Why Are Nilpotent Elements Significant in Ring Theory?

In summary, we discussed the definitions of a nilpotent element and a commutative ring. We then proved that if $r$ is nilpotent, then $1-r$ is invertible in $R$. Furthermore, we showed that if $R$ is commutative, then the set $N(R)$ of nilpotent elements is an ideal of $R$. We also discussed an example of a non-commutative ring where this is not true. Finally, we explored the properties of the set $N(\mathbb{Z}_m)$ and showed that $N(\mathbb{Z}_m)=0$ if and only if $m$ is not divided by a square of any prime.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

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)
 
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  • #2
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 $r = \dfrac{m}{p}$?
 
  • #3
Deveno said:
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...

Deveno said:
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)

Deveno said:
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)
 
  • #4
mathmari said:
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?
 
  • #5
Deveno said:
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.
Deveno said:
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.
Deveno said:
If a (square) matrix $A$ over a field is invertible then $\det(A) \neq 0$. But if a matrix is nilpotent, then:

$A^n = 0 \implies \det(A^n) = 0 \implies (\det(A))^n = 0 \implies \det(A) = 0$.
Deveno said:
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)
Deveno said:
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 $r^n\equiv 0\pmod m$.
Deveno said:
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)
 
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Related to Why Are Nilpotent Elements Significant in Ring Theory?

1. What is a set of nilpotent elements?

A set of nilpotent elements is a subset of a mathematical structure (such as a group or ring) that consists of elements whose powers eventually become equal to the identity element. In other words, for every element in the set, there exists a positive integer n such that the nth power of the element is equal to the identity element.

2. What is the significance of nilpotent elements?

Nilpotent elements are significant because they provide a way to classify elements in a mathematical structure based on their behavior. They are also useful in studying and understanding the properties of the structure as a whole.

3. How are nilpotent elements different from zero divisors?

Nilpotent elements and zero divisors are two different concepts. While nilpotent elements have powers that eventually become equal to the identity element, zero divisors are elements that, when multiplied together, result in the zero element. In other words, zero divisors produce a product of zero, while nilpotent elements do not necessarily have this property.

4. Can a set of nilpotent elements be empty?

Yes, a set of nilpotent elements can be empty. In fact, for some structures such as fields, the set of nilpotent elements is always empty because the only element that has a power equal to the identity element is the identity element itself.

5. How do nilpotent elements relate to other types of elements in a mathematical structure?

Nilpotent elements can coexist with other types of elements in a mathematical structure. For example, in a ring, a nilpotent element can be both a zero divisor and a unit. However, in some structures, such as fields, there can be no nilpotent elements, as every element has a multiplicative inverse and powers of any element cannot be equal to the identity element unless the element itself is the identity element.

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