Assume that the ring $R$ has a commutative multiplication. Prove that the set of nilpotent elements $N$ forms a subring of $R$

Question

Assume that the ring R has a commutative multiplication. Prove that the set of nilpotent elements N form a subring of R.

So I know that an element $x$ in a ring R is called nilpotent if there exists a positive integer n such that $x^n$ = $0$.

1) I would imagine identity element $0$ is contained in N because the nilpoint elements equal 0.

2) The inverse of $x^n$ would still equal $0$ so inverses are contained

3) Closure for addition: two elements $x^n$ + $x^n$ is still 0 so closure in addition is achieved.

4) Closure for multiplication: two elements $x^n$ * $x^n$ is still 0 so closure in multiplication is achieved.

Is my logic for this proof even close to being right? I feel like I am looking at this wrong and it's much too simplistic. I don't even use that commutative multiplication.

Thank you for any and all help.

Answer 1

You need to be more careful definitely.

For example "the inverse of $x^n$ is equal to $0$" makes no sense. In fact you don't need to work with $x^n$, but with $x$ instead. Anyway you need to prove that $-x$ is nilpotent, which is obviously true, as $(-x)^n = (-1)^nx^n = 0$

Now to let $y$ and $x$ be nilpotent s.t. $x^n = y^m = 0$. Then by the Binomial Formula we have that $(x+y)^{m+n} = 0$, so the sum is nilpotent too.

Similarly you can see that $(xy)^{n} = x^{n}y^{n} = 0$.

Hence the proof.

Answer 2

Is my logic for this proof even close to being right? I feel like I am looking at this wrong and it's much too simplistic. I don't even use that commutative multiplication.

No, the logic isn't even close to being right. It's not a matter of being too simple, it's just that you haven't even interpreted the definitions correctly, nor attempted to verify the salient properties.

1. Here is the aforementioned problem with interpreting the definition: nilpotent elements aren't "equal to zero," but rather they have a power equal to zero. You should have no problem concluding $0$ has a power that is $0$.

2. The point is to show that $-x$ has a power that is zero if $x$ has a power that is zero.

3. The point is to show that if $x$ and $y$ are nilpotent, then $x+y$ is nilpotent.

4. The point is to show that if $x$ and $y$ are nilpotent, then $xy$ is nilpotent.

Answer 3

To verify the set of nilpotent elements form a subring of $ R $, we need to check:

$1$. $ N $ is a subgroup of the abelian group of $ R $;

$2$. $ N $ is closed under ring multiplication.

Seeing that $ R $ is commutative, for $ 1 $:

$ \forall n_1, n_2\in N $, there exist $ m_1, m_2 $, such that $ n_1^{m_1}=n_2^{m_2}=0 $. $ (n_1-n_2)^{m_1+m_2}=0 $;

As for $2$, $ (n_1n_2)^{m_1}=0 $. We are done.

Answer 4

Let

$N \subset R \tag 1$

be the set of nilpotent elements of the commutative ring $R$; that is,

$N = \{ s \in R \mid \exists n \in \Bbb N, \; s^n = 0 \}; \tag 2$

to show $N$ is a subring of $R$, we need to verify that (i.) $N$ forms an abelian group under the ring operation "$+$"; and (ii.) is closed under "$\cdot$", the "multiplication" operation of the ring $R$; the other ring axioms which these operations obey, such as associativity, commutativity, and distributivity, hold in $N$ "automatically" since the are stipulated to bind in $R$.

Before proceeding further, we note the fairly recent trend to reserve the word "ring" for the algebraic systems which have more traditionally been referred to as "unital rings", that is, which have a multiplicative unit $1_R$ such that

$1_R r = r = r 1_R, \; \forall r \in R; \tag 3$

in the absence of such a $1_R$, the modern trend is to refer to $R$ as a "rng". Now it is clear that $1_R \notin N$, since for $n \in \Bbb N$ we have

$1_R^n = 1_R \ne 0; \tag 4$

so at this point we are forced to recognize that $N$ can't be a unital ring, so to be in accord with the most modern parlance we should probably say, "$N$ is a sub-rng of the ring $R$"; I mention this point to clarify the fact that, though $N$ doesn't contain $1_R$, we will still allow ourselves to call it a "ring" for the present discussion, since (as will be seen) it satisfies all the other ring axioms.

We observe that since $0^1 = 0$,

$0 \in \Bbb N, \tag 5$

and if

$x \in N, \tag 6$

then choosing $m \in \Bbb N$ such that

$x^m = 0, \tag 7$

we have

$(-x)^m = (-1)^m x^m = (-1)^m \cdot 0 = 0, \tag 8$

whence

$-x \in N \tag 9$

whenever $x$ is; now suppose

$x, y \in N; \tag{10}$

then we have $m, n \in \Bbb N$ with

$x^m = y^n = 0; \tag{11}$

we compute $(x + y)^{m + n}$; since $R$ is commutative, the ordinary binomial theorem applies and so

$(x + y)^{m + n} = \displaystyle \sum_0^{m + n} \dfrac{(m +n)!}{i!(m + n - i)!} x^{m + n - i}y^i; \tag{12}$

now a term-by-term inspection of (12) reveals that when $i < n$, $m + n - i > m$, so

$x^{m + n - i} = x^m x^{n - i} = 0 \cdot x^{n - i} = 0; \tag{13}$

likewise with $i \ge n$,

$y^i = y^n y^{i - n} = 0 \cdot y^{i - n} = 0; \tag{14}$

we thus see that every term in (12) contains a factor of $0$, hence

$(x + y)^{m + n} = 0, \tag{15}$

or

$x + y \in N. \tag{16}$

We have now shown that $N$ is closed under $+$ (15), has an identity for this operation (5) and also has the additive inverse $-x$ of every $x \in N$ (9); therefore, $N$ is an abelian group with respect to the $+$ operation, as was required. Now as far as concerns multiplication, again using (11) we have

$(xy)^m = x^m y^m = 0 \cdot y^m = 0 = x^n \cdot 0 = x^n y^n = (xy)^n; \tag{17}$

thus

$xy \in \Bbb N; \tag{18}$

$N$ is thus closed under both the ring operations, so we see that it is, in fact, a subring of $R$.

Finally, another property of $N$ worth knowing: it is not only a subring of $R$, but in fact $N$ is an ideal; indeed, if $x \in N$ and $r \in R$, then

$(rx)^m = r^m x^m = r^m \cdot 0 = 0, \tag{19}$

whence

$rx \in N, \tag{20}$

which shows $N$ is in fact an ideal in $R$.