## Ring homomorphisms

Maps between commutative rings are just as important as these between abelian groups. As you should anticipate, our maps should preserve ring operations.

A map $\phi: R\to S$ between commutative rings $R$ and $S$ is called a homomorphism if $\phi: (R,+)\to (S,+)$ is a group homomorphism and $\phi(a\cdot b)=\phi(a)\cdot\phi(b)$ for any $a,b\in R.$

Here, as usual, we use $+$ and $\cdot$ to denote addition and multiplications in respective rings. You should notice that the second property is essentially the same as for the addition in the group homomorphism definition. Analogously, we define the kernel and the image of a homomorphism: $Ker(\phi)=\{a\in R\mid \phi(a)=0\},$ $Im(\phi)=\{s\in S\mid \exists r\in R: \phi(r)=s\}.$ As well, a ring isomorphism is a homomorphism that is a bijection. Again, we would be interested in the properties of rings that are not changing under an isomorphism.
The following is why the ideals of $R$ are so important.

The kernel of a homomorphism $\phi: R\to S$ is an ideal in $R.$

To see this, recall first that the kernel $I=Ker(\phi)$ of the group homomorphism $\phi: (R,+)\to (S,+)$ is a subgroup of $(R,+).$ Let $x\in I.$ We have to show that $r\cdot x\in I$ for any $r\in R.$ By the homomorphism property, $\phi(r\cdot x)=\phi(r)\cdot\phi(x)=\phi(r)\cdot 0=0.$

Given an ideal $I\subset R,$ we can define a multiplication on the abelian group $A=R/I,$ by setting $(x+I)(y+I)=xy+I.$ To see that this is well-defined, we take $s,t\in I$ and compute $(x+s+I)(y+t+I)=xy+xt+ys+st+I=xy+I,$ i.e. the product does not depend upon a particular choice of coset representatives of $I$ in $R.$ We should also check the distributivity, certainly.
Unlike for quotient groups, the equation $(x+I)(y+I)=xy+I$ does not mean an equality of sets. Indeed, it can already happen that $I\cdot I=\{0\},$ while $I\neq\{0\}$ (Such an example can already be obtained by taking $I=\{0,2\}\subset \mathbb{Z}_4.$)

It is a good point to discuss this phenomenon a bit more.

Nilpotent elements and ideals
A ring element $r\in R$ is called nilpotent if there exists a natural number $k$ such that $r^k=0,$ and the minimal such $k$ is called the degree of nilpotency. For instance $2\in \mathbb{Z}_4$ is nilpotent, and has nilpotency degree 2. More generally, we can talk about a subset $S\subset R$ satisfying $S^k=S\cdot S\cdot\dots\cdot S=\{0\}.$ Such subsets, and in particular, ideals, are also called nilpotent.

Exercise. Prove that the sum of two nilpotent elements of a commutative ring $R$ is nilpotent. Derive that the set of all nilpotent elements of $R$ is an ideal.

Here I wrote more about various other operations on ideals.

Just as for abelian groups, we have an isomorphism theorem:

Let $\phi: R\to S$ be a ring homomorphism, then $\pi: Im(\phi)\to R/Ker(\phi)$ s.t. $\phi(r)\mapsto r+Ker(\phi)$ is a ring isomorphism.

By the analogous theorem for groups, we already know that $\pi$ is a bijection. To prove the rest, we only need to check that $\pi(st)=\pi(s)\pi(t)$ for any $s,t\in Im(\phi).$ Let $\phi(x)=s$ and $\phi(y)=t.$ Then, as $\phi$ is a homomorphism, $\phi(xy)=st$ and so $\pi(st)=xy+Ker(\phi)=(x+Ker(\phi))(y+Ker(\phi))=\pi(s)\pi(t).$

Examples.

1. Let $\pi:\mathbb{Z}[t]\to\mathbb{Z}_p[t]$ be given by $\pi(\sum_k a_k t^k)=\sum_k (a_k \mod p) t^k,$ for a fixed prime $p.$ This is a ring homomorphism, and thus $\mathbb{Z}_p[t]\cong \mathbb{Z}[t]/p\mathbb{Z}[t].$
2. Let $c\in \mathbb{R}$ and $\mathbb{R}[t]\to \mathbb{R}$ be given by $f(t)\mapsto f(c).$ Then this is a ring homomorphism, and thus $\mathbb{R} \cong \mathbb{R}[t]/(t-c)\mathbb{R}.$
3. Let $c\in \mathbb{C}$ and $f(t)=(t-c)(t-\overline{c})\in \mathbb{R}[t].$ Then $\mathbb{R}[t]/f(t) \mathbb{R}[t]\cong \mathbb{C}.$

Multiplicative inverses. (A bit of number theory.)
Let $R$ be a commutative ring with identity. How far is it from a field? In a field $\mathbb{F},$ all the non-zero elements form an abelian (or better said, a commutative) group (w.r.t. to the multiplication), denoted $\mathbb{F}^*.$
For instance, for $\mathbb{Z}_p$ one has $\mathbb{Z}_p^*$ isomorphic to the cyclic group of order $p-1.$ This follows from a more general result, that we state here.

Exercise Any finite subgroup of $\mathbb{F}^*$ is cyclic. Prove this using the observation that the polynomial equation $X^m=1$ has at most $m$ solutions in $\mathbb{F}.$

More generally, for $\mathbb{Z}_n$ one has $|\mathbb{Z}^*_n|=\phi(n),$ the number of integers between 1 and $n-1$ that are relatively prime to $n.$ Thus in particular we have Euler Theorem $a^{\phi(n)}=1\mod n$ for any $a$ such that $(a,n)=1.$

The isomorphism of the abelian groups $\mathbb{Z}_{mn}=\mathbb{Z}_m\oplus \mathbb{Z}_n$ is in fact also a ring isomorphism: $\mathbb{Z}_{mn}=\mathbb{Z}_m\times \mathbb{Z}_n.$In particular, this implies $\mathbb{Z}^*_{mn}=\mathbb{Z}^*_m\times \mathbb{Z}^*_n.$

Excercise. Show that $\phi(n)=n(1-1/p_1)\dots (1-1/p_\ell),$ where $p_k$‘s are the distinct prime divisors of $n.$