Mixing
prove two rings are the same
$begingroup$
Let $R$ be a commutative ring with identity and $b ∈ R$. Let $T$ be the subring of all multiples of $b$. If $u$ is a unit in $R$ and $u∈T$, prove that $T=R$.
I think the two rings are like $R={text{identity}, b,u,text{other elements}}$, and $T={abmid a∈mathbb{Z}}$, do I make something wrong?
$u$ is a unit in $R$ means that $ux=1$ has solution in $R$, but I saw some comments that $u=ab$, where $a, b∈R$, I know that $b∈R$, but how do we get $a∈R$?
Some comments show that $u$ is also a unit in $T$ (how?)
abstract-algebra
edited Jan 15 at 9:04
egreg
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring with identity and $b ∈ R$. Let $T$ be the subring of all multiples of $b$. If $u$ is a unit in $R$ and $u∈T$, prove that $T=R$.
I think the two rings are like $R={text{identity}, b,u,text{other elements}}$, and $T={abmid a∈mathbb{Z}}$, do I make something wrong?
$u$ is a unit in $R$ means that $ux=1$ has solution in $R$, but I saw some comments that $u=ab$, where $a, b∈R$, I know that $b∈R$, but how do we get $a∈R$?
Some comments show that $u$ is also a unit in $T$ (how?)
abstract-algebra
edited Jan 15 at 9:04
egreg
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
$endgroup$
$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22
add a comment |
$begingroup$
Let $R$ be a commutative ring with identity and $b ∈ R$. Let $T$ be the subring of all multiples of $b$. If $u$ is a unit in $R$ and $u∈T$, prove that $T=R$.
I think the two rings are like $R={text{identity}, b,u,text{other elements}}$, and $T={abmid a∈mathbb{Z}}$, do I make something wrong?
$u$ is a unit in $R$ means that $ux=1$ has solution in $R$, but I saw some comments that $u=ab$, where $a, b∈R$, I know that $b∈R$, but how do we get $a∈R$?
Some comments show that $u$ is also a unit in $T$ (how?)
abstract-algebra
edited Jan 15 at 9:04
egreg
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
$endgroup$
Let $R$ be a commutative ring with identity and $b ∈ R$. Let $T$ be the subring of all multiples of $b$. If $u$ is a unit in $R$ and $u∈T$, prove that $T=R$.
I think the two rings are like $R={text{identity}, b,u,text{other elements}}$, and $T={abmid a∈mathbb{Z}}$, do I make something wrong?
$u$ is a unit in $R$ means that $ux=1$ has solution in $R$, but I saw some comments that $u=ab$, where $a, b∈R$, I know that $b∈R$, but how do we get $a∈R$?
Some comments show that $u$ is also a unit in $T$ (how?)
abstract-algebra
abstract-algebra
edited Jan 15 at 9:04
egreg
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
edited Jan 15 at 9:04
egreg
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
edited Jan 15 at 9:04
egreg
182k1485204
edited Jan 15 at 9:04
egreg
182k1485204
edited Jan 15 at 9:04
egreg
182k1485204
182k1485204
asked Jan 15 at 4:04
M ZM Z
61
asked Jan 15 at 4:04
M ZM Z
61
asked Jan 15 at 4:04
M ZM Z
61
61
$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22
add a comment |
$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22
$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The language is a bit misleading; however, ${mb:minmathbb{Z}}$ is generally not a subring of $R$, so it's very likely that the text means
$$
T={ab:ain R}
$$
which is an ideal of $R$, rather than just a subring.
Saying that $T$ contains a unit $u$ means:
$u=ab$ for some $ain R$;- there exists $vin R$ with $vu=1$.
As a consequence, $1=vu=v(ab)=(va)b$ implies $b$ is a unit. Can you finish now?
answered Jan 15 at 9:08
egregegreg
182k1485204
$endgroup$
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
add a comment |
$begingroup$
Let $u in Tsubset R$ be aunit, and $T$ as you defined, then we want to prove that $T=R$ now to show this we do the clssical, we prove $R subset T$, since then we are done, since we already have by definition the other inclusion.
