GCD in a PID persists in extension domains Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)GCD question in PID'sWhy “characteristic zero” and not “infinite characteristic”?$A+A^2B+B=0$ implies $A^2+I$ invertible?An irreducible polynomial cannot share a root with a polynomial without dividing itGCD in a subring is GCD in a bigger ring?Understanding of a theorem about criterion for multiple zerosOn Bezout's identityIs there another meaning of this notation?Integer coprime over $mathbb Z$ but have common factor over an GCD DomainGcd of two elementsGCD in a subring is GCD in a bigger ring?A question on ringsA problem on $textUFD$Greatest common divisors in Integral DomainA problem on $textACCP$$X^4-5X^2+X+1$ is irreducible in $mathbbQ[X]$If $M$ is Noetherian, then $R/textAnn(M)$ is Noetherian, where $M$ is an $R$-moduleSuppose $R$ is a ring and $exists n in mathbbZ_> 0$ such that $(ab)^n=ab, forall a,b in R.$ Then $ab = 0 $ iff $ba=0 ?$If $R$ is $textPID$ and $x in R$ is irreducible, then $R/(x^k)$ is a local ring.Annihilator of an element in a ring
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GCD in a PID persists in extension domains
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)GCD question in PID'sWhy “characteristic zero” and not “infinite characteristic”?$A+A^2B+B=0$ implies $A^2+I$ invertible?An irreducible polynomial cannot share a root with a polynomial without dividing itGCD in a subring is GCD in a bigger ring?Understanding of a theorem about criterion for multiple zerosOn Bezout's identityIs there another meaning of this notation?Integer coprime over $mathbb Z$ but have common factor over an GCD DomainGcd of two elementsGCD in a subring is GCD in a bigger ring?A question on ringsA problem on $textUFD$Greatest common divisors in Integral DomainA problem on $textACCP$$X^4-5X^2+X+1$ is irreducible in $mathbbQ[X]$If $M$ is Noetherian, then $R/textAnn(M)$ is Noetherian, where $M$ is an $R$-moduleSuppose $R$ is a ring and $exists n in mathbbZ_> 0$ such that $(ab)^n=ab, forall a,b in R.$ Then $ab = 0 $ iff $ba=0 ?$If $R$ is $textPID$ and $x in R$ is irreducible, then $R/(x^k)$ is a local ring.Annihilator of an element in a ring
$begingroup$
Let $R$ be subring of integral domain $S.$ Suppose $R$ is $textPID.$ Let $ain R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$
Hints will suffice, thank you.
ring-theory
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
$endgroup$
|
show 1 more comment
$begingroup$
Let $R$ be subring of integral domain $S.$ Suppose $R$ is $textPID.$ Let $ain R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$
Hints will suffice, thank you.
ring-theory
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
$endgroup$
$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22
|
show 1 more comment
$begingroup$
Let $R$ be subring of integral domain $S.$ Suppose $R$ is $textPID.$ Let $ain R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$
Hints will suffice, thank you.
ring-theory
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
$endgroup$
Let $R$ be subring of integral domain $S.$ Suppose $R$ is $textPID.$ Let $ain R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$
Hints will suffice, thank you.
ring-theory
ring-theory
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
edited Jul 22 '17 at 16:08
Bill Dubuque
214k29197660
214k29197660
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
asked Mar 15 '14 at 15:45
Alexy VincenzoAlexy Vincenzo
2,1953926
2,1953926
$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22
|
show 1 more comment
$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22
$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint $ $ gcds in PIDs D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $rm,R supset D.:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$begineqnarray rmgcd(a,b) = c &iff&rm (a,b) = (c)\
&iff&rm a = c:color#C00x , b= c:color#C00 y ,, a:color#C00 u + b: color#C00v = c has roots color#C00x,y,u,vin Dendeqnarray$$
Proof $ (Leftarrow):$ In any ring $rm R,:$ $rm:a: x = c, b: y = c:$ have roots $rm:x,yin R$ $iff$ $rm c | a,b:$ in $rm R.$ Further if $rm:c = a: u + b: v:$ has roots $rm:u,vin R:$ then $rm:d | a,b$ $:Rightarrow:$ $rm:d | a:u+b:v = c:$ in $rm: R.:$ Hence we infer $rm:c = gcd(a,b):$ in $rm: R,:$ being a common divisor divisible by every common divisor. $ (Rightarrow) $ If $rm:c = gcd(a,b):$ in D then the Bezout identity implies the existence of such roots $rm:u,vin D. $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $rm:gcd(2,x) = 1:$ in $rm:mathbb Z[x]:$ but the gcd is the nonunit $:2:$ in $rm:mathbb Z[x/2]subset mathbb Q[x]$.
