The set ${\frak R}$ of all nilpotent elements in ring $A$ is an ideal, and $A / {\frak R}$ has no nilpotent elements. To see that ${\frak R}$ form an ideal, convince yourself that the addition of two nilpotents is still a nilpotent.

Definition. The ideal ${\frak R}$ of nilpotent elements of $A$ is called the nilradical. The nilradical is also the intersection of all the prime ideals of $A$.

Proof. Let ${\frak R}’$ be the intersection of all the prime ideals of $A$. If $f\in A$ is nilpotent, then for some $n$ we have $f^n=0\in {\frak p}$, where ${\frak p}$ is a prime ideal. hence ${\frak p}$ is prime, $f$ must be in ${\frak p}$ as well, i.e., a nilpotent must be an element all the prime ideals, $f\in {\frak R}’$. Conversely, we shall show that if $f$ is not nilpotent, $f$ is not in ${\frak R}’$. Suppose $f$ is not nilpotent, let $\Sigma$ be all the ideals that does not contain multiples of $f$, each element of $\Sigma$ is an ideal ${\frak a}$ such that

\[n>0 \implies f^n \notin {\frak a}.\]

Then $\Sigma$ is no empty since $0\in \Sigma$, and Zorn’s lemma says that there is a maximal element, order by inclusion. Let ${\frak p}$ be a maximal element of $\Sigma$, we shall show that ${\frak p}$ is indeed prime, as the notation suggests. Let $x,y \notin {\frak p}$, then the ideals ${\frak p}+(x),{\frak p}+(y)$ strictly contain ${\frak p}$ and therefore do not belong to $\Sigma$. Hence, by construction, multiples of $f$ is an element of these larger ideals,

\[f^m\in {\frak p}+(x),\quad f^n\in {\frak p}+(y)\]

for some $m,n$. It follows that $f^{m+n}\in {\frak p}+(xy)$, which means ${\frak p}+(xy)$ is not in $\Sigma$, thus $xy\notin {\frak p}$. To sum up,

\[x,y\notin {\frak p} \implies xy\notin {\frak p},\]

it follows that ${\frak p}$ is prime. Q.E.D.

Definition. The Jacobson radical ${\frak R}$ of $A$ is defined to be the intersection of all the maximal ideals.

Proposition. Leg ${\frak R}$ be Jacobson radical of ring $R$.

\[x\in {\frak R} \Leftrightarrow 1-xy \text{ is a unit } \forall y \in R\]

The proof can be found in Atiyah’s textbook.

Next we discuss operations allowed on ideals. Ideals are sets, naively we would think that all the operations defined for ideals are naturally defined for ideals as well, however, the question is whether the result is still an ideal?

Sum

Let ${\frak a}{\frak b}$ be two ideals of ring $R$. The sum ${\frak a}+{\frak b}$ is defined element-wise, namely

\[{\frak a}+{\frak b} := \{ x+y \mid x\in {\frak a}, \,y\in {\frak b} \}.\]

Note ${\frak a}+{\frak b}$ is different from ${\frak a} \cup {\frak b}$. The sum of ${\frak a},{\frak b}$ is the smallest ideal containing ${\frak a}$ and ${\frak b}$. This definition can be generalized to the sum of arbitrary numbers of ideals.

Intersection The intersection of any family of ideals are also ideals. Thus the ideals of $A$ form a complete lattice with respect to inclusion. Recall that a partially ordered set is called a complete lattice if every subset has a supremum and infimum.

Product

The product of two ideals ${\frak a},{\frak b}$ is defined similarly to the sum,

\[{\frak a}{\frak b} := \{ xy \mid x\in {\frak a}, \,y\in {\frak b} \}.\]

This definition can be generalized to finite products of ideals. Especially we can define the power of an ideal, ${\frak a}^n$, conventionally ${\frak a}^0=1$.

Examples are important to understand these operations. If $A = \mathbb{Z}$, $m,n\in \mathbb{Z}$, then $(m)+(n)$ is the ideal generated by the highest common factor (h.c.f. for short) of $m$ and $n$; $(m) \cap(n)$ is the ideal generated by their lowest common multiple (l.c.m.); and $(a)(b)=(ab)$.

The three operations defined so far (sum, intersection and product) are all commutative and associative. Also there is the distributive law,

\[{\frak a}({\frak b}+{\frak c}) = {\frak a}{\frak b}+{\frak a}{\frak c}.\]

In the integer ring $\mathbb{Z}$, intersection $\cap$ and addition $+$ are distributive over each other. This is not the case in general, the best we have in this direction is the modular law,

\[{\frak a} \cap ({\frak b}+{\frak c}) = {\frak a}\cap {\frak b} + {\frak a}\cap {\frak c} \quad\text{if}\quad {\frak b}\subseteq {\frak a}\text{ or }{\frak c}\subseteq {\frak a},\]

since there might be elements in ${\frak b}+{\frak c}$ that is neither in ${\frak b}$ and ${\frak c}$.