Now Let $r in R$ , this means that $$r=rcdot 1.$$ As you already observed: $1= x cdot u$ (the other direction is here more convenient and would not even make a problem in noncommutative rings, as left inverses are right inverses, but here we are in particular save thanks to commutativity) for some $x$, so lets throw that in:
$$r=r cdot 1= r cdot xcdot u= $$
but since now $u in T$ we have that $u$ is a scalar multiple of $b$, so there is a $y in R$ such that $ycdot b =u$. Throwing that in again yields now:
$$r=r cdot 1= r cdot xcdot u= r cdot xcdot ycdot b =(r cdot xcdot y)cdot b $$
but since $r cdot xcdot ycdot b$ now defines an element in $R$ we get that the left end of the above equation is in $T$, but so has to be the right side (as they are linked by equalities). So we get $r in T$ for all $rin R$. In particular we get $R subset T$, and hence $R=T$
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
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
$begingroup$
The language is a bit misleading; however, ${mb:minmathbb{Z}}$ is generally not a subring of $R$, so it's very likely that the text means
$$
T={ab:ain R}
$$
which is an ideal of $R$, rather than just a subring.
Saying that $T$ contains a unit $u$ means:
$u=ab$ for some $ain R$;- there exists $vin R$ with $vu=1$.
As a consequence, $1=vu=v(ab)=(va)b$ implies $b$ is a unit. Can you finish now?
answered Jan 15 at 9:08
egregegreg
182k1485204
$endgroup$
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
add a comment |
$begingroup$
The language is a bit misleading; however, ${mb:minmathbb{Z}}$ is generally not a subring of $R$, so it's very likely that the text means
$$
T={ab:ain R}
$$
which is an ideal of $R$, rather than just a subring.
Saying that $T$ contains a unit $u$ means:
$u=ab$ for some $ain R$;- there exists $vin R$ with $vu=1$.
As a consequence, $1=vu=v(ab)=(va)b$ implies $b$ is a unit. Can you finish now?
answered Jan 15 at 9:08
egregegreg
182k1485204
$endgroup$
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
add a comment |
$begingroup$
The language is a bit misleading; however, ${mb:minmathbb{Z}}$ is generally not a subring of $R$, so it's very likely that the text means
$$
T={ab:ain R}
$$
which is an ideal of $R$, rather than just a subring.
Saying that $T$ contains a unit $u$ means:
$u=ab$ for some $ain R$;- there exists $vin R$ with $vu=1$.
As a consequence, $1=vu=v(ab)=(va)b$ implies $b$ is a unit. Can you finish now?
answered Jan 15 at 9:08
egregegreg
182k1485204
$endgroup$
The language is a bit misleading; however, ${mb:minmathbb{Z}}$ is generally not a subring of $R$, so it's very likely that the text means
$$
T={ab:ain R}
$$
which is an ideal of $R$, rather than just a subring.
Saying that $T$ contains a unit $u$ means:
$u=ab$ for some $ain R$;- there exists $vin R$ with $vu=1$.
As a consequence, $1=vu=v(ab)=(va)b$ implies $b$ is a unit. Can you finish now?
answered Jan 15 at 9:08
egregegreg
182k1485204
answered Jan 15 at 9:08
egregegreg
182k1485204
answered Jan 15 at 9:08
egregegreg
182k1485204
answered Jan 15 at 9:08
egregegreg
182k1485204
182k1485204
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
add a comment |
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
$begingroup$
Thanks for the hints, I got really confused about the sentence' multiples of b' before.
$endgroup$
– M Z
Jan 15 at 19:00
add a comment |
$begingroup$
Let $u in Tsubset R$ be aunit, and $T$ as you defined, then we want to prove that $T=R$ now to show this we do the clssical, we prove $R subset T$, since then we are done, since we already have by definition the other inclusion.
Now Let $r in R$ , this means that $$r=rcdot 1.$$ As you already observed: $1= x cdot u$ (the other direction is here more convenient and would not even make a problem in noncommutative rings, as left inverses are right inverses, but here we are in particular save thanks to commutativity) for some $x$, so lets throw that in:
$$r=r cdot 1= r cdot xcdot u= $$
but since now $u in T$ we have that $u$ is a scalar multiple of $b$, so there is a $y in R$ such that $ycdot b =u$. Throwing that in again yields now:
$$r=r cdot 1= r cdot xcdot u= r cdot xcdot ycdot b =(r cdot xcdot y)cdot b $$
but since $r cdot xcdot ycdot b$ now defines an element in $R$ we get that the left end of the above equation is in $T$, but so has to be the right side (as they are linked by equalities). So we get $r in T$ for all $rin R$. In particular we get $R subset T$, and hence $R=T$
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
add a comment |
$begingroup$
Let $u in Tsubset R$ be aunit, and $T$ as you defined, then we want to prove that $T=R$ now to show this we do the clssical, we prove $R subset T$, since then we are done, since we already have by definition the other inclusion.