edited Nov 20 '17 at 6:33
yu qian
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $emid r_1$ and $emid r_2$ in $S$, then $(r_1,r_2)_S subset (e)_S$. Obviously, $(r_1,r_2)_R subset (r_1,r_2)_S$, so we get $(a)_R subset (e)_S$, thus $a in (e)_S$ so $e mid a$ in $S$. And of course $amid r_1$ and $amid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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active
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$begingroup$
Hint $ $ gcds in PIDs D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $rm,R supset D.:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$begineqnarray rmgcd(a,b) = c &iff&rm (a,b) = (c)\
&iff&rm a = c:color#C00x , b= c:color#C00 y ,, a:color#C00 u + b: color#C00v = c has roots color#C00x,y,u,vin Dendeqnarray$$
Proof $ (Leftarrow):$ In any ring $rm R,:$ $rm:a: x = c, b: y = c:$ have roots $rm:x,yin R$ $iff$ $rm c | a,b:$ in $rm R.$ Further if $rm:c = a: u + b: v:$ has roots $rm:u,vin R:$ then $rm:d | a,b$ $:Rightarrow:$ $rm:d | a:u+b:v = c:$ in $rm: R.:$ Hence we infer $rm:c = gcd(a,b):$ in $rm: R,:$ being a common divisor divisible by every common divisor. $ (Rightarrow) $ If $rm:c = gcd(a,b):$ in D then the Bezout identity implies the existence of such roots $rm:u,vin D. $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $rm:gcd(2,x) = 1:$ in $rm:mathbb Z[x]:$ but the gcd is the nonunit $:2:$ in $rm:mathbb Z[x/2]subset mathbb Q[x]$.
edited Nov 20 '17 at 6:33
yu qian
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Hint $ $ gcds in PIDs D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $rm,R supset D.:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$begineqnarray rmgcd(a,b) = c &iff&rm (a,b) = (c)\
&iff&rm a = c:color#C00x , b= c:color#C00 y ,, a:color#C00 u + b: color#C00v = c has roots color#C00x,y,u,vin Dendeqnarray$$
Proof $ (Leftarrow):$ In any ring $rm R,:$ $rm:a: x = c, b: y = c:$ have roots $rm:x,yin R$ $iff$ $rm c | a,b:$ in $rm R.$ Further if $rm:c = a: u + b: v:$ has roots $rm:u,vin R:$ then $rm:d | a,b$ $:Rightarrow:$ $rm:d | a:u+b:v = c:$ in $rm: R.:$ Hence we infer $rm:c = gcd(a,b):$ in $rm: R,:$ being a common divisor divisible by every common divisor. $ (Rightarrow) $ If $rm:c = gcd(a,b):$ in D then the Bezout identity implies the existence of such roots $rm:u,vin D. $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $rm:gcd(2,x) = 1:$ in $rm:mathbb Z[x]:$ but the gcd is the nonunit $:2:$ in $rm:mathbb Z[x/2]subset mathbb Q[x]$.
edited Nov 20 '17 at 6:33
yu qian
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
add a comment |
$begingroup$
Hint $ $ gcds in PIDs D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $rm,R supset D.:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$begineqnarray rmgcd(a,b) = c &iff&rm (a,b) = (c)\
&iff&rm a = c:color#C00x , b= c:color#C00 y ,, a:color#C00 u + b: color#C00v = c has roots color#C00x,y,u,vin Dendeqnarray$$
Proof $ (Leftarrow):$ In any ring $rm R,:$ $rm:a: x = c, b: y = c:$ have roots $rm:x,yin R$ $iff$ $rm c | a,b:$ in $rm R.$ Further if $rm:c = a: u + b: v:$ has roots $rm:u,vin R:$ then $rm:d | a,b$ $:Rightarrow:$ $rm:d | a:u+b:v = c:$ in $rm: R.:$ Hence we infer $rm:c = gcd(a,b):$ in $rm: R,:$ being a common divisor divisible by every common divisor. $ (Rightarrow) $ If $rm:c = gcd(a,b):$ in D then the Bezout identity implies the existence of such roots $rm:u,vin D. $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $rm:gcd(2,x) = 1:$ in $rm:mathbb Z[x]:$ but the gcd is the nonunit $:2:$ in $rm:mathbb Z[x/2]subset mathbb Q[x]$.
