Units and Nilpotents
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If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.
I've been working on this problem for an hour that I tried to construct an element $x in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
abstract-algebra ring-theory faq
$endgroup$
add a comment |
$begingroup$
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.
I've been working on this problem for an hour that I tried to construct an element $x in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
abstract-algebra ring-theory faq
$endgroup$
4
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Dear Shannon, Try the case $u = 1$ first. Regards,
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– Matt E
Mar 14 '12 at 2:23
1
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See also here.
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– Bill Dubuque
May 3 '12 at 14:43
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See also here for the commutative case.
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– Bill Dubuque
Aug 6 '17 at 15:21
add a comment |
$begingroup$
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.
I've been working on this problem for an hour that I tried to construct an element $x in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
abstract-algebra ring-theory faq
$endgroup$
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.
I've been working on this problem for an hour that I tried to construct an element $x in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
abstract-algebra ring-theory faq
abstract-algebra ring-theory faq
edited Jun 26 '17 at 14:20
Trevor Gunn
14.9k32047
14.9k32047
asked Mar 14 '12 at 2:18
ShannonShannon
5433620
5433620
4
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Dear Shannon, Try the case $u = 1$ first. Regards,
$endgroup$
– Matt E
Mar 14 '12 at 2:23
1
$begingroup$
See also here.
$endgroup$
– Bill Dubuque
May 3 '12 at 14:43
$begingroup$
See also here for the commutative case.
$endgroup$
– Bill Dubuque
Aug 6 '17 at 15:21
add a comment |
4
$begingroup$
Dear Shannon, Try the case $u = 1$ first. Regards,
$endgroup$
– Matt E
Mar 14 '12 at 2:23
1
$begingroup$
See also here.
$endgroup$
– Bill Dubuque
May 3 '12 at 14:43
$begingroup$
See also here for the commutative case.
$endgroup$
– Bill Dubuque
Aug 6 '17 at 15:21
4
4
$begingroup$
Dear Shannon, Try the case $u = 1$ first. Regards,
$endgroup$
– Matt E
Mar 14 '12 at 2:23
$begingroup$
Dear Shannon, Try the case $u = 1$ first. Regards,
$endgroup$
– Matt E
Mar 14 '12 at 2:23
1
1
$begingroup$
See also here.
$endgroup$
– Bill Dubuque
May 3 '12 at 14:43
$begingroup$
See also here.
$endgroup$
– Bill Dubuque
May 3 '12 at 14:43
$begingroup$
See also here for the commutative case.
$endgroup$
– Bill Dubuque
Aug 6 '17 at 15:21
$begingroup$
See also here for the commutative case.
$endgroup$
– Bill Dubuque
Aug 6 '17 at 15:21
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $u=1$, then you could do it via the identity
$$(1+a)(1-a+a^2-a^3+cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$
by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
$endgroup$
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
add a comment |
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Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that
$$(u+a)cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$
See if you can generalize this.