Now Let $r in R$ , this means that $$r=rcdot 1.$$ As you already observed: $1= x cdot u$ (the other direction is here more convenient and would not even make a problem in noncommutative rings, as left inverses are right inverses, but here we are in particular save thanks to commutativity) for some $x$, so lets throw that in:
$$r=r cdot 1= r cdot xcdot u= $$
but since now $u in T$ we have that $u$ is a scalar multiple of $b$, so there is a $y in R$ such that $ycdot b =u$. Throwing that in again yields now:
$$r=r cdot 1= r cdot xcdot u= r cdot xcdot ycdot b =(r cdot xcdot y)cdot b $$
but since $r cdot xcdot ycdot b$ now defines an element in $R$ we get that the left end of the above equation is in $T$, but so has to be the right side (as they are linked by equalities). So we get $r in T$ for all $rin R$. In particular we get $R subset T$, and hence $R=T$
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
add a comment |
$begingroup$
Let $u in Tsubset R$ be aunit, and $T$ as you defined, then we want to prove that $T=R$ now to show this we do the clssical, we prove $R subset T$, since then we are done, since we already have by definition the other inclusion.
Now Let $r in R$ , this means that $$r=rcdot 1.$$ As you already observed: $1= x cdot u$ (the other direction is here more convenient and would not even make a problem in noncommutative rings, as left inverses are right inverses, but here we are in particular save thanks to commutativity) for some $x$, so lets throw that in:
$$r=r cdot 1= r cdot xcdot u= $$
but since now $u in T$ we have that $u$ is a scalar multiple of $b$, so there is a $y in R$ such that $ycdot b =u$. Throwing that in again yields now:
$$r=r cdot 1= r cdot xcdot u= r cdot xcdot ycdot b =(r cdot xcdot y)cdot b $$
but since $r cdot xcdot ycdot b$ now defines an element in $R$ we get that the left end of the above equation is in $T$, but so has to be the right side (as they are linked by equalities). So we get $r in T$ for all $rin R$. In particular we get $R subset T$, and hence $R=T$
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
$endgroup$
Let $u in Tsubset R$ be aunit, and $T$ as you defined, then we want to prove that $T=R$ now to show this we do the clssical, we prove $R subset T$, since then we are done, since we already have by definition the other inclusion.
Now Let $r in R$ , this means that $$r=rcdot 1.$$ As you already observed: $1= x cdot u$ (the other direction is here more convenient and would not even make a problem in noncommutative rings, as left inverses are right inverses, but here we are in particular save thanks to commutativity) for some $x$, so lets throw that in:
$$r=r cdot 1= r cdot xcdot u= $$
but since now $u in T$ we have that $u$ is a scalar multiple of $b$, so there is a $y in R$ such that $ycdot b =u$. Throwing that in again yields now:
$$r=r cdot 1= r cdot xcdot u= r cdot xcdot ycdot b =(r cdot xcdot y)cdot b $$
but since $r cdot xcdot ycdot b$ now defines an element in $R$ we get that the left end of the above equation is in $T$, but so has to be the right side (as they are linked by equalities). So we get $r in T$ for all $rin R$. In particular we get $R subset T$, and hence $R=T$
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
answered Jan 15 at 8:32
EnkiduEnkidu
1,36719
1,36719
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
add a comment |
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
$begingroup$
thanks! I get it now
$endgroup$
– M Z
Jan 15 at 19:01
add a comment |
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$begingroup$
Multiples of $b$ does not mean $a *b$ for $a in mathbb Z$. It refers to the set formed by multiplying $b$ by every element of the ring $R$ (and not just $mathbb Z$). So, $T = {br : r in R}$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 4:11
$begingroup$
Is the key of the question to show that b is a unit in R? then for any r∈R, r=(r*b-)*b which is the multiples of b, so that r=T? is that right?
$endgroup$
– M Z
Jan 15 at 4:48
$begingroup$
Yes, you are right.
$endgroup$
– Bonbon
Jan 15 at 4:52
$begingroup$
Yes, what you are saying is correct. Alternately, there are answers, so you can view and upvote/accept them.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 15 at 15:22