edited Nov 20 '17 at 6:33
yu qian
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
$endgroup$
Hint $ $ gcds in PIDs D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $rm,R supset D.:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$begineqnarray rmgcd(a,b) = c &iff&rm (a,b) = (c)\
&iff&rm a = c:color#C00x , b= c:color#C00 y ,, a:color#C00 u + b: color#C00v = c has roots color#C00x,y,u,vin Dendeqnarray$$
Proof $ (Leftarrow):$ In any ring $rm R,:$ $rm:a: x = c, b: y = c:$ have roots $rm:x,yin R$ $iff$ $rm c | a,b:$ in $rm R.$ Further if $rm:c = a: u + b: v:$ has roots $rm:u,vin R:$ then $rm:d | a,b$ $:Rightarrow:$ $rm:d | a:u+b:v = c:$ in $rm: R.:$ Hence we infer $rm:c = gcd(a,b):$ in $rm: R,:$ being a common divisor divisible by every common divisor. $ (Rightarrow) $ If $rm:c = gcd(a,b):$ in D then the Bezout identity implies the existence of such roots $rm:u,vin D. $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $rm:gcd(2,x) = 1:$ in $rm:mathbb Z[x]:$ but the gcd is the nonunit $:2:$ in $rm:mathbb Z[x/2]subset mathbb Q[x]$.
edited Nov 20 '17 at 6:33
yu qian
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
edited Nov 20 '17 at 6:33
yu qian
52
edited Nov 20 '17 at 6:33
yu qian
52
edited Nov 20 '17 at 6:33
yu qian
52
52
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
answered Mar 15 '14 at 16:33
Bill DubuqueBill Dubuque
214k29197660
214k29197660
add a comment |
add a comment |
$begingroup$
Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $emid r_1$ and $emid r_2$ in $S$, then $(r_1,r_2)_S subset (e)_S$. Obviously, $(r_1,r_2)_R subset (r_1,r_2)_S$, so we get $(a)_R subset (e)_S$, thus $a in (e)_S$ so $e mid a$ in $S$. And of course $amid r_1$ and $amid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
$endgroup$
add a comment |
$begingroup$
Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $emid r_1$ and $emid r_2$ in $S$, then $(r_1,r_2)_S subset (e)_S$. Obviously, $(r_1,r_2)_R subset (r_1,r_2)_S$, so we get $(a)_R subset (e)_S$, thus $a in (e)_S$ so $e mid a$ in $S$. And of course $amid r_1$ and $amid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
$endgroup$
add a comment |
$begingroup$
Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $emid r_1$ and $emid r_2$ in $S$, then $(r_1,r_2)_S subset (e)_S$. Obviously, $(r_1,r_2)_R subset (r_1,r_2)_S$, so we get $(a)_R subset (e)_S$, thus $a in (e)_S$ so $e mid a$ in $S$. And of course $amid r_1$ and $amid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
$endgroup$
Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $emid r_1$ and $emid r_2$ in $S$, then $(r_1,r_2)_S subset (e)_S$. Obviously, $(r_1,r_2)_R subset (r_1,r_2)_S$, so we get $(a)_R subset (e)_S$, thus $a in (e)_S$ so $e mid a$ in $S$. And of course $amid r_1$ and $amid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
answered Jul 22 '17 at 16:54
CauchyCauchy
2,583923
2,583923
add a comment |
add a comment |
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$begingroup$
Hint: there exists $s_1,s_2in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$.
$endgroup$
– user119882
Mar 15 '14 at 15:49
$begingroup$
Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) subseteq S$ such that $r_1,r_2 in (z).$ For any $x,y in S, r_1x+r_2y in (z),$ so $(a) subseteq (z)$ in $S.$ Now, if $R$ is $textUFD$ but not $textPID,$ how do I disprove the statement?
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:02
$begingroup$
I'm sorry, how you define $k[r_1, r_2]?$
$endgroup$
– Alexy Vincenzo
Mar 15 '14 at 16:14
$begingroup$
So you need a counterexample, $k[xy,xz]subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$.
$endgroup$
– user119882
Mar 15 '14 at 16:20
$begingroup$
$k[r_1,r_2]=f(r_1,r_2)$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$.
$endgroup$
– user119882
Mar 15 '14 at 16:22