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1
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The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
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– Shannon
Mar 14 '12 at 2:55
add a comment |
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Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $Msubset R$. Since $a$ is nilpotent, $ain M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-ain M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $Ssubseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-ain I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+binom{n}{2}v^2(u+a)^2+dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)left(nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}right)$$ and so $$-sum_{k=1}^n binom{n}{k}(-v)^k(u+a)^{k-1}= nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
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add a comment |
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Note that since $u$ is a unit and
$ua = au, tag 1$
we may write
$a = u^{-1}au, tag 2$
and thus
$au^{-1} = u^{-1}a; tag 3$
also, since $a$ is nilpotent there is some $0 < n in Bbb N$ such that
$a^n = 0, tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; tag 5$
we observe that
$u + a = u(1 + u^{-1}a), tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) displaystyle sum_0^{n - 1} (-u^{-1}a)^k = sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}asum_0^{n - 1} (-u^{-1}a)^k$
$= displaystyle sum_0^{n - 1} (-1)^k(u^{-1}a)^k + sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + displaystyle sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; tag 7$
this shows that
$(1 + u^{-1}a)^{-1} = displaystyle sum_0^{n - 1} (-u^{-1}a)^k, tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $u=1$, then you could do it via the identity
$$(1+a)(1-a+a^2-a^3+cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$
by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
$endgroup$
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
add a comment |
$begingroup$
If $u=1$, then you could do it via the identity
$$(1+a)(1-a+a^2-a^3+cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$
by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
$endgroup$
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
add a comment |
$begingroup$
If $u=1$, then you could do it via the identity
$$(1+a)(1-a+a^2-a^3+cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$
by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
$endgroup$
If $u=1$, then you could do it via the identity
$$(1+a)(1-a+a^2-a^3+cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$
by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
edited Mar 25 '15 at 13:28
AMPerrine
2,90822026
2,90822026
answered Mar 14 '12 at 2:26
Arturo MagidinArturo Magidin
265k34590919
265k34590919
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
add a comment |
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
$begingroup$
I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a?
$endgroup$
– Shannon
Mar 14 '12 at 2:51
1
1
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent?
$endgroup$
– Arturo Magidin
Mar 14 '12 at 2:56
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
oh, I see. I will work on it. Thank you so much.
$endgroup$
– Shannon
Mar 14 '12 at 3:10
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
$begingroup$
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises.
$endgroup$
– Derek Elkins
Jan 21 '16 at 4:01
add a comment |
$begingroup$
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that
$$(u+a)cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$
See if you can generalize this.
$endgroup$
1
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
add a comment |
$begingroup$
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that
$$(u+a)cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$
See if you can generalize this.
$endgroup$
1
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
add a comment |
$begingroup$
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that
$$(u+a)cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$
See if you can generalize this.
$endgroup$
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that
$$(u+a)cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$
See if you can generalize this.
answered Mar 14 '12 at 2:23
Zev ChonolesZev Chonoles
110k16230429
110k16230429
1
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
add a comment |
1
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
1
1
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
$begingroup$
The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from?
$endgroup$
– Shannon
Mar 14 '12 at 2:55
add a comment |
$begingroup$
Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $Msubset R$. Since $a$ is nilpotent, $ain M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-ain M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $Ssubseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-ain I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+binom{n}{2}v^2(u+a)^2+dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)left(nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}right)$$ and so $$-sum_{k=1}^n binom{n}{k}(-v)^k(u+a)^{k-1}= nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
$endgroup$
add a comment |
$begingroup$
Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $Msubset R$. Since $a$ is nilpotent, $ain M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-ain M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $Ssubseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-ain I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+binom{n}{2}v^2(u+a)^2+dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)left(nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}right)$$ and so $$-sum_{k=1}^n binom{n}{k}(-v)^k(u+a)^{k-1}= nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
$endgroup$
add a comment |
$begingroup$
Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $Msubset R$. Since $a$ is nilpotent, $ain M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-ain M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $Ssubseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-ain I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+binom{n}{2}v^2(u+a)^2+dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)left(nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}right)$$ and so $$-sum_{k=1}^n binom{n}{k}(-v)^k(u+a)^{k-1}= nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
$endgroup$
Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $Msubset R$. Since $a$ is nilpotent, $ain M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-ain M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $Ssubseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-ain I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+binom{n}{2}v^2(u+a)^2+dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)left(nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}right)$$ and so $$-sum_{k=1}^n binom{n}{k}(-v)^k(u+a)^{k-1}= nv-binom{n}{2}v^2(u+a)+dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
edited Nov 2 '17 at 8:40
answered Dec 16 '15 at 8:28
Eric WofseyEric Wofsey
190k14216348
190k14216348
add a comment |
add a comment |
$begingroup$
Note that since $u$ is a unit and
$ua = au, tag 1$
we may write
$a = u^{-1}au, tag 2$
and thus
$au^{-1} = u^{-1}a; tag 3$
also, since $a$ is nilpotent there is some $0 < n in Bbb N$ such that
$a^n = 0, tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; tag 5$
we observe that
$u + a = u(1 + u^{-1}a), tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) displaystyle sum_0^{n - 1} (-u^{-1}a)^k = sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}asum_0^{n - 1} (-u^{-1}a)^k$
$= displaystyle sum_0^{n - 1} (-1)^k(u^{-1}a)^k + sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + displaystyle sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; tag 7$
this shows that
$(1 + u^{-1}a)^{-1} = displaystyle sum_0^{n - 1} (-u^{-1}a)^k, tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.
