Why Are Nilpotent Elements Significant in Ring Theory?

[mathmari]

Hey!

I am looking at the following exercise:

An element $r\in R$ is called nilpotent if $r^n=0$ for some integer $n=1,2, \dots $.

1. Show that if $r$ is nilpotent then $1-r$ is invertible in $R$.
2. Show that if $R$ is commutative then the set $N(R)$ of nilpotent elements is an ideal of $R$. Give an example where that is not true when the ring is not commutative.
3. Show that $N(\mathbb{Z}_m)=0$ if and only if $m$ is not divided by a square of any prime.

I have done the following:

1. Since $r\in R$ is nilpotent we have that $r^n=0$ for some $n=1, 2, \dots $.
   Then $$1=1-r^n=(1-r)(r^{n-1}+\dots +1)$$
   So, $1-r$ is invertible in $R$.
   
   $$$$
   
2. To show that $N(R)$ is an ideal of $R$ we have to show that $N(R)$ is a left and a right ideal, right? (Wondering)
   So, we have to show that $ra\in N(R)$ and $ar\in N(R)$, for $r\in R$ and $a\in N(R)$.
   Since $R$ is commutative, we have that $(ra)^n=r^na^n$. Since $a\in N(R)$ we have that $a^n=0$. Therefore, $(ra)^n=0$.
   Since $R$ is commutative, we have that $(ar)^n=a^nr^n$. Since $a\in N(R)$ we have that $a^n=0$. Therefore, $(ar)^n=0$.
   So, $N(R)$ is an ideal of $R$.
   
   Is this correct? (Wondering)
   
   How could we find an example where that is not true when the ring is not commutative? (Wondering)
   
   $$$$
   
3. Suppose that $N(\mathbb{Z}_m)=0$, then $r^n\neq 0$, for all $n\in \mathbb{N}$, where $r\in \mathbb{Z}_m$, right? (Wondering)
   How could we continue? (Wondering)

 

[Deveno]

Your proof for 2 is missing a crucial ingredient: you have to show that if $a$ is nilpotent that $a+b$ is also nilpotent (its obvious that $-a$ is nilpotent, since if:

$a^n = 0$, then $(-a)^n = (-1)^na^n = (-1)^n0 = 0$).

The trick to this is to use the binomial theorem, which holds when $R$ is commutative.

For a non-commutative example, look at $\text{Mat}_2(\Bbb F_2)$. This has but 16 elements, and you can straight-away eliminate the 6 invertible elements (why?). Show there exists two nilpotent matrices with a non-nilpotent sum, so they cannot form an ideal.

The key to number 3 lies with $m$, not $n$. You want to show:

a) If $m$ is square-free, then $r^n = 0 \implies r = 0$ (mod $m$). Try factoring $r$ into primes.

b) If $p^2|m$ for a prime $p$, what can you say about $r = \dfrac{m}{p}$?

 

[mathmari]

Your proof for 2 is missing a crucial ingredient: you have to show that if $a$ is nilpotent that $a+b$ is also nilpotent (its obvious that $-a$ is nilpotent, since if:

$a^n = 0$, then $(-a)^n = (-1)^na^n = (-1)^n0 = 0$).

The trick to this is to use the binomial theorem, which holds when $R$ is commutative.

Why do we have to show also that $a+b$ and $-a$ are nilpotent? (Wondering)
I got stuck right now...

For a non-commutative example, look at $\text{Mat}_2(\Bbb F_2)$. This has but 16 elements, and you can straight-away eliminate the 6 invertible elements (why?).

I haven't really understood that... (Wondering)

The key to number 3 lies with $m$, not $n$. You want to show:

a) If $m$ is square-free, then $r^n = 0 \implies r = 0$ (mod $m$). Try factoring $r$ into primes.

b) If $p^2|m$ for a prime $p$, what can you say about $r = \dfrac{m}{p}$?

a) Suppose that $m$ is square-free. Why does it follow then from $r^n=0$ that $r=0\pmod m$ ? (Wondering)

b) Suppose that $p^2\mid m$ for a prime $p$. Why do we take $=\frac{m}{p}$ ? (Wondering)

 

[Deveno]

Why do we have to show also that $a+b$ and $-a$ are nilpotent? (Wondering)
I got stuck right now...

Because ideals of a ring must *also* be additive subgroups...

I haven't really understood that... (Wondering)

If a (square) matrix $A$ over a field is invertible then $\det(A) \neq 0$. But if a matrix is nilpotent, then:

$A^n = 0 \implies \det(A^n) = 0 \implies (\det(A))^n = 0 \implies \det(A) = 0$.

a) Suppose that $m$ is square-free. Why does it follow then from $r^n=0$ that $r=0\pmod m$ ? (Wondering)

Factor $r$ into primes, say: $r = p_1^{k_1}\cdots p_t^{k_t}$.

What can you say if $r^n = p_1^{nk_1}\cdots p_t^{nk_t} = am$?

b) Suppose that $p^2\mid m$ for a prime $p$. Why do we take $=\frac{m}{p}$ ? (Wondering)

That's a good question-why would I say that?

 

[mathmari]

Your proof for 2 is missing a crucial ingredient: you have to show that if $a$ is nilpotent that $a+b$ is also nilpotent (its obvious that $-a$ is nilpotent, since if:

$a^n = 0$, then $(-a)^n = (-1)^na^n = (-1)^n0 = 0$).

The trick to this is to use the binomial theorem, which holds when $R$ is commutative.

Because ideals of a ring must *also* be additive subgroups...

Ah ok...

Suppose that $a,b\in N(R)$. So, there are $x,y\in \mathbb{N}$ such that $a^x=b^y=0$.

Then for $n=x+y$ we have the following:
$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$$
If $k\geq x$ then $a^k=0$ and if $k<x$ then $n-k=x+y-k>x+y-x=y \Rightarrow n-k>y$ and so $b^{n-k}=0$.
Therefore, every term of $(a+b)^n$ is equal to $0$, right? (Wondering)

So, $a+b$ is nilpotent.

If a (square) matrix $A$ over a field is invertible then $\det(A) \neq 0$. But if a matrix is nilpotent, then:

$A^n = 0 \implies \det(A^n) = 0 \implies (\det(A))^n = 0 \implies \det(A) = 0$.

For a non-commutative example, look at $\text{Mat}_2(\Bbb F_2)$. This has but 16 elements, and you can straight-away eliminate the 6 invertible elements (why?). Show there exists two nilpotent matrices with a non-nilpotent sum, so they cannot form an ideal.

So, can we take for example the following two matrices? $$A=\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix} \ \ , \ \ \ \ \ B=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}$$

We have that $$A^2=\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix} \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix} =\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}\ \ , \ \ \ \ \ B^2=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}$$

The sum of these two matrices is the matrix $A+B=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$. Then $(A+B)^2=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} \begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}=I$ so $(A+B)^n=I^n=I, \forall n\in \mathbb{N}_{\geq 2}$, right? (Wondering)

Therefore, $A$ and $B$ are nilpotent but the matrix $A+B$ is not nilpotent.

Is this correct? (Wondering)

Factor $r$ into primes, say: $r = p_1^{k_1}\cdots p_t^{k_t}$.

What can you say if $r^n = p_1^{nk_1}\cdots p_t^{nk_t} = am$?

In that case we have $r^n\equiv 0\pmod m$.

b) If $p^2|m$ for a prime $p$, what can you say about $r = \dfrac{m}{p}$?

(Thinking)

Then we have that $m=rp$.

So, $rp=m\equiv 0\pmod m$.

Then $rp\in N(\mathbb{Z}_m)$, right? (Wondering)