core25 min studyLesson 143 in Mathematics

Ring homomorphisms and isomorphisms

Use kernels and images to turn a structure-preserving map into an exact description of a quotient ring.

Before you begin

Ideals and quotient rings

On this page

The question for today

What you’ll understand

Check addition, multiplication, and identity preservation for a ring map.
Identify the kernel and image of evaluation and reduction maps.
Build the first isomorphism theorem from a well-defined bijection.

Substituting a value into a polynomial respects addition and multiplication. Reducing an integer modulo five does too. These maps can simplify a calculation while preserving its algebraic meaning, but they can also forget information. Kernels describe exactly what disappears, and images describe exactly what remains reachable. The first isomorphism theorem joins those two descriptions: after identifying the information a map cannot distinguish, the resulting quotient is the same ring structure as its image.

Preserve the operations, not merely the elements

A ring homomorphism from R to S preserves sums, products, and the multiplicative identity under our convention. These conditions force it to preserve zero and additive negatives as well. Checking only addition is insufficient: a map can preserve sums while changing products. Evaluation at a fixed field element is a ring homomorphism from the polynomial ring to that field because expanding a sum or product and then substituting gives the same answer as substituting first.

An isomorphism is a bijective ring homomorphism. Its inverse also preserves the operations, so it provides two names for the same algebraic structure. A bijection of underlying sets is not enough; the operation tables must correspond. Likewise, a map into a large ring need not cover that ring. Always distinguish a map's specified codomain from its image, which is the actual subset reached by its values.

\varphi(a+b)=\varphi(a)+\varphi(b),\quad \varphi(ab)=\varphi(a)\varphi(b),\quad \varphi(1_R)=1_S

These are the three preservation requirements used throughout this module.↔ Scroll wide equations horizontally.

The kernel records every collision

The kernel consists of elements sent to zero. It is an ideal because differences of zero-valued inputs remain zero-valued, and multiplying such an input by any ring element still maps to zero. More generally, two inputs have the same output precisely when their difference lies in the kernel. This gives a practical injectivity test: the map is injective exactly when its kernel contains only zero.

For evaluation of real polynomials at two, the kernel is the ideal generated by x minus two. Polynomial division writes every polynomial as a multiple of x minus two plus a constant remainder; evaluating at two identifies that remainder. The image is all real numbers because constant polynomials already produce every real output. Thus infinitely many different input polynomials can produce a given output, but their differences are controlled by one simple ideal.

\ker\varphi=\{a:\varphi(a)=0\},\qquad \varphi(a)=\varphi(b)\iff a-b\in\ker\varphi

Kernel membership characterizes indistinguishable inputs, not just inputs that literally equal zero.↔ Scroll wide equations horizontally.

Factor through the information you kept

Send the class of a modulo the kernel to the value of the original map at a. The collision criterion makes this induced map well-defined and injective. It is surjective onto the image by the definition of image, and it preserves the operations because the original map does. This constructs the first isomorphism theorem rather than treating its notation as a cancellation of symbols.

The theorem does not identify the quotient with the entire codomain unless the original map is surjective. The inclusion from integers into rational numbers has zero kernel but reaches only the integers. Its induced isomorphism therefore identifies the integer quotient by zero with that integer image, not with all rationals. When applying the theorem, name the actual map, verify its kernel, and justify its image before writing an isomorphism.

R/\ker\varphi\ \cong\ \operatorname{im}\varphi,\qquad a+\ker\varphi\longmapsto\varphi(a)

The induced map is a bijective ring homomorphism onto the image. Surjectivity of the original map is needed to replace the image by its codomain.↔ Scroll wide equations horizontally.

Work it through

Describe a polynomial quotient by evaluation

Consider $\varphi:\mathbb{Q}[x]\to\mathbb{Q}$ given by $\varphi(f)=f(3)$ . Identify the kernel and the quotient it determines.

Substitution at three preserves sums, products, and the constant one, so this is an identity-preserving ring homomorphism. Every rational number is attained by the corresponding constant polynomial, which proves surjectivity.
Divide f by $x-3$ to write $f=(x-3)q+r$ with rational constant r. Evaluating gives $f(3)=r$ , so f is in the kernel exactly when its remainder is zero, equivalently when f is a multiple of $x-3$ .
Apply the first isomorphism theorem to obtain $\mathbb{Q}[x]/(x-3)\cong\mathbb{Q}$ . The class of $x^2+1$ corresponds to ten, since evaluating its representative at three gives ten.

The result

The kernel is $(x-3)$ , the image is $\mathbb{Q}$ , and the induced isomorphism evaluates each class at three.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect evaluation as a homomorphism

The lab displays $f(x)=ax^2$ . Fix position x at two and predict how changing coefficient a changes the evaluation output: a becomes four times a. Then set x to zero and observe that this entire restricted family maps to zero. The lab shows evaluation of one polynomial family, not a complete kernel calculator.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forEvaluation preserves arithmetic while often collapsing different polynomials to the same output.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The induced isomorphism always has the image as its target. A zero kernel makes the original map injective, but does not enlarge its image.

Your turn

Try it yourself

For the reduction map $\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}$ , find the kernel and explain why the map is surjective.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Ask which integers represent the zero class, and where an arbitrary residue class comes from.

Compare with the solution

The kernel is the set of integers divisible by seven, the ideal (7). Every class is represented by an integer, so every codomain element has a preimage. Consequently the quotient of the integers by (7) is isomorphic to the entire codomain in this example. Two integers map to the same class precisely when their difference is divisible by seven.

Keep the idea

Take it with you

A ring homomorphism preserves both operations and the identity under our convention.
The kernel is an ideal and describes every pair of inputs with the same output.
The quotient by the kernel is isomorphic to the image; the codomain may be larger.

Read further

Original sources and deeper explanations to follow at your own pace.

James McKernan, MIT — Ring homomorphisms and ideals

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