$endgroup$
add a comment |
$begingroup$
Note that since $u$ is a unit and
$ua = au, tag 1$
we may write
$a = u^{-1}au, tag 2$
and thus
$au^{-1} = u^{-1}a; tag 3$
also, since $a$ is nilpotent there is some $0 < n in Bbb N$ such that
$a^n = 0, tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; tag 5$
we observe that
$u + a = u(1 + u^{-1}a), tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) displaystyle sum_0^{n - 1} (-u^{-1}a)^k = sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}asum_0^{n - 1} (-u^{-1}a)^k$
$= displaystyle sum_0^{n - 1} (-1)^k(u^{-1}a)^k + sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + displaystyle sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; tag 7$
this shows that
$(1 + u^{-1}a)^{-1} = displaystyle sum_0^{n - 1} (-u^{-1}a)^k, tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.
$endgroup$
add a comment |
$begingroup$
Note that since $u$ is a unit and
$ua = au, tag 1$
we may write
$a = u^{-1}au, tag 2$
and thus
$au^{-1} = u^{-1}a; tag 3$
also, since $a$ is nilpotent there is some $0 < n in Bbb N$ such that
$a^n = 0, tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; tag 5$
we observe that
$u + a = u(1 + u^{-1}a), tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) displaystyle sum_0^{n - 1} (-u^{-1}a)^k = sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}asum_0^{n - 1} (-u^{-1}a)^k$
$= displaystyle sum_0^{n - 1} (-1)^k(u^{-1}a)^k + sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + displaystyle sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; tag 7$
this shows that
$(1 + u^{-1}a)^{-1} = displaystyle sum_0^{n - 1} (-u^{-1}a)^k, tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.
$endgroup$
Note that since $u$ is a unit and
$ua = au, tag 1$
we may write
$a = u^{-1}au, tag 2$
and thus
$au^{-1} = u^{-1}a; tag 3$
also, since $a$ is nilpotent there is some $0 < n in Bbb N$ such that
$a^n = 0, tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; tag 5$
we observe that
$u + a = u(1 + u^{-1}a), tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) displaystyle sum_0^{n - 1} (-u^{-1}a)^k = sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}asum_0^{n - 1} (-u^{-1}a)^k$
$= displaystyle sum_0^{n - 1} (-1)^k(u^{-1}a)^k + sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + displaystyle sum_1^{n - 1} (-1)^k (u^{-1}a)^k + sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + displaystyle sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; tag 7$
this shows that
$(1 + u^{-1}a)^{-1} = displaystyle sum_0^{n - 1} (-u^{-1}a)^k, tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.
edited Dec 29 '18 at 0:34
answered Dec 29 '18 at 0:20
Robert LewisRobert Lewis
48.3k23167
48.3k23167
add a comment |
add a comment |
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$begingroup$
Dear Shannon, Try the case $u = 1$ first. Regards,
$endgroup$
– Matt E
Mar 14 '12 at 2:23
1
$begingroup$
See also here.
$endgroup$
– Bill Dubuque
May 3 '12 at 14:43
$begingroup$
See also here for the commutative case.
$endgroup$
– Bill Dubuque
Aug 6 '17 at 15:21