Abstract algebra/Ideals

Note: For the purposes of developing the concept of an ideal, we will only work with commutative rings. The development can be done with non-commutative rings but it is more complicated than is necessary at the moment.

Motivation

In [[../Rings, fields and modules/]] we saw that the set of even integers $2\mathbb{Z}\subset \mathbb{Z}$ was a subring of the integers.

We can also see very easily that the integers $\mathbb{Z}\subset \mathbb{Q}$ are a subring of the rational numbers under the usual operations of addition and multiplication.

The even integers, when taken as a subring of the integers have a property that the integers when taken as a subring of the rationals do not. The even integers taken as a subring of the rationals also lack this property.

The property is that the even integers, taken as a subring of the integers, absorb multiplication. Let's call the even integers $I=2\mathbb{Z}$ for ease of notation.

Consider the following: For all $i\in I$, we can see by the definition of $I$ that $i=2k$ for some $k\in \mathbb{Z}$.

For all $a\in \mathbb{Z}$ see that $ai=a(2k)=2ak\in I$.

In English, regardless of which even integer is chosen, multiplying it by any integer will give us a different even integer.

Definition of an Ideal

Definition: Given a commutative ring $R$, a subring $I\subseteq R$ is said to be an ideal of $R$ if it absorbs multiplication; that is, if $\mathrm{\forall }i\in I,\mathrm{\forall }r\in R,ir=ri\in I$.

We write $I◃R$ as shorthand for this.

To verify that a subset of a commutative ring is an ideal, it is only necessary to check that it is closed under subtraction and that it absorbs multiplication.

Definition: An ideal $I◃R$ is proper if $I⊊R$.

Lemma: An ideal $I$ is proper if and only if $1\notin I$.

Proof: If $1\in I$ then $r=r1\in I$ so $I=R$.

The converse is obvious.

Congruence Modulo an Ideal

Definition: Two elements $a,b$ are said to be congruent in an ideal $I$ if and only if $b-a\in I$. Write $a\sim b$ to mean that $a$ is congruent to $b$ modulo $I$.

Theorem: $\sim$ an equivalence relation.

Proof:

• $\sim$ is reflexive. $\mathrm{\forall }i\in I,i\sim i\iff (i-i)\in I\iff 0\in I$. This follows from the fact that $I$ is an ideal (and hence, a ring).
• $\sim$ is symmetric. $\mathrm{\forall }i,j\in I,i\sim j\iff j-i\in I$. Since ideals are rings, they are closed under additive inversion. Thus $(j-i)\in I\iff -(j-i)=(i-j)\in I\iff j\sim i$.
• $\sim$ is transitive. $\mathrm{\forall }i,j,k\in I,i\sim j$ and $j\sim k$ implies $(j-i)\in I$ and $(k-j)\in I$. Since I is an ideal, $(j-i)+(k-j)=(k-i)\in I$. Therefore, $i\sim j$ and $j\sim k$ imply $i\sim k$.

Definition: Since $\sim$ is an equivalence relation, its equivalence classes, called the cosets of $R$ modulo $I$ are distinct and partition $R$.

Given a ring $R$ and an ideal of that ring $I\subseteq R$, the coset of an element $r\in R$ is written $r+I$ and is defined as follows: $[r]=r+I:=\{r+i\mid i\in I\}$.

Lemma: Given an ideal $I\subseteq R$, a subset of a ring $R$, the congruence class $[r]$ modulo $I$ of an element $r\in R$ is $[0]$ if and only if $r\in I$. To see this, simply observe that $r\in [r]$, that $[r]=[0]$ therefore requires $r\in [0]$, and that the definition $[0]=\{0+i\mid i\in I\}$ requires $r\in I$.

Ideals as kernels of homomorphisms

One ready source of ideals is as kernels of homomorphisms. Namely if $\varphi :R\to S$ is a ring homomorphism then $\ker \varphi =\{r\mid \varphi (r)=0\}$ is an ideal of $R$. To see this note that $\varphi (i)=0⟹\varphi (ri)=\varphi (r)\varphi (i)=0$. The next construction show s that all ideals arise in this way.

Factor Rings

Notation: Given a commutative ring $R$ and an ideal $I\subseteq R$, the set of cosets of $R$ under the equivalence relation $\sim$ defined by $I$ is written $R/I$.

Definition/Theorem: This set of cosets, called the factor ring (or quotient ring) of $R$ modulo $I$ is a ring with operations $(+,\cdot )$ as defined below. Let $(a+I),(b+I)$ be two elements of $R/I$.

• $(a+I)+(b+I):=(a+b)+I$.
• $(a+I)\cdot (b+I):=(ab)+I$.

