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If D is integral domain, then $b_m$ is unit
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Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and
$b_m ne 0$ satisfies for all $a(x) in D[x]$, $q(x)$ and $r(x)$ $in D[x]$ exist such that
$a(x) = b(x)q(x) + r(x)$ with $r(x) = 0$ or $deg(r(x)) < deg(b(x))$; $Rightarrow b_m$ is unit.
I'm not sure how to begin, in case $r(x)=0$ then $b(x)mid a(x)$ but I think this path doesn´t help me to prove $b_m$ is a unit. Is there any theorem I can use? Can you provide me with tips?
polynomials ring-theory integral-domain
asked Jul 19 at 21:07
Jorge
426
add a comment |Â
up vote
0
down vote
favorite
Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and
$b_m ne 0$ satisfies for all $a(x) in D[x]$, $q(x)$ and $r(x)$ $in D[x]$ exist such that
$a(x) = b(x)q(x) + r(x)$ with $r(x) = 0$ or $deg(r(x)) < deg(b(x))$; $Rightarrow b_m$ is unit.
I'm not sure how to begin, in case $r(x)=0$ then $b(x)mid a(x)$ but I think this path doesn´t help me to prove $b_m$ is a unit. Is there any theorem I can use? Can you provide me with tips?
polynomials ring-theory integral-domain
asked Jul 19 at 21:07
Jorge
426
1
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
1
Yes,is the degree
– Jorge
Jul 19 at 21:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and
$b_m ne 0$ satisfies for all $a(x) in D[x]$, $q(x)$ and $r(x)$ $in D[x]$ exist such that
$a(x) = b(x)q(x) + r(x)$ with $r(x) = 0$ or $deg(r(x)) < deg(b(x))$; $Rightarrow b_m$ is unit.
I'm not sure how to begin, in case $r(x)=0$ then $b(x)mid a(x)$ but I think this path doesn´t help me to prove $b_m$ is a unit. Is there any theorem I can use? Can you provide me with tips?
polynomials ring-theory integral-domain
asked Jul 19 at 21:07
Jorge
426
Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and
$b_m ne 0$ satisfies for all $a(x) in D[x]$, $q(x)$ and $r(x)$ $in D[x]$ exist such that
$a(x) = b(x)q(x) + r(x)$ with $r(x) = 0$ or $deg(r(x)) < deg(b(x))$; $Rightarrow b_m$ is unit.
I'm not sure how to begin, in case $r(x)=0$ then $b(x)mid a(x)$ but I think this path doesn´t help me to prove $b_m$ is a unit. Is there any theorem I can use? Can you provide me with tips?
polynomials ring-theory integral-domain
asked Jul 19 at 21:07
Jorge
426
edited Jul 19 at 21:29
user401938
asked Jul 19 at 21:07
Jorge
426
asked Jul 19 at 21:07
Jorge
426
asked Jul 19 at 21:07
Jorge
426
426
1
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
1
Yes,is the degree
– Jorge
Jul 19 at 21:11
add a comment |Â
1
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
1
Yes,is the degree
– Jorge
Jul 19 at 21:11
1
1
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
1
1
Yes,is the degree
– Jorge
Jul 19 at 21:11
Yes,is the degree
– Jorge
Jul 19 at 21:11
add a comment |Â
1 Answer
1
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1
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Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.
answered Jul 19 at 21:18
Sage Phreak
2189
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.
answered Jul 19 at 21:18
Sage Phreak
2189
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
add a comment |Â
up vote
1
down vote
accepted
Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.
answered Jul 19 at 21:18
Sage Phreak
2189
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.
answered Jul 19 at 21:18
Sage Phreak
2189
Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.
answered Jul 19 at 21:18
Sage Phreak
2189
edited Jul 19 at 21:29
answered Jul 19 at 21:18
Sage Phreak
2189
answered Jul 19 at 21:18
Sage Phreak
2189
answered Jul 19 at 21:18
Sage Phreak
2189
2189
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
add a comment |Â
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
1
1
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
@gd1035 Your answer was only missing to say that because $D$ is integral domain the $q$ cannot have positive degree for your choice of $a$.
– user577471
Jul 19 at 21:20
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
True. I've added that now.
– Sage Phreak
Jul 19 at 21:29
add a comment |Â
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1
What is $gr(r(x))$, is this the degree of the polynomial?
– gd1035
Jul 19 at 21:09
1
Yes,is the degree
– Jorge
Jul 19 at 21:11