Homomorphisms

Ring Homomorphisms

We have established a category of objects called rings, the natural question is how should we define suitable maps between these objects that interact well with the ring structures. We have seen one such example:

$$\begin{array}{rl}\phi :\mathbb{Z}& \to {\mathbb{Z}}_n\\ k& \mapsto [k].\end{array}$$

This map let us interchange between two ring structures when doing addition and multiplication. The interchangeability makes working with congruence classes as easy as working with integers, we can feel free to write:

$$[2]+[3]=[2+3]=[5],[2]\cdot [3]=[2\cdot 3]=[6].$$

without worrying about errors. We introduced these two rules as a compatibility of change order of operations, see the diagram.

Now if we review the definition of a ring, we can think a ring is a set equipped with two operations: addition and multiplication, and these operations satisfy some properties. When we are dealing with maps between rings, we would like these maps to respect the ring structures, and ring structure only relies on the two operations. So a reasonable definition is:

Definition

A map $\phi :R\to S$ between two rings $R$ and $S$ is called a ring homomorphism if it satisfies the following properties:

1. $\phi (a+b)=\phi (a)+\phi (b)$ for all $a,b\in R$.
2. $\phi (a\cdot b)=\phi (a)\cdot \phi (b)$ for all $a,b\in R$.

The quotient map $\phi :\mathbb{Z}\to {\mathbb{Z}}_n$ is a ring homomorphism.

Examples of Ring Homomorphisms

Identity Map

The identity map $\text{id}:R\to R$ is a ring homomorphism for any ring $R$.

Zero Map

The zero map $\phi :R\to S$ defined by $\phi (r)=0$ for all $r\in R$ is a ring homomorphism.

Inclusion Map of Subring

If $S$ is a subring of $R$, then the inclusion map $\iota :S\to R$ defined by $\iota (s)=s$ for all $s\in S$ is a ring homomorphism.

So we have list of ring homomorphisms coming from subrings:

1. The inclusion $R\times 0_S\to R\times S$ and $0_R\times S\to R\times S$.
2. The inclusion $R\to M_n(R)$ by map to constant diagonal matrices.
3. The inclusion $R\to R[x]$ by map to constant polynomials.

Projection Map from Product Ring

Given a product ring $R\times S$, the projection map ${\pi }_1:R\times S\to R$ defined by ${\pi }_1(r,s)=r$ is a ring homomorphism. Similarly, the projection map ${\pi }_2:R\times S\to S$ defined by ${\pi }_2(r,s)=s$ is a ring homomorphism.

Determinant Map on Matrix Ring (Not a Ring Homomorphism)

Given a commutative and unitary ring $R$, let $M_n(R)$ be the ring of matrices with entries in $R$. There is a natural map $\text{det}:M_n(R)\to R$ defined by taking the determinant of a matrix. This map is NOT a ring homomorphism. From linear algebra:

1. $\text{det}(A+B)=\text{det}(A)+\text{det}(B)$ for any two matrices $A,B\in M_n(R)$. 🚫
2. $\text{det}(AB)=\text{det}(A)\text{det}(B)$ for any two matrices $A,B\in M_n(R)$. ✅

For example, we can take:

$$A=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),$$

then $\text{det}(A+B)=0\ne 2=\text{det}(A)+\text{det}(B)$.

Evaluation Map on Polynomial Ring

Given a commutative and unitary ring $R$, let $R[x]$ be the ring of polynomials with coefficients in $R$. Let $a\in R$ be any element in the base ring, the evaluation map ${\text{ev}}_a:R[x]\to R$ defined by ${\text{ev}}_a(f(x))=f(a)$ is a ring homomorphism.

Quotient Map between ${\mathbb{Z}}_n$ and ${\mathbb{Z}}_m$

Now if we consider two integers $m,n\in {\mathbb{Z}}_{+}$, we have two rings ${\mathbb{Z}}_m$ and ${\mathbb{Z}}_n$. The elements in the two rings are represented by congruence classes, to distinguish, we write:

$$\begin{array}{rl}{\mathbb{Z}}_m& =\{[k{]}_m\mid k\in \mathbb{Z}\},\\ {\mathbb{Z}}_n& =\{[k{]}_n\mid k\in \mathbb{Z}\}.\end{array}$$

One may wonder if we just define:

$$\begin{array}{rl}\phi :{\mathbb{Z}}_m& \to {\mathbb{Z}}_n\\ [k{]}_m& \mapsto [k{]}_n.\end{array}$$

Do we have a ring homomorphism? The answer is no in general, but in the following special case, we have:

Proposition

Let $m,n\in {\mathbb{Z}}_{+}$ be two positive integers such that $n\mid m$. Then the map $\phi :{\mathbb{Z}}_m\to {\mathbb{Z}}_n$ defined by:

$$\begin{array}{rl}\phi :{\mathbb{Z}}_m& \to {\mathbb{Z}}_n\\ [k{]}_m& \mapsto [k{]}_n\end{array}$$

is well-defined and a ring homomorphism.

Proof of Proposition

We need to show $\phi$ is well-defined and a ring homomorphism.

Well-defined: Suppose $[k{]}_m=[k^{\prime }{]}_m$, then $m\mid (k-k^{\prime })$. Since $n\mid m$, we have $n\mid (k-k^{\prime })$, so $[k{]}_n=[k^{\prime }{]}_n$.

Ring Homomorphism: We need to show:

1. $\phi ([k{]}_m+[k^{\prime }{]}_m)=\phi ([k{]}_m)+\phi ([k^{\prime }{]}_m)$.
2. $\phi ([k{]}_m\cdot [k^{\prime }{]}_m)=\phi ([k{]}_m)\cdot \phi ([k^{\prime }{]}_m)$.

All of them are straightforward. $■$

Warning

As one can see, the ring homomorphism property almost does not need proof in above case. So one may have the illusion that we can drop the condition $n\mid m$ and still have a ring homomorphism. But this is not true, without the condition $n\mid m$ , the map $\phi :{\mathbb{Z}}_m\to {\mathbb{Z}}_n$ defined by $\phi ([k{]}_m)=[k{]}_n$ is NOT even a map between sets.

Properties of Ring Homomorphisms

The definition of ring homomorphisms is very minimalistic if compare with the definition of rings. But remember when we talking about ring homomorphisms, we are talking about maps between rings, so in fact we not only have two rules from definition of ring homomorphisms, we still have the ring properties on both domain and codomain. So we are able to derive some properties of ring homomorphisms. Let $\phi :R\to S$ be a ring homomorphism, we have:

1. $\phi (0_R)=0_S$.
2. $\phi (-a)=-\phi (a)$ for all $a\in R$.
3. The image of $\phi$ is a subring of $S$.
4. If $R$ has an identity $1_R$ and $\phi$ is surjective, then $S$ has an identity $1_S$ and $\phi (1_R)=1_S$.
5. If $R$ has an identity $1_R$, $\phi$ is surjective, the for any unit $u\in R$, $\phi (u)$ is a unit in $S$ and $\phi (u^{-1})=\phi (u{)}^{-1}$.
6. If $R$ is commutative, then $\phi (R)$ is commutative.

Kernel and Image of Ring Homomorphisms

In every ring $S$, we always have an distinguished element $0_S$ which is the additive identity. By the first property of ring morphisms $\phi :R\to S$, we have $\phi (0_R)=0_S$, namely ${\phi }^{-1}(0_S)$ is never empty. We can define:

Definition

The kernel of a ring homomorphism $\phi :R\to S$ is the set:

$$\text{ker}(\phi )=\{a\in R\mid \phi (a)=0_S\}.$$

The kernel of a morphism is more than just a set, it is a subring of $R$.

Proposition

The kernel of a ring homomorphism $\phi :R\to S$ is a subring of $R$.

Proof of Proposition

By the definition of a subring, we need to show:

1. Closure under Addition: Suppose $a,b\in \text{ker}(\phi )$, then $\phi (a+b)=\phi (a)+\phi (b)=0_S+0_S=0_S$, so $a+b\in \text{ker}(\phi )$.
2. Closure under multiplication: Suppose $a\in \text{ker}(\phi )$, then $\phi (ra)=r\phi (a)=r\cdot 0_S=0_S$, so $ra\in \text{ker}(\phi )$.
3. Additive Identity: $\phi (0_R)=0_S$, so $0_R\in \text{ker}(\phi )$.
4. Additive Inverse: Suppose $a\in \text{ker}(\phi )$, then $\phi (-a)=-\phi (a)=-0_S=0_S$, so $-a\in \text{ker}(\phi )$. $■$

Using the notion of kernel, we can characterize the injectivity of a ring homomorphism:

Proposition

A ring homomorphism $\phi :R\to S$ is injective if and only if $\text{ker}(\phi )=\{0_R\}$.