Again, in $\mathbb{Z}$ we have

\[({\frak a} + {\frak b})({\frak a} \cap {\frak b}) = ({\frak a}{\frak b}),\]

for example $(4)+(6)=(2)$, $(4)\cap(6)=(12)$, thus $((4)+(6))((4)\cap(6))=(24)$, which equals $(4)(6)$. However, in general we have only

\[({\frak a} + {\frak b})({\frak a}\cap {\frak b}) \subseteq ({\frak a}{\frak b}),\]

On the other hand, clearly $({\frak ab})\subseteq({\frak a})\cap({\frak b})$, thus if $({\frak a}+{\frak b})=(1)$ then we have both $({\frak a}\cap {\frak b}) \subseteq ({\frak ab})$ and $({\frak ab})\subseteq({\frak a})\cap({\frak b})$, hence $({\frak ab}) = ({\frak a})\cap({\frak b})$.

Definition. Two ideals ${\frak a},{\frak b}$ are said to be coprime if ${\frak a} + {\frak b}=(1)$.

If ${\frak a}, {\frak b}$ are coprime then $({\frak ab}) = ({\frak a}) \cap ({\frak b})$.

Definition. The direct product of rings $A_{1},\dots,A_{n}$ is denoted by

\[A = \prod_{i=1}^n A_{i}\]

which is comprised of element $x = (x_{1},\dots,x_{n})$ where $x_{i} \in A_{i}$. Addition and multiplication is defined componentwise. A is a commutative ring with identity $(1,\dots,1)$. We have projections $p_{i}:A\to A_{i}$ which projects $x$ to its $i$-th component.

Let $A$ be a ring (any ring, not necessarily a product ring) and ${\frak a}_ {i}$ be ideals of $A$, we can first make a list of rings ${A} / {\frak a}_ i$ then multiply them together to form a product ring $\prod_{i} A / {\frak a}_ {i}$. There is a ring homomorphism from $A$ to this product ring, denoted by $\phi$,

\[\begin{align} \phi : &A \to \prod_{i=1}^n A / {\frak a}_ {i}, \\ & x \mapsto (x+{\frak a}_{1},\dots,x+{\frak a}_{n}), \end{align}\]

which maps each element of $A$ to a n-tuple.

Let’s take a look at the kernel of $\phi$, which is comprised of the elements that are send to zero in the product ring. Take a look at the first component, for $x+{\frak a} _ {1}$ to be zero in $A / {\frak a} _ {1}$, $x$ must be an element of ${\frak a}_ {1}$. Then,

\[\text{ker}(\phi) = \{ x\in A \mid x\in {\frak a}_ {i},\,i=1,\dots,n \} \implies \text{ker}(\phi)={\frak a}_ {1}\cap\dots \cap {\frak a}_ {n}.\]

If $\phi$ is injective, meaning the kernel has only one element, then $\cap_ {i}{\frak a}_ {i}=0$.

Definition. Let ${\frak a},{\frak b}$ be ideals of a ring $A$. Their ideal quotient is

\[({\frak a}: {\frak b})=\{ x\in A \mid x{\frak b}\subset {\frak a} \}\]

which is an ideal, as can be easily verified, and justifies the usage of parenthesis.

$({\frak a}: {\frak b})$ is useful when ${\frak a}<{\frak b}$, where the ordering is given by inclusion. Particularly, $(0: {\frak b})$ is called the annihilator of ${\frak b}$ and is sometimes denoted by $\text{Ann}({\frak b})$: it is the set of all $x\in A$ such that $x{\frak b}=0$. In this notation the zero-divisors of $A$ are denoted by

\[D = \bigcup_{x\neq 0} \text{Ann}(x).\]

To simplify the notations, if $(x)$ is a principal ideal, instead of $({\frak a}:(x))$ we simply write $({\frak a}:x)$.

Take the ring of integers for example, we have

\[(8:2)=\{ x\in \mathbb{Z} \mid 2x\in 8\mathbb{Z} \} =\left\{ 4\mathbb{Z} \right\} =(4).\]

Various relation concerning ideal quotient can be found in textbooks, we will not list them here.

Definition. Let ${\frak a}$ be any ideal of $A$, the radical of ${\frak a}$ is

\[\text{rad}({\frak a}):=\{ x\in A \mid x^n\in{\frak a} \text{ for some }n\in\mathbb{Z}^+ \}.\]

Let $\phi:A \to A / {\frak a}$ be the standard ring homomorphism, what would $\phi( \text{rad}(\frak a))$ be like? Since some power of $x$ is in ${\frak a}$, in the quotient ring $A / {\frak a}$, some power of $x$ is zero, thus

\[\phi(\text{rad} ({\frak a})) = \text{nilradical of } A / {\frak a} = {\frak R}(A) \implies \text{rad}({\frak a}) = \phi^{-1}({\frak R}_{A / {\frak a}}),\]

which also shows that $\text{rad}({\frak a})$ is an ideal.

The radical of an ideal has the following properties:

  • ${\frak a}\subseteq\text{rac}({\frak a})$
  • $\text{rad}(\text{rad}{(\frak a)})=\text{rad}({\frak a})$
  • $\text{rad}({\frak ab})=\text{rad}({\frak a})\cap \text{rad}({\frak b})$
  • if ${\frak p}$ is prime, $\text{rad}({\frak p}^n)=\text{rad}({\frak p})$ for all $n>0$.

Proposition. The radical of an ideal ${\frak a}$ is the intersection of all the prime ideals containing ${\frak a}$.

More generally, we can define the radical of any subset $E$ of ring $A$ in the same way. It is not necessarily an ideal.