Mixing
A mild version of Bezout's identity in any commutative ring
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Let $A$ be any commutative ring (with $1$) and $x,y in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b in A$ such that $ax^k+by^l = 1$.
(Proof: Suppose otherwise. Then, $(x^,k,y^l) subset mathfrak p$ for some prime ideal $mathfrak p$. But then this implies that $x,y in mathfrak p$ and therefore $1 = x+y in mathfrak p$, contradiction.)
My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?
algebraic-geometry commutative-algebra
asked Jul 24 at 23:40
ArithmeticGeometer
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up vote
1
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Let $A$ be any commutative ring (with $1$) and $x,y in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b in A$ such that $ax^k+by^l = 1$.
(Proof: Suppose otherwise. Then, $(x^,k,y^l) subset mathfrak p$ for some prime ideal $mathfrak p$. But then this implies that $x,y in mathfrak p$ and therefore $1 = x+y in mathfrak p$, contradiction.)
My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?
algebraic-geometry commutative-algebra
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be any commutative ring (with $1$) and $x,y in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b in A$ such that $ax^k+by^l = 1$.
(Proof: Suppose otherwise. Then, $(x^,k,y^l) subset mathfrak p$ for some prime ideal $mathfrak p$. But then this implies that $x,y in mathfrak p$ and therefore $1 = x+y in mathfrak p$, contradiction.)
My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?
algebraic-geometry commutative-algebra
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
Let $A$ be any commutative ring (with $1$) and $x,y in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b in A$ such that $ax^k+by^l = 1$.
(Proof: Suppose otherwise. Then, $(x^,k,y^l) subset mathfrak p$ for some prime ideal $mathfrak p$. But then this implies that $x,y in mathfrak p$ and therefore $1 = x+y in mathfrak p$, contradiction.)
My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?
algebraic-geometry commutative-algebra
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
asked Jul 24 at 23:40
ArithmeticGeometer
3,15021329
3,15021329
add a comment |Â
add a comment |Â
2 Answers
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3
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accepted
Assuming $k,l$ non-negative integers.
Write $$1=(x+y)^k+l=sum_i=0^k+lbinomk+lix^iy^k+l-i$$
Now, for $i=0,dots,k+l$ either $igeq k$ or $k+l-igeq l.$
So we just separate the terms. If we set:
$$beginaligna&=sum_i=k^k+lbinomk+lix^i-ky^k+l-i\
b&=sum_i=0^k-1binomk+lix^iy^k-iendalign$$
Then $ax^k+by^l=1.$
This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0in R.$
For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
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up vote
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You have $(x+y)^k+l=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $sum binommr x^ry^m-r=1$. Thus, you have $sum_rleq kbinommr x^ry^m-r+sum_r>k binommr x^ry^m-r=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.
answered Jul 25 at 0:03
Mohan
11k1816
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Assuming $k,l$ non-negative integers.
Write $$1=(x+y)^k+l=sum_i=0^k+lbinomk+lix^iy^k+l-i$$
Now, for $i=0,dots,k+l$ either $igeq k$ or $k+l-igeq l.$
So we just separate the terms. If we set:
$$beginaligna&=sum_i=k^k+lbinomk+lix^i-ky^k+l-i\
b&=sum_i=0^k-1binomk+lix^iy^k-iendalign$$
Then $ax^k+by^l=1.$
This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0in R.$
For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
add a comment |Â
up vote
3
down vote
accepted
Assuming $k,l$ non-negative integers.
Write $$1=(x+y)^k+l=sum_i=0^k+lbinomk+lix^iy^k+l-i$$
Now, for $i=0,dots,k+l$ either $igeq k$ or $k+l-igeq l.$
So we just separate the terms. If we set:
$$beginaligna&=sum_i=k^k+lbinomk+lix^i-ky^k+l-i\
b&=sum_i=0^k-1binomk+lix^iy^k-iendalign$$
Then $ax^k+by^l=1.$
This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0in R.$
For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Assuming $k,l$ non-negative integers.
Write $$1=(x+y)^k+l=sum_i=0^k+lbinomk+lix^iy^k+l-i$$
Now, for $i=0,dots,k+l$ either $igeq k$ or $k+l-igeq l.$
So we just separate the terms. If we set:
$$beginaligna&=sum_i=k^k+lbinomk+lix^i-ky^k+l-i\
b&=sum_i=0^k-1binomk+lix^iy^k-iendalign$$
Then $ax^k+by^l=1.$
This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0in R.$
For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
Assuming $k,l$ non-negative integers.
Write $$1=(x+y)^k+l=sum_i=0^k+lbinomk+lix^iy^k+l-i$$
Now, for $i=0,dots,k+l$ either $igeq k$ or $k+l-igeq l.$
So we just separate the terms. If we set:
$$beginaligna&=sum_i=k^k+lbinomk+lix^i-ky^k+l-i\
b&=sum_i=0^k-1binomk+lix^iy^k-iendalign$$
Then $ax^k+by^l=1.$
This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0in R.$
For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
answered Jul 24 at 23:58
Thomas Andrews
128k10144285
128k10144285
add a comment |Â
add a comment |Â
up vote
3
down vote
You have $(x+y)^k+l=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $sum binommr x^ry^m-r=1$. Thus, you have $sum_rleq kbinommr x^ry^m-r+sum_r>k binommr x^ry^m-r=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.
answered Jul 25 at 0:03
Mohan
11k1816
add a comment |Â
up vote
3
down vote
You have $(x+y)^k+l=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $sum binommr x^ry^m-r=1$. Thus, you have $sum_rleq kbinommr x^ry^m-r+sum_r>k binommr x^ry^m-r=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.
answered Jul 25 at 0:03
Mohan
11k1816
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You have $(x+y)^k+l=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $sum binommr x^ry^m-r=1$. Thus, you have $sum_rleq kbinommr x^ry^m-r+sum_r>k binommr x^ry^m-r=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.
answered Jul 25 at 0:03
Mohan
11k1816
You have $(x+y)^k+l=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $sum binommr x^ry^m-r=1$. Thus, you have $sum_rleq kbinommr x^ry^m-r+sum_r>k binommr x^ry^m-r=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.
answered Jul 25 at 0:03
Mohan
11k1816
answered Jul 25 at 0:03
Mohan
11k1816
answered Jul 25 at 0:03
Mohan
11k1816
answered Jul 25 at 0:03
Mohan
11k1816
11k1816
add a comment |Â
add a comment |Â
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