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# 数学代写|抽象代数代写Abstract Algebra代考|MATH412 Nilpotent elements

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## 数学代写|抽象代数代写Abstract Algebra代考|Nilpotent elements

Example 29.13. Consider $3 \in \mathbb{Z}_{81}$. We have
$$3^1=3,3^2=9,3^3=27,3^4=\mathbf{0} .$$
After that, we have
\begin{aligned} & 3^5=3^4 \cdot 3=\mathbf{0} \cdot 3=0 \text {, } \ & 3^6=\mathbf{3}^5 \cdot 3=\mathbf{0} \cdot 3=0 \text {, } \ & 3^7=\mathbf{3}^6 \cdot 3=\mathbf{0} \cdot 3=0 \text {, } \ & \end{aligned}
Thus, we conclude that $3^n=0$ for all positive integers $n \geq 4$.

Example $29.13$ motivates the following definition.
Definition 29.14 (Nilpotent). A ring element $r$ is said to be nilpotent if $r^n=0$ for some positive integer $n$.

Example 29.15. In Example 29.13, we saw that $3 \in \mathbb{Z}{81}$ is nilpotent. In $\mathbb{Z}{81}$, we also have $0^1=0,6^4=0$, and $9^2=0$, so that 0,6 , and 9 are nilpotent in $\mathbb{Z}{81}$. In an exercise at the end of the chapter, you’ll find all other nilpotent elements in $\mathbb{Z}{81}$.
Example 29.16. In any ring, $0^1=0$. Thus, the additive identity 0 is nilpotent.
Example 29.17. In $\mathbb{Z}$, the only nilpotent element is 0 . For $a \neq 0$, we have $a^n \neq 0$ for all positive integers $n$.

Example $29.17$ illustrates the following theorem, whose proof is left for you as an exercise.

## 数学代写|抽象代数代写Abstract Algebra代考|Examples and definition

Example 30.1. We might say that the polynomial $x^2+1$ is unfactorable; i.e., it cannot be factored. Indeed, it’s true that $x^2+1$ is unfactorable in $\mathbb{R}[x]$. But if we work in $\mathbb{Z}_5[x]$, then
$$(x+2) \cdot(x+3)=x^2+5 x+6=x^2+0 x+1=x^2+1,$$
as we reduce the coefficients $5=0$ and $6=1$ in $\mathbb{Z}_5$. Thus, $x^2+1=(x+2) \cdot(x+3)$ is factorable in $\mathbb{Z}_5[x]$. The lesson here is that we need to specify the polynomial ring in which we’re working (e.g., $\mathbb{R}[x]$ or $\mathbb{Z}_5[x]$ ).

Example 30.2. We might say that $x^2+1$ is actually factorable in $\mathbb{R}[x]$, because $x^2+1=$ $3 \cdot\left(\frac{1}{3} x^2+\frac{1}{3}\right)$. But this is not a legitimate factorization. When we factor a polynomial, we must write it as a product of two “smaller” polynomials, i.e., polynomials of lower degree. However, $\frac{1}{3} x^2+\frac{1}{3}$ is not smaller than $x^2+1$, because they both have degree 2 .

These examples motivate the following definition. Note how factorable and unfactorable polynomials are analogous to composite and prime integers.

Definition $30.3$ (Factorable/unfactorable polynomials). Let $F$ be a field. Suppose $f(x) \in F[x]$ with $\operatorname{deg} f(x) \geq 1$; i.e., $f(x)$ is not a constant polynomial.

• We say that $f(x)$ is factorable in $F[x]$ when $f(x)=p(x) \cdot q(x)$ for some $p(x), q(x) \in$ $F[x]$ with $\operatorname{deg} p(x), \operatorname{deg} q(x)<\operatorname{deg} f(x)$.
• Otherwise, we say that $f(x)$ is unfactorable in $F[x]$.

## 数学代写|抽象代数代写Abstract Algebra代考|Nilpotent elements

$$3^1=3,3^2=9,3^3=27,3^4=\mathbf{0} .$$

$$3^5=3^4 \cdot 3=\mathbf{0} \cdot 3=0, \quad 3^6=\mathbf{3}^5 \cdot 3=\mathbf{0} \cdot 3=0,3^7=\mathbf{3}^6 \cdot 3=\mathbf{0} \cdot 3=0,$$

## 数学代写|抽象代数代写Abstract Algebra代考|Examples and definition

$$(x+2) \cdot(x+3)=x^2+5 x+6=x^2+0 x+1=x^2+1,$$

• 我们说 $f(x)$ 可以考虑 $F[x]$ 什么时候 $f(x)=p(x) \cdot q(x)$ 对于一些 $p(x), q(x) \in F[x]$ 和 $\operatorname{deg} p(x), \operatorname{deg} q(x)<\operatorname{deg} f(x)$.
• 否则，我们说 $f(x)$ 在 $F[x]$.

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