Question
Let $R$ be a commutative ring and let $A$ be any ideal of $R$. Show that the nil radical of $A, N(A)=\left\{r \in R \mid r^{n} \in A\right.$ for some positive integer $n(n$ depends on $r)\}$, is an ideal of $R$. $[N(\langle 0\rangle)$ is called the nil radical of $R .]$
Step 1
First, we need to show that $N(A)$ is non-empty. Since $A$ is an ideal of $R$, it must contain the zero element of $R$. Thus, $0^1 \in A$, and $0 \in N(A)$. So, $N(A)$ is non-empty. Show more…
Show all steps
Your feedback will help us improve your experience
Amy Jiang and 82 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Assume that $a$ is a positive integer that is not a perfect square. State whether the expression represents a rational number or an irrational number. $$\sqrt{\sqrt{81 a^{8}}}$$
Radical Expressions
Introduction to Radical Expressions
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD