Exploration 2: Ring homomorphisms

• By induction on the rational expression.
• $\forall a,b\in R\ldotp a+b\in R\implies \theta(a)+\theta(b)\in R$ since rings are closed under addition by definition, and homomorphisms preserve addition.
• $\forall a,b\in R\ldotp a-b\in R\implies \theta(a)-\theta(b)\in R$ since rings are closed under addition and additive inverse by definition, and homomorphisms preserve addition and additive inverse.
• $\forall a,b\in R\ldotp ab\in R\implies \theta(a)\theta(b)\in R$ since rings are closed under multiplication by definition, and homomorphisms preserve multiplication. This also implies that units are preserved, because $ab=1$ implies $\theta(a)\theta(b)=\theta(1)$.
• $\forall a\in R,k\in\ZZ\ldotp ka\in R\implies k\theta(a)\in R$ since this is just repeated addition, and rings are closed under addition by definition, and homomorphisms preserve addition.
• $\forall a\in R,k\in\ZZ\ldotp a^k\in R\implies \theta(a)^k\in R$ since this is just repeated multiplication, and rings are closed under multiplication by definition, and homomorphisms preserve multiplication.
• $\forall a\in R,b\in R^\times\ldotp a/b\in R\implies \theta(a)/\theta(b)\in R$ since this is just multiplication by multiplicative inverse, and rings are closed under multiplication by definition, and homomorphisms preserve multiplication and units.

• Since a ring homomorphism must preserve equations composed of only rational expressions, it preserves the equation $ab=1$. But that’s just the definition of being a unit.
• The same goes for $r^2=r$ (idempotents) and $r^n=0$ (nilpotents).

• This is a direct consequence of preserving addition and the multiplicative identity $1$: if $n\cdot 1_R=0$ in $R$, then that maps to $n\cdot 1_S=0$ in $S$.

Kernel is always an additive subgroup of $R$, like from group theory. To show it absorbs product and is therefore an ideal, note that $\theta(\ker\theta)=\{0\}$ by definition, and $0$ absorbs all products.

Since $\theta$ is a ring homomorphism, this is implied. $\im\theta$ is an additive subgroup closed under multiplication, and $\theta$ preserves unity.

• ($\to$) If the kernel is trivial (only $0$ maps to $0$), then $\theta(r)=\theta(s)$ i.e. $\theta(r-s)=0$ implies $r-s$ must be $0$, which means $r=s$.
• ($\from$) If $\theta(r)=\theta(s)$ implies $r=s$, then in particular $\theta(r)=\theta(0)=0$ implies $r=0$, i.e. only $0$ can map to $0$, which means the kernel is trivial.

• Since the kernel of a ring homomorphism $\theta:F\to K$ is an ideal of $F$, and the only ideals in a field $F$ are $0$ and $F$ itself, the kernel is either trivial ($0$) or $F$ itself.
• But since ring homomorphisms must preserve the multiplicative identity $1$, the kernel cannot include $1$ (since it doesn’t get mapped to zero).
• Therefore the kernel is trivial, so $\theta$ is injective.

• Just like we did in proving the First Isomorphism Theorem for groups, we prove that the map $\bar{\theta}=[r]\mapsto\theta(r):R/\ker\theta\to\im\theta$ is a ring isomorphism.
• $\bar{\theta}$ is well defined: $\ker\theta$ is an ideal of $R$ (proof) so $R/\ker\theta$ is a well-defined factor ring. Then from the universal property of the group quotient, we know that $\bar{\theta}$ is a well-defined group homomorphism.
• $\bar{\theta}$ preserves unity and product: they are basically inherited from $\theta$.
  • $\bar{\theta}([1])=\theta(1)=1$
  • $\bar{\theta}([a][b])=\bar{\theta}([ab])=\theta(ab)=\theta(a)\theta(b)$
• Therefore, $\bar{\theta}$ is also a ring homomorphism.
• $\bar{\theta}$ is bijective: $\bar{\theta}$ is onto, since the output is $\theta(r)$ for all $r\in R$, which encompasses $\im\theta$. The reverse direction of the proof above for $[a]=[b]\iff\theta(a)=\theta(b)$ shows $\bar{\theta}$ is one-to-one.
• Therefore, $\bar{\theta}$ is an isomorphism.

• Define the map $\theta:\ZZ\to\ZZ_n$, which is just the residue map from the integers to their corresponding equivalence class in $\ZZ_n$. This map is obviously surjective, since each class in $\ZZ_n$ is typically represented by some integer in $\ZZ$.
• The kernel of $\theta$ is exactly every multiple of $n$, i.e. the principal ideal generated by $n$.
• By the First Isomorphism Theorem, we have $R/\ker\theta\iso\im\theta$. With $\ker\theta=(n)$ and $\im\theta=\ZZ/(n)$ (since it is surjective), we get $R/(n)\iso\ZZ/(n)$.

• The map $(r,s)\mapsto ([r]_A,[s]_B)$ has kernel $A\times B$ and image $R/A\times S/B$.
• By the First Isomorphism Theorem, $(R\times S)/\ker\theta\iso\im\theta$, so we get $(R\times S)/(A\times B)\iso R/A\times S/B$ immediately.

Let $\psi=r\mapsto ([r]_A,[r]_B)$, which you can check is a surjective homomorphism. Then the result is provided by the First Isomorphism Theorem, since $\ker\psi=A\cap B$, and by surjectivity, $\im\psi=R/A\times R/B$.