Proof: We first show that the above operations are independent of the choice of a and b (or, the operations are well-defined). Suppose there is another choice of coset representatives $a^{\prime },b^{\prime }$ such that $(a^{\prime }+I)=(a+I)$ and $(b^{\prime }+I)=(b+I)$. Then this implies $(a-a^{\prime })\in I$ and $(b-b^{\prime })\in I$. Since $I$ is closed under addition $(a+b)-(a^{\prime }+b^{\prime })\in I$ which shows that addition is well-defined (that is $(a+b+I)=(a^{\prime }+b^{\prime }+I)$). Moreover $(ab-a^{\prime }{b}^{\prime })=a(b-b^{\prime })+b^{\prime }(a-a^{\prime })\in I$, where the last inclusion follows because $I$ is a two-sided ideal. This shows that multiplication is well-defined.

To show that $R/I$ is a ring with the operations defined above, we need to show that it is a set with additive identity and that it satisfies the closure, associativity, commutativity, and distributivity axioms.

• Existence of Additive Identity: Consider the element $0_R\in R$. The element $0_R+I$ (which happens to be identical to $I$ considered as a set) is the additive identity of $R/I$.
• Closure, Associativity, and Commutativity of Addition: For all $a,b,r,s\in R$ satisfying $r-a\in I$ and $s-b\in I$, $(r-a)+(s-b)=(r+s)-(a+b)\in I$. Additive commutativity follows directly from the fact that addition in $R$ is always commutative. Additive associativity is not shown.
• Closure, Associativity, and Commutativity of Multiplication: For all $a,b,r,s\in R$ satisfying $r-a\in I$ and $s-b\in I$, $(r-a)(s-b)=(rs-as-rb)+(ab)\in I$. This follows from the fact that $I$, assumed to be a two-sided ideal, absorbs multiplication. Multiplicative commutativity follows from the assumption that $R$ is a commutative ring.
• Distributivity of Multiplication over Addition: $(a+I)((b+I)+(c+I))=(a+I)((b+c)+I)=(a(b+c))+I=(ab+ac)+I$ as expected.
• Closure under Additive Inversion: For all $r,a\in R$, $r-a\in I\iff a-r=-r-(-a)\in I$ since $I$ is a subring of $R$, a ring.

Observe that there is a canonical ring homomorphism $\pi :R\to R/I$ determined by $\pi (r)=[r]$, called the projection map. We record some trivial properties:

Lemma: Let $I$ be a two-sided ideal of a ring $R$. The map $\pi :R\to R/I$ determined by $r\to [r]$ is a surjective ring homomorphism whose kernel is exactly I.

One can also show that the ideal structure of $R/I$ is closely related to that of $R$. Namely

Proposition: The projection map $\pi :R\to R/I$ induces a bijection between the collection $\{J◃R\mid I\subseteq J\}$ and the collection $\{\overline{J}◃R/I\}$. Moreover this is order-preserving and determined by $\overline{J}\to {\pi }^{-1}(\overline{J})$.

Mapping Property

Observe that if $\tilde{\varphi }:R/I\to S$ is a ring homomorphism, then the composition $\varphi =\tilde{\varphi }\circ \pi :R\to R/I\to S$ is a ring homomorphism such that $I\subset \ker \varphi$ (because $x\in I⟹\varphi (x)=\tilde{\varphi }\circ \pi (x)=\tilde{\varphi }(0_{R/I})=0_S$). This characterizes all such morphisms in the following sense

Lemma: Let $\varphi :R\to S$ be a ring homomorphism such that $I\subset \ker \varphi$. Then there is a unique homomorphism $\tilde{\varphi }:R/I\to S$ such that $\varphi =\tilde{\varphi }\circ \pi$.

Principal Ideals

For this section assume $R$ is commutative with a one. Given an element $r\in R$ one can construct the principal ideal $(r)=\{rs\mid s\in R\}$. One can check that this is an ideal and it is the smallest ideal in $R$ containing $r$ - that is if $r\in I◃R$ then $rR\subset I$.

Example: Let $R=\mathbb{Z}$ be the ring of integers. The principal ideal $(n)$ is the the subset of $\mathbb{Z}$ consisting of positive and negative multiples of $n$. For example $(2)$ is the subset of even integers. Then one can view the factor ring $\mathbb{Z}/(n)$ simply as the set $\{0,1,\dots ,n-1\}$ under addition and multiplication modulo $n$.

Operations on Ideals

Given a collection of ideals we can other ideals. For instance it is easy to check that the intersection of any family of ideals is again an ideal. We write this simply as ${\cap }_{j\in J}{I}_j$

Given any set $S\subset R$ we can construct the smallest ideal of $R$ containing $S$ which we denote by $⟨S⟩$. It is determined by $⟨S⟩={\cap }_{S\subset I◃R}I$, though often we can be more explicit than this.

If $I_{j\in J}$ is a collection of ideals we can determine the sum, written $\sum _{j\in J}{I}_j$, as the smallest ideal containing all the ideals $I_j$. One can check explicitly that its elements are finite sums of the form $\sum _{j\in J}{x}_j$.

