Integral Domain, PID and gcd
$begingroup$
Let $R$ be a PID and $R'$ a Integral Domain and $Rsubseteq R'$. Let $a,b,d in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$.
Proof:
$d$ is a gcd of $a$ and $b$ in $R$ $Leftrightarrow$ $(a,b)=(d)$. It is clear that $d$ is a divisor of of $a$ and $b$ in $R'$, we have to show that it is the Greatest divisor. Let $ein R'$ be a divisor of $a$ and $b$ in $R'$. This means $exists , v,w in R' , : a=ve , , b=we$. Because $(a,b)=(d) $ we can assume $exists , s,tin R' , : d=sa+tb = sev + tew = esv + etw = e(sv+tw)$ and $sv+twin R'$ since $s,t,v,win R'$. Let $c:=sv+tw$, then we found a $cin R'$ with $ce=d$ so $e$ is a divisor of $d$ in $R'$.
Is this correct?
abstract-algebra greatest-common-divisor principal-ideal-domains integral-domain
edited Jan 10 at 12:22
darij grinberg
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
$endgroup$
add a comment |
$begingroup$
Let $R$ be a PID and $R'$ a Integral Domain and $Rsubseteq R'$. Let $a,b,d in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$.
Proof:
$d$ is a gcd of $a$ and $b$ in $R$ $Leftrightarrow$ $(a,b)=(d)$. It is clear that $d$ is a divisor of of $a$ and $b$ in $R'$, we have to show that it is the Greatest divisor. Let $ein R'$ be a divisor of $a$ and $b$ in $R'$. This means $exists , v,w in R' , : a=ve , , b=we$. Because $(a,b)=(d) $ we can assume $exists , s,tin R' , : d=sa+tb = sev + tew = esv + etw = e(sv+tw)$ and $sv+twin R'$ since $s,t,v,win R'$. Let $c:=sv+tw$, then we found a $cin R'$ with $ce=d$ so $e$ is a divisor of $d$ in $R'$.
Is this correct?
abstract-algebra greatest-common-divisor principal-ideal-domains integral-domain
edited Jan 10 at 12:22
darij grinberg
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
$endgroup$
add a comment |
$begingroup$
Let $R$ be a PID and $R'$ a Integral Domain and $Rsubseteq R'$. Let $a,b,d in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$.
Proof:
$d$ is a gcd of $a$ and $b$ in $R$ $Leftrightarrow$ $(a,b)=(d)$. It is clear that $d$ is a divisor of of $a$ and $b$ in $R'$, we have to show that it is the Greatest divisor. Let $ein R'$ be a divisor of $a$ and $b$ in $R'$. This means $exists , v,w in R' , : a=ve , , b=we$. Because $(a,b)=(d) $ we can assume $exists , s,tin R' , : d=sa+tb = sev + tew = esv + etw = e(sv+tw)$ and $sv+twin R'$ since $s,t,v,win R'$. Let $c:=sv+tw$, then we found a $cin R'$ with $ce=d$ so $e$ is a divisor of $d$ in $R'$.
Is this correct?
abstract-algebra greatest-common-divisor principal-ideal-domains integral-domain
edited Jan 10 at 12:22
darij grinberg
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
$endgroup$
Let $R$ be a PID and $R'$ a Integral Domain and $Rsubseteq R'$. Let $a,b,d in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$.
Proof:
$d$ is a gcd of $a$ and $b$ in $R$ $Leftrightarrow$ $(a,b)=(d)$. It is clear that $d$ is a divisor of of $a$ and $b$ in $R'$, we have to show that it is the Greatest divisor. Let $ein R'$ be a divisor of $a$ and $b$ in $R'$. This means $exists , v,w in R' , : a=ve , , b=we$. Because $(a,b)=(d) $ we can assume $exists , s,tin R' , : d=sa+tb = sev + tew = esv + etw = e(sv+tw)$ and $sv+twin R'$ since $s,t,v,win R'$. Let $c:=sv+tw$, then we found a $cin R'$ with $ce=d$ so $e$ is a divisor of $d$ in $R'$.
Is this correct?
abstract-algebra greatest-common-divisor principal-ideal-domains integral-domain
abstract-algebra greatest-common-divisor principal-ideal-domains integral-domain
edited Jan 10 at 12:22
darij grinberg
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
edited Jan 10 at 12:22
darij grinberg
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
edited Jan 10 at 12:22
darij grinberg
11k33167
edited Jan 10 at 12:22
darij grinberg
11k33167
edited Jan 10 at 12:22
darij grinberg
11k33167
11k33167
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
asked Jan 10 at 9:05
KingDingelingKingDingeling
1607
1607
add a comment |
add a comment |
1 Answer
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$begingroup$
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $left(xright)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
$endgroup$
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
add a comment |
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$begingroup$
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $left(xright)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
$endgroup$
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
add a comment |
$begingroup$
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $left(xright)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
$endgroup$
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
add a comment |
$begingroup$
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $left(xright)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
$endgroup$
I've made a few edits to correct typos and confusing notations (be careful: if you use the old-fashioned notation $left(xright)$ for the ideal generated by a single element $x$, then you should not put extra parentheses around elements when you mean the elements themselves!). Now your proof is correct.
Note that you never use that $R'$ is an integral domain. It is a redundant assumption.
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
answered Jan 10 at 12:24
darij grinbergdarij grinberg
11k33167
11k33167
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
add a comment |
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
$begingroup$
Thanks a lot, very kind :)
$endgroup$
– KingDingeling
Jan 10 at 12:56
add a comment |
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