Proof of Proposition

Suppose $\phi$ is injective, then $\phi (a)=\phi (b)$ implies $a=b$ for all $a,b\in R$. In particular, $\phi (a)=0_S$ implies $a=0_R$ since $\phi (0_R)=\phi (0_R)$.

Conversely, suppose $\text{ker}(\phi )=\{0_R\}$, then $\phi (a)=\phi (b)$ implies $\phi (a-b)=0_S$, so $a-b=0_R$ and $a=b$. $■$

Now we find our morphism $\phi$ gives a subring in the domain $R$, we can also find a subring in the codomain $S$, simply by taking the image of $\phi$.

Definition

The image of a ring homomorphism $\phi :R\to S$ is the set:

$$\text{im}(\phi )=\{\phi (a)\mid a\in R\}.$$

The image of a morphism is more than just a set, it is a subring of $S$.

Proposition

The image of a ring homomorphism $\phi :R\to S$ is a subring of $S$.

The proof is similar to the proof of the kernel, we omit it here.

Isomorphisms

When we have a bijection between two sets, we sometimes think the two sets are the same up to some renaming. In the ring theory, we have a similar concept.

Definition

A ring homomorphism $\phi :R\to S$ is called an isomorphism if it is bijective as a set map. Two rings $R$ and $S$ are called isomorphic if there exists an isomorphism $\phi :R\to S$.

Iso- means “same”, so isomorphic rings should be understood as “the same rings”. We can think of isomorphic rings as two rings that are essentially the same, except that the elements are named differently. We can write $R\cong S$ to denote that $R$ and $S$ are isomorphic.

To check whether two rings are isomorphic, one can check the set-theoritc bijectivity. But we can also take advantage of the ring homomorphism properties:

Proposition

A ring homomorphism $\phi :R\to S$ is an isomorphism if and only if:

1. $\text{ker}(\phi )=\{0_R\}$, which is equivalent to $\phi$ is injective.
2. $\text{im}(\phi )=S$, which is equivalent to $\phi$ is surjective.

We have the following important example of isomorphism:

Proposition

Let $m,n\in {\mathbb{Z}}_{+}$ be two positive integers such that $\gcd (m,n)=1$. Then the map $\phi :{\mathbb{Z}}_{mn}\to {\mathbb{Z}}_m\times {\mathbb{Z}}_n$ defined by:

$$\begin{array}{rl}\phi :{\mathbb{Z}}_{mn}& \to {\mathbb{Z}}_m\times {\mathbb{Z}}_n\\ [k{]}_{mn}& \mapsto ([k{]}_m,[k{]}_n)\end{array}$$

is an isomorphism.

Proof of Proposition

We need to show $\phi$ is a bijection and a ring homomorphism. The proof of the ring homomorphism is straightforward by morphisms between zn and zm. We now use the above criterion to check the isomorphism:

1. $\text{ker}(\phi )=\{[k{]}_{mn}\mid \phi ([k{]}_{mn})=([k{]}_m,[k{]}_n)=([0{]}_m,[0{]}_n)\}$ which implies $m\mid k$ and $n\mid k$, since $\gcd (m,n)=1$, we have $mn\mid k$, so $[k{]}_{mn}=[0{]}_{mn}$. Now we know the kernel is $\{[0{]}_{mn}\}$, so $\phi$ is injective.
2. To check surjectivity, we simply count the elements in the domain and codomain, we have $mn$ elements in ${\mathbb{Z}}_{mn}$ and also $mn$ elements in ${\mathbb{Z}}_m\times {\mathbb{Z}}_n$, so an injective map between them must be surjective. $■$

Warning

The coprime condition $\gcd (m,n)=1$ is essential in the above proposition. If we take $m=n=2$, then the map $\phi :{\mathbb{Z}}_4\to {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ defined by $\phi ([k{]}_4)=([k{]}_2,[k{]}_2)$ is NOT an isomorphism. The kernel is $\{[0{]}_4,[2{]}_4\}$, so the map is not injective.