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Ring with unity. Show that if ab units, then a and b are units.
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Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.
I know that a ring with unity means that
$$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$
For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.
For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.
Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.
I have no starting point thus far so any hint(s) are greatly appreciated.
abstract-algebra ring-theory
asked 49 mins ago
salad salad
585
add a comment |Â
up vote
3
down vote
favorite
Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.
I know that a ring with unity means that
$$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$
For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.
For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.
Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.
I have no starting point thus far so any hint(s) are greatly appreciated.
abstract-algebra ring-theory
asked 49 mins ago
salad salad
585
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.
I know that a ring with unity means that
$$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$
For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.
For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.
Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.
I have no starting point thus far so any hint(s) are greatly appreciated.
abstract-algebra ring-theory
asked 49 mins ago
salad salad
585
Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.
I know that a ring with unity means that
$$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$
For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.
For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.
Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.
I have no starting point thus far so any hint(s) are greatly appreciated.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked 49 mins ago
salad salad
585
asked 49 mins ago
salad salad
585
asked 49 mins ago
salad salad
585
asked 49 mins ago
salad salad
585
asked 49 mins ago
salad salad
585
585
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?
This is where the zero divisor condition comes in. Consider the following product:
$$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.
A very similar method works to show that $b^-1 = (ab)^-1a$.
answered 35 mins ago
Theo Bendit
15k12045
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up vote
2
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$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$
answered 35 mins ago
Tsemo Aristide
53.1k11244
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up vote
1
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Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).
answered 43 mins ago
Cronus
879415
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
add a comment |Â
up vote
0
down vote
Hint $ $ Use the special case $,d=1,$ of the following
Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$
Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$
answered 10 mins ago
Bill Dubuque
204k29188614
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?
This is where the zero divisor condition comes in. Consider the following product:
$$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.
A very similar method works to show that $b^-1 = (ab)^-1a$.
answered 35 mins ago
Theo Bendit
15k12045
add a comment |Â
up vote
4
down vote
You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?
This is where the zero divisor condition comes in. Consider the following product:
$$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.
A very similar method works to show that $b^-1 = (ab)^-1a$.
answered 35 mins ago
Theo Bendit
15k12045
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?
This is where the zero divisor condition comes in. Consider the following product:
$$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.
A very similar method works to show that $b^-1 = (ab)^-1a$.
answered 35 mins ago
Theo Bendit
15k12045
You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?
This is where the zero divisor condition comes in. Consider the following product:
$$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.
A very similar method works to show that $b^-1 = (ab)^-1a$.
answered 35 mins ago
Theo Bendit
15k12045
answered 35 mins ago
Theo Bendit
15k12045
answered 35 mins ago
Theo Bendit
15k12045
answered 35 mins ago
Theo Bendit
15k12045
15k12045
add a comment |Â
add a comment |Â
up vote
2
down vote
$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$
answered 35 mins ago
Tsemo Aristide
53.1k11244
add a comment |Â
up vote
2
down vote
$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$
answered 35 mins ago
Tsemo Aristide
53.1k11244
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$
answered 35 mins ago
Tsemo Aristide
53.1k11244
$(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$
answered 35 mins ago
Tsemo Aristide
53.1k11244
answered 35 mins ago
Tsemo Aristide
53.1k11244
answered 35 mins ago
Tsemo Aristide
53.1k11244
answered 35 mins ago
Tsemo Aristide
53.1k11244
53.1k11244
add a comment |Â
add a comment |Â
up vote
1
down vote
Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).
answered 43 mins ago
Cronus
879415
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
add a comment |Â
up vote
1
down vote
Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).
answered 43 mins ago
Cronus
879415
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).
answered 43 mins ago
Cronus
879415
Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).
answered 43 mins ago
Cronus
879415
answered 43 mins ago
Cronus
879415
answered 43 mins ago
Cronus
879415
answered 43 mins ago
Cronus
879415
879415
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
add a comment |Â
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
3
3
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
– Theo Bendit
41 mins ago
add a comment |Â
up vote
0
down vote
Hint $ $ Use the special case $,d=1,$ of the following
Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$
Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$
answered 10 mins ago
Bill Dubuque
204k29188614
add a comment |Â
up vote
0
down vote
Hint $ $ Use the special case $,d=1,$ of the following
Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$
Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$
answered 10 mins ago
Bill Dubuque
204k29188614
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint $ $ Use the special case $,d=1,$ of the following
Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$
Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$
answered 10 mins ago
Bill Dubuque
204k29188614
Hint $ $ Use the special case $,d=1,$ of the following
Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$
Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$
answered 10 mins ago
Bill Dubuque
204k29188614
edited 4 mins ago
answered 10 mins ago
Bill Dubuque
204k29188614
answered 10 mins ago
Bill Dubuque
204k29188614
answered 10 mins ago
Bill Dubuque
204k29188614
204k29188614
add a comment |Â
add a comment |Â
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