📂Abstract Algebra

Radicals and Nilradicals in Abstract Algebra

Definition ¹

Let $N$ be an ideal of $R$.

1. $\text{rad} N := \left\{ a \in R \ | \ a^n \in N \right\}$ is called the radical of $N$.
2. If there exists a $n \in \mathbb{N}$ that satisfies $a^{n} = 0$, then $a$ is called nilpotent.
3. The set of nilpotent elements $\text{nil} R := \left\{ a \in R \ | \ a^n = 0 \right\}$ is called the nilradical of $R$.

Explanation

The radical of $N$ is denoted as $\sqrt{N}$, and the nilradical of $R$ is denoted as $\sqrt{0}$. It makes sense to think of $\sqrt{N}$ in such a way that an element of $\sqrt{N}$, when raised to a certain power, produces an element of $N$.

The following two theorems are useful because they specify that $\sqrt{N}$ and $\sqrt{0}$ can be concretely defined when an ideal is needed. The radical and nilradical satisfy rather strong conditions, making them easy to handle.

Theorems

Let $R$ be a commutative ring.

• [1]: $\sqrt{N}$ is an ideal of $R$.
• [2]: $\sqrt{0}$ is an ideal of $R$.

Proof

[1]

Since $R$ is a commutative ring and $N$ is an ideal, for $r \in R$, $a \in N$ $$ra \in N \\ r^{n} \in R$$ and, for $a^{n} \in N$ $$r^{n} a^{n} = (ra)^{n} \in \sqrt{N}$$ Therefore, $$r \sqrt{N} = \sqrt{N} r \subset \sqrt{N}$$ It remains to show that $( \sqrt{N} , + )$ is a subgroup of $R$, by checking the existence of the identity and inverses.

• (ii): Since $0^{n} \in N$, $0$ exists as the identity of $\sqrt{n}$.
• (iii): For all $a$, since $(-a)^{n} = (-1)^{n} a^{n} \in N$, $-a \in \sqrt{N}$ exists as the inverse of $a$.

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[2]

Since $R$ is a commutative ring, for $r \in R$, $a \in \sqrt{0}$ $$(ra)^{n} = r^{n} a^{n} = 0$$ and, since $ra \in r \sqrt{0}$ $$r \sqrt{0} = \sqrt{0} r \subset \sqrt{0}$$ It remains to show that $( \sqrt{0} , + )$ is a subgroup of $R$, by checking the existence of the identity and inverses.

• (ii): Since $0^{1} = 0$, $0$ exists as the identity of $\sqrt{0}$.
• (iii): For all $a$, since $(-a)^{n} = (-1)^{n} a^{n} = 0$, $-a \in \sqrt{0}$ exists as the inverse of $a$.

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