
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
add a comment |
$begingroup$
Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example
$$ n {xchoose n}, =, (x-n+1){xchoose n-1}$$
The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.
For completeness, below is a proof of the irreducibility of $xchoose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This can even be done with one variable:
$$
2cdot left(frac{x(x+1)}{2}right)=big(xbig)cdotbig(x+1big).
$$
If you prefer to avoid irreducibles that become units in $mathbb{Q}$:
$$
left(frac{x(x+1)}{2}right)cdotleft(frac{(x+2)(x+3)}{2}right) = left(frac{x(x+3)}{2}right)cdotleft(frac{(x+1)(x+2)}{2}right).
$$
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
$endgroup$
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
add a comment |
$begingroup$
This can even be done with one variable:
$$
2cdot left(frac{x(x+1)}{2}right)=big(xbig)cdotbig(x+1big).
$$
If you prefer to avoid irreducibles that become units in $mathbb{Q}$:
$$
left(frac{x(x+1)}{2}right)cdotleft(frac{(x+2)(x+3)}{2}right) = left(frac{x(x+3)}{2}right)cdotleft(frac{(x+1)(x+2)}{2}right).
$$
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
$endgroup$
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
add a comment |
$begingroup$
This can even be done with one variable:
$$
2cdot left(frac{x(x+1)}{2}right)=big(xbig)cdotbig(x+1big).
$$
If you prefer to avoid irreducibles that become units in $mathbb{Q}$:
$$
left(frac{x(x+1)}{2}right)cdotleft(frac{(x+2)(x+3)}{2}right) = left(frac{x(x+3)}{2}right)cdotleft(frac{(x+1)(x+2)}{2}right).
$$
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
$endgroup$
This can even be done with one variable:
$$
2cdot left(frac{x(x+1)}{2}right)=big(xbig)cdotbig(x+1big).
$$
If you prefer to avoid irreducibles that become units in $mathbb{Q}$:
$$
left(frac{x(x+1)}{2}right)cdotleft(frac{(x+2)(x+3)}{2}right) = left(frac{x(x+3)}{2}right)cdotleft(frac{(x+1)(x+2)}{2}right).
$$
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
answered Dec 7 '18 at 16:06
Julian RosenJulian Rosen
11.8k12348
11.8k12348
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
add a comment |
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
$begingroup$
Nice! Thanks a lot!
$endgroup$
– Lehs
Dec 7 '18 at 17:11
add a comment |
$begingroup$
Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example
$$ n {xchoose n}, =, (x-n+1){xchoose n-1}$$
The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.
For completeness, below is a proof of the irreducibility of $xchoose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
add a comment |
$begingroup$
Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example
$$ n {xchoose n}, =, (x-n+1){xchoose n-1}$$
The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.
For completeness, below is a proof of the irreducibility of $xchoose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
add a comment |
$begingroup$
Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example
$$ n {xchoose n}, =, (x-n+1){xchoose n-1}$$
The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.
For completeness, below is a proof of the irreducibility of $xchoose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
Worth emphasis is how extremely different the lengths of irreducible factorizations can be in such nonunique factorizations in the ring of integer-valued polynomials. For example
$$ n {xchoose n}, =, (x-n+1){xchoose n-1}$$
The RHS is a product of two irreducibles, but the LHS can have an arbitrarily large number of irreducible factors by choosing $n$ with many prime factors.
For completeness, below is a proof of the irreducibility of $xchoose n$ from the 2016 Monthly paper What You Should Know About Integer-Valued Polynomials by Cahen and Chabert.
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
edited Dec 7 '18 at 19:47
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
answered Dec 7 '18 at 19:41
Bill DubuqueBill Dubuque
209k29191634
209k29191634
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
add a comment |
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
1
1
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
$begingroup$
Very interesting! Thank you!
$endgroup$
– Lehs
Dec 7 '18 at 20:03
add a comment |
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