📂Abstract Algebra

Radical and Nilradical in Abstract Algebra

Definition¹

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

1. $\text{rad} N := \left\{ a \in R \ | \ a^n \in N \right\}$ is called the radical of $N$.
2. If there exists $n \in \mathbb{N}$ satisfying $a^{n} = 0$, then $a$ is said to be 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

As a very simple example, in the ring of integers $\mathbb{Z}$, the set of multiples of $4$, namely $4 \mathbb{Z}$, is an ideal of $\mathbb{Z}$, and the set of even numbers $2 \mathbb{Z}$ is the radical of $4 \mathbb{Z}$.

The radical of $N$ is denoted by $\sqrt{N}$, and the nilradical of $R$ by $\sqrt{0}$. Considering that raising an element of $\sqrt{N}$ to some power produces an element of $N$, this can be seen as a notation that makes plenty of sense.

The following two theorems are useful in that, when an ideal is needed, one can concretely take $\sqrt{N}$ and $\sqrt{0}$. The radical and the nilradical satisfy rather strong conditions, which also makes them easy to handle.

Theorem

For a commutative ring $R$ with unity, let $N$ be an ideal of $R$.

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

Proof

[1]

Since $N$ is an ideal, for every $r \in R$ and $a \in N$, $$ ra \in N $$ and since $R$ is a ring and therefore closed under multiplication, $r^{n} \in R$. Here once more, since $N$ is an ideal, $r^{n} N \subset N$ must hold for $r^{n} \in R$ as well, and therefore, for $a^{n} \in N$, $$ r^{n} a^{n} \in N $$ must hold too. By assumption $R$ is a commutative ring, so $r^{n} a^{n} = (ra)^{n}$, and $$ (ra)^{n} \in N \implies ra \in \sqrt{N} \implies a \in r \sqrt{N} $$ Therefore, $$ r \sqrt{N} = \sqrt{N} r \subseteq \sqrt{N} $$ Now it remains to show that $( \sqrt{N} , + )$ is a subgroup of $R$; let us only check that the identity and inverses exist.

• (ii): Since $0^{n} \in N$, $0$ exists as the identity of $\sqrt{n}$.
• (iii): For every $a$, 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 $r \in R$ and $a \in \sqrt{0}$, $$ (ra)^{n} = r^{n} a^{n} = 0 $$ and since $ra \in r \sqrt{0}$, $$ r \sqrt{0} = \sqrt{0} r \subset \sqrt{0} $$ Now it remains to show that $( \sqrt{0} , + )$ is a subgroup of $R$; let us check that the identity and inverses exist.

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

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