Finally if $I,J$ are two ideals in $R$ one can determine the ideal-theoretic product as the smallest ideal containing the set-theoretic product - $\{ij\mid i\in Ij\in J\}$. Note that the ideal-theoretic product is in general strictly larger than the set-theoretic product, and that it simply consists of finite sums of the form $\sum i_r{j}_r$ where $i_r\in I{j}_r\in J$

Example: Let $(m)$ and $(n)$ the principal ideals in $\mathbb{Z}$ just given. Then one can check explicitly that $(m)\cap (n)=(r)$, where r is the lcm of m and n. Moreover $(m)(n)=(mn)$, and $(m)+(n)=(s)$ where s is the hcf of m and n. Observe that $(m)(n)=(m)\cap (n)$ if and only if s = mn if and only if m and n are co-prime if and only if $(m)+(n)=(1)$.

Prime and Maximal Ideals

There are two important classes of ideals in a commutative ring - Prime and Maximal.

Definition: An ideal $I◃R$ is prime if it satisfies:

$xy\in I⟹x\in I\text{ or }y\in I$

Definition: An ideal $I◃R$ is maximal if it is proper (i.e. $I⊊R$ and it satisfies:

$I\subset J◃R⟹I=J\text{ or }J=R$

That is, there are no proper ideals between $I$ and $R$. One can easily check that

Lemma: A maximal ideal is also prime.

Proof: Suppose $I◃R$ is a maximal ideal, and $xy\in I$. Suppose further that $x\notin I$. Then the ideal $I+(x)$ is an ideal containing $I$ and $x$, so is strictly larger than $I$. By maximality $I+(x)=R\ni 1$. So $1=i+rx⟹y=iy+rxy\in I$.

The following Lemma is important for many results, and it makes essential use of Zorn's Lemma (or equivalently the Axiom of Choice)

Lemma: Every non-invertible element of a ring is contained in some maximal ideal

Proof: Suppose $x$ is the non-invertible element. Then the first observation is that $(x)$ is a proper ideal, for if $(x)=R$, then in particular $1\in (x)$ so $1=ax$ contradiction the assumption. Let $𝒮$ be the set of proper ideals in $R$ containing $x$ ordered by inclusion. The first observation implies that $𝒮$ is non-empty, so to apply Zorn's Lemma we need only show that every increasing set of ideals contains an upper-bound. Suppose $\{I_j{\}}_j$ is such an increasing set, then the least upper bound is $\sum _{j\in j}{I}_j$ as this is the smallest ideal containing each ideal. If one checks that the union ${\cup }_{j\in J}{I}_j$ is an ideal, then this must be $\sum _{j\in J}{I}_j$. To show it's proper, we need only show $1\notin {\cup }_{j\in J}{I}_j$ for all $j$. But this follows precisely because each $I_j$ is proper.

Therefore by Zorn's Lemma there is a maximal element $J$ of $𝒮$. It is clearly maximal for if $J^{\prime }$ were any ideal satisfying $J\subset J^{\prime }⊊R$ then $J^{\prime }$ would be an element of $𝒮$, and by maximality of $J$ we would have $J^{\prime }\subset J$ whence $J=J^{\prime }$.

Properties of rings may be naturally restated in terms of the ideal structure. For instance

Proposition: A commutative ring $R$ is an Integral Domain if and only if $(0)$ is a prime ideal.

Proof: This follows simply because $x=0⟺x\in (0)$.

This explains why an Integral Domain is also referred to as a Prime Ring. Similarly, we may give a necessary and sufficient condition for a ring to be a field :

Proposition: A commutative ring $R$ is a Field if and only if $(0)$ is a maximal ideal (that is there are no proper ideals)

Proof: We only need to show that every element $0\ne x\in R$ is invertible. Suppose not then by Lemma ... $x$ is contained in some (proper) maximal ideal, a contradiction.

Corollary: An ideal $I◃R$ is maximal if and only if $R/I$ is a field.

Proof: By the previous Proposition we know $R/I$ is a field if and only if its only proper ideal is $(0)$. By the correspondence theorem (...) this happens if and only if there are no proper ideals containing $I$.

It's also clear that

Lemma: An ideal $I◃R$ is prime if and only if $R/I$ is an integral domain.

Proof: Write $\overline{x}$ for the element of $R/I$ corresponding to the equivalence class $[x]$. Clearly every element of $R/I$ can be written in this form.

$\Rightarrow )$ $\overline{xy}=0⟺xy\in I⟺x\in I\text{ or }y\in I⟺\overline{x}=0\text{ or }\overline{y}=0$ where the second equivalence follows directly because $I$ is prime.

$\Leftarrow )$ This follows in exactly the same way.

Glossary

Please see the extensive Wikipedia:Glossary of ring theory.