Ring Homomorphisms

Ring homomorphisms are the working language of algebraic number theory, where reduction maps $\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ (as in Example 2) let number-theorists study Diophantine equations "modulo $p$" for many primes $p$ simultaneously, then reassemble information via the Chinese Remainder Theorem — the direct-product target ring in Example 2 is exactly this phenomenon in miniature. In algebraic geometry, ring homomorphisms between coordinate rings correspond (contravariantly) to maps of geometric spaces, the foundational dictionary of the subject. In cryptography, the RSA cryptosystem's correctness depends on the ring homomorphism properties of exponentiation modulo $n = pq$, and homomorphic encryption schemes are literally built to be ring (or partial-ring) homomorphisms between plaintext and ciphertext rings, so that computations on encrypted data correspond to computations on the underlying data. In computer algebra and coding theory, evaluation homomorphisms like the one in Example 1 underlie polynomial interpolation and Reed–Solomon codes, and finite-field homomorphisms are the backbone of error-correcting code constructions.

Ask the tutor: How does the First Isomorphism Theorem use the kernel-is-an-ideal and image-is-a-subring facts to build the explicit isomorphism $R/\ker\varphi \cong \varphi(R)$?

Think of a ring homomorphism as a translator between two arithmetic worlds that never garbles a sentence: if a statement in $R$ says "$a$ plus $b$ equals $c$," the translated statement in $S$ says "$\varphi(a)$ plus $\varphi(b)$ equals $\varphi(c)$," and likewise for products. The translator may lose information — many different elements of $R$ might translate to the same word in $S$ — but it never contradicts itself. The set of things that get translated to "zero," the kernel, is exactly the set of information the translator irretrievably discards. Why must the kernel be an ideal rather than merely a subring? A subring only needs to be closed under the ring's own multiplication. An ideal needs the much stronger property of absorbing multiplication by anything in the ambient ring. The homomorphism axioms force exactly this absorption: if $\varphi(k) = 0$ and $r \in R$ is arbitrary, then $\varphi(rk) = \varphi(r)\varphi(k) = \varphi(r)\cdot 0 = 0$, so $rk$ is again in the kernel — no matter how wild $r$ is. This is the one-line miracle at the heart of the theory. The image, by contrast, only inherits closure under multiplication by other image elements, which is precisely subring closure, nothing more.

Example 1: The evaluation homomorphism $\mathbb{Z}[x] \to \mathbb{Z}[i]$. Let $R = \mathbb{Z}[x]$, the ring of polynomials with integer coefficients, and let $S = \mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}$, the Gaussian integers, with $i^2 = -1$. Define

the evaluation map at $x = i$. Checking it is a homomorphism. For $f, g \in \mathbb{Z}[x]$: $\varphi(f+g) = (f+g)(i) = f(i) + g(i) = \varphi(f) + \varphi(g)$, and $\varphi(fg) = (fg)(i) = f(i)g(i) = \varphi(f)\varphi(g)$, since evaluation of a product of polynomials is the product of the evaluations (this is a direct computation from the definition of polynomial multiplication, term by term). Also $\varphi(1) = 1$. So $\varphi$ is a ring homomorphism. Computing the image. Every $f(i)$ has the form $a + bi$ with $a, b \in \mathbb{Z}$, because any power $i^n$ reduces (via $i^2 = -1$) to $\pm 1$ or $\pm i$. Conversely, the constant-degree polynomial $a + bx$ maps to $a + bi$ for any chosen $a, b \in \mathbb{Z}$. So $\varphi$ is surjective: $\varphi(\mathbb{Z}[x]) = \mathbb{Z}[i]$. This confirms part (b) of the theorem concretely: the image is exactly the subring $\mathbb{Z}[i] \subseteq \mathbb{Z}[i]$ (all of it, here). Computing the kernel. We claim

($\supseteq$): if $f(x) = (x^2+1)h(x)$, then $\varphi(f) = (i^2+1)h(i) = 0 \cdot h(i) = 0$, so $f \in \ker\varphi$. ($\subseteq$): suppose $f(i) = 0$. Perform polynomial division of $f(x)$ by the monic polynomial $x^2+1$ in $\mathbb{Z}[x]$ (division by a monic polynomial works with integer, not just rational, coefficients, since the leading coefficient $1$ is invertible at every stage): $f(x) = (x^2+1)q(x) + r(x)$ with $\deg r < 2$, so $r(x) = c + dx$ for some $c, d \in \mathbb{Z}$. Evaluating at $x = i$:

