---- > [!definition] Definition. ([[nilpotent element of a ring]]) > An element $a$ of a [[ring]] $R$ is called **nilpotent** if $a^{n}=0$ for some $n$. > > If $R$ has *no* nonzero nilpotents, we say it is **reduced**. ^definition > [!basicproperties] > 1. If $a$ and $b$ are nilpotent in $R$ with $ab=ba$, then so is $a+b$. ^properties > [!basicexample] > $[m]$ is nilpotent in $\mathbb{Z} / n\mathbb{Z}$ if and only if $m$ is [[divides|divisible]] by all [[prime number|prime]] factors of $n$. > > $\to.$ Suppose $[m]$ is nilpotent; fix $k \in \mathbb{N}$ such that $m^{k} \text{ mod }n = 0$, i.e., such that $n | m^{k}$. [[Fundamental Theorem of Arithmetic|Prime-factorize]] (here we use that $\mathbb{Z}/n\mathbb{Z}$ is [[commutative ring|commutative]]) as $n=p_{1}^{i_{1}}\dots p_{z}^{i_{z}}$. If $p_{1}^{i_{1}}\dots p_{z}^{i_{z}} | m^{k}$ then each $p_{j}$ [[divides]] $m^{k}$. [[Euclid's Lemma]] then says in particular that each $p_{j}$ [[divides]] $m$. > > $\leftarrow.$ Conversely, suppose $m$ is [[divides|divisible]] by all prime factors of $n$. We claim $n | m^{\max(i_{1,\dots,i_{z}})}=m^{k}$, $k=\max(i_{1},\dots,i_{z})$. This is because we know that $m$ [[Fundamental Theorem of Arithmetic|factorizes]] like $m=p_{1}^{j_{1}} \dots p_{z}^{j_{z}} \cdot q_{1}^{\ell_{1}}\dots q_{y}^{\ell_{y}}$ for suitable exponents, where the $q$ are comprised of the prime factors of $m$ not among the $p$. From this we can write $\begin{align} > m^{\max(i_{1},\dots,i_{z})}=&m^{k} \\ > = &(p_{1}^{k})^{j_{1}} \dots (p_{z}^{k})^{j_{z}} \cdot q_{1}^{k\ell_{1}}\dots q_{y}^{k\ell_{y}} \\ > = & (p_{1}^{i_{1}})p_{1}^{k-i_{1}}p_{1}^{k(j_{1}-1)} \dots p_{z}^{i_{z}}p_{z}^{k-i_{z}}p_{z}^{k(j_{z}-1)} \cdot q_{1}^{k\ell_{1}} \dots q_{y}^{k\ell_{y}} \\ > = & \underbrace{p_{1}^{i_{1}}\dots p_{z}^{i_{z}}}_{=n} p_{1}^{k-i_{1}}p_{1}^{k(j_{1}-1)} \dots p_{z}^{k-i_{z}}p_{z}^{k(j_{z}-1)} \cdot q_{1}^{k\ell_{1}} \dots q_{y}^{k\ell_{y}} . > \end{align}$ > From this it follows that $n$ divides $m^{k}$, thus $m^{k} \text{ mod }n=0$, thereby demonstrating $[m]$ is nilpotent. > [!proof] Proof of Basic Properties. > Let $a,b \in R$ be nilpotent; write $a^{m}=0$ and $b^{n}=0$ for some integers $m,n \geq 1$. Then $(a+b)^{2nm}=\sum_{k=0}^{2mn} {n \choose k} a^{2mn-k}b^{k}$ (note that the derivation of such a [[binomial expansion]] relies on the fact that $ab=ba$ in order to rearrange powers). The terms in this sum when $k < mn$ vanish, since $2mn-k>mn \geq m$ and $a^{m}=0$ (hence $a^{M}=0$ for all $M \geq m$). The terms where $k \geq mn$ vanish also, for $mn \geq n$ and $b^{n}=0$. Hence $(a+b)^{2mn}=0$ (probably there is a smaller number that suffices, but $2mn$ seems to work cleanly... in particular $m+n$ loos like it would work). > The hypothesis $ab=ba$ is generally necessary. For example, let $a=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \text{ and } b=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}.$ Then $a^{2}=\boldsymbol 0_{2 \times 2}=b^{2}$, but $(a+b)=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ is involutive. ^proof ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```