Since $c, d \in \mathbb{Z}$ and $\{1, i\}$ are $\mathbb{Z}$-linearly independent in $\mathbb{Z}[i]$ (as $a + bi = 0 \Rightarrow a=b=0$ by comparing real and imaginary parts), we get $c = d = 0$, so $r(x) = 0$ and $f(x) = (x^2+1) q(x) \in (x^2+1)$. This confirms $\ker\varphi = (x^2+1)$, a two-sided ideal (in fact $\mathbb{Z}[x]$ is commutative, so every ideal is automatically two-sided) — exactly as part (a) of the theorem guarantees, without our needing to verify absorption by hand. Conclusion, foreshadowing. Since $\varphi$ is surjective with kernel $(x^2+1)$, this example is the natural setting for the First Isomorphism Theorem for Rings, which will show

We do not need that theorem to see, right now, exactly what the kernel and image are and why the theorem's shape must hold: everything not in the kernel maps to a genuinely distinct residue, and the quotient by the kernel should "be" the image.

Example 2: A non-surjective, non-injective homomorphism $\mathbb{Z} \to \mathbb{Z}_6 \times \mathbb{Z}_{10}$, and why reduction mod $n$ is a homomorphism. Let $R = \mathbb{Z}$ and $S = \mathbb{Z}_6 \times \mathbb{Z}_{10}$ (ring direct product, with componentwise addition and multiplication, identity $(1,1)$). Define

Checking it is a homomorphism. Reduction mod $m$, $\pi_m: \mathbb{Z} \to \mathbb{Z}_m$, $\pi_m(n) = n \bmod m$, is a ring homomorphism: this is because addition and multiplication in $\mathbb{Z}_m$ are defined by reducing the integer sum/product mod $m$, so $\pi_m(a+b) = (a+b)\bmod m = (\pi_m(a) + \pi_m(b))$ computed in $\mathbb{Z}_m$, and similarly for products; also $\pi_m(1) = 1$. Then $\varphi = (\pi_6, \pi_{10})$, i.e. $\varphi(n) = (\pi_6(n), \pi_{10}(n))$, and a direct check shows a pair of homomorphisms into a product assembles into a homomorphism into the product ring: $\varphi(a+b) = (\pi_6(a+b), \pi_{10}(a+b)) = (\pi_6(a)+\pi_6(b), \pi_{10}(a)+\pi_{10}(b)) = \varphi(a) + \varphi(b)$ in $S$ (componentwise addition), and likewise for products. So $\varphi$ is a ring homomorphism. Computing the kernel. $\varphi(n) = (0,0)$ iff $6 \mid n$ and $10 \mid n$, iff $\operatorname{lcm}(6,10) \mid n$. Since $\operatorname{lcm}(6,10) = 30$, we get

the principal ideal generated by $30$ — an ideal of $\mathbb{Z}$, consistent with part (a). Note $\ker\varphi \neq \{0\}$, so by part (c), $\varphi$ is not injective: for instance $\varphi(0) = \varphi(30) = (0,0)$. Computing the image, and showing it is a proper subring. We claim $\varphi$ is not surjective. Consider the target element $(1, 0) \in \mathbb{Z}_6 \times \mathbb{Z}_{10}$. If $(1,0) = \varphi(n) = (n \bmod 6, n\bmod 10)$ for some $n \in \mathbb{Z}$, then $n \equiv 1 \pmod 6$ and $n \equiv 0 \pmod{10}$. The second congruence gives $n \in \{0, 10, 20, 30, \ldots\} \cup \{-10, -20, \ldots\}$; check these mod $6$: $10 \equiv 4$, $20 \equiv 2$, $30 \equiv 0$, $40 \equiv 4$, $50 \equiv 2$, $60 \equiv 0 \pmod 6$ — the residues cycle through $\{0, 2, 4\}$ only (since $10 \equiv 4 \pmod 6$ and $\gcd(10,6)=2$, the multiples of $10$ hit only residues in the subgroup of index $2$ generated by $4$ in $\mathbb{Z}_6$, namely $\{0,2,4\}$), never $1$. So no such $n$ exists, and $(1,0) \notin \varphi(\mathbb{Z})$. This is the point of the example: $\varphi(\mathbb{Z})$ is a proper subring of $\mathbb{Z}_6\times\mathbb{Z}_{10}$ (it is exactly the "diagonal image," isomorphic by the First Isomorphism Theorem to $\mathbb{Z}/30\mathbb{Z}$, a ring of size $30$, strictly smaller than $|\mathbb{Z}_6\times\mathbb{Z}_{10}| = 60$), demonstrating concretely that a homomorphism's image can be much smaller than the codomain, exactly as part (b) permits — a subring, with no surjectivity promised.