core29 min studyLesson 149 in Mathematics

Chinese remainders for rings

Split a quotient into compatible local components, prove the comaximal condition, and reconstruct an element explicitly.

Before you begin

On this page

The question for today

What you’ll understand

Recognize comaximal ideals and construct their Bézout-style certificate.
Identify the kernel and image of the simultaneous reduction map.
Use the Chinese remainder theorem to expose zero divisors in a product decomposition.

A number known modulo three and modulo five can be recovered modulo fifteen. The deeper reason is that those two divisibility conditions are independent in a precise ideal-theoretic sense. The Chinese remainder theorem extends this decomposition to polynomial rings and other commutative rings. It tells us when a single quotient is a product of simpler rings, how to reconstruct an element from its components, and why decompositions often create zero divisors rather than fields.

Comaximal means that the conditions can be separated

Two ideals I and J are comaximal when their sum is the whole ring. Equivalently, there are elements e in I and f in J whose sum is one. That short identity is a certificate of independence: modulo I, f behaves like one while e behaves like zero; modulo J, their roles reverse. For integer principal ideals, this condition says the two positive generators are relatively prime.

Consider the map sending a ring element to its residue class modulo I and its residue class modulo J. It preserves both operations componentwise. An input maps to zero exactly when it belongs to both ideals, so its kernel is their intersection regardless of comaximality. To reach arbitrary target classes represented by a and b, use af plus be. The certificate makes this expression congruent to a modulo I and to b modulo J, proving surjectivity in the comaximal case.

e+f=1,\ e\in I,\ f\in J\quad\Longrightarrow\quad c=af+be\equiv\begin{cases}a\pmod I,\\b\pmod J.\end{cases}

The elements e and f separate the two residue conditions. Changing a or b within its class changes c only by an element of the intersection.↔ Scroll wide equations horizontally.

The intersection becomes a product under this hypothesis

The product ideal IJ consists of finite sums of products of elements from I and J. Such sums always lie in the intersection. Conversely, if z belongs to both ideals and e plus f equals one as above, then z equals ze plus zf. In the first term z can be viewed as belonging to J, and in the second it belongs to I, so both terms belong to IJ. Therefore intersection and product agree for comaximal ideals.

The first isomorphism theorem now gives the Chinese remainder isomorphism. Without comaximality, the same map still has the intersection as its kernel, but it may fail to reach all pairs. For example, reducing an integer modulo four and modulo six cannot produce one component that is even and another that is odd. Both components must agree modulo two. Merely having two different ideals does not make their residue conditions independent.

I+J=R\Longrightarrow I\cap J=IJ,\qquad R/IJ\cong R/I\times R/J

The isomorphism is induced by simultaneous reduction. The full product is the image because the ideals are comaximal.↔ Scroll wide equations horizontally.

A product ring has informative zero divisors

Over the rationals, the ideals generated by x minus one and x plus one are comaximal: their generators differ by two, which is a unit. Simultaneous evaluation at one and negative one therefore identifies the quotient by x squared minus one with a product of two rational fields. In that product, the elements with components one-zero and zero-one are nonzero but multiply to zero. Their corresponding polynomial classes expose why the quotient is not itself a field.

A proper ideal is prime if a product belonging to it forces one factor to belong to it; it is maximal if no proper ideal lies strictly between it and the whole ring. These definitions match quotient behavior: the quotient is a domain exactly for prime ideals, and a field exactly for maximal ideals. The principal ideal generated by x squared minus one is not prime over the rationals because its quotient has those zero divisors. Over characteristic two, the two linear factors coincide instead of being comaximal. The decomposition into two field components cannot be imported unchanged; the coefficient field's characteristic controls the argument.

Work it through

Reconstruct a class from two evaluations

In $\mathbb{Q}[x]/(x^2-1)$ , find the degree-at-most-one representative whose values at one and negative one are two and four respectively.

The polynomial $u=(x+1)/2$ has evaluation pair $(1,0)$ at $(1,-1)$ , while $v=(1-x)/2$ has pair $(0,1)$ . Their sum is one, providing component selectors.
Multiply each selector by its desired component: $2u+4v=(x+1)+2(1-x)=3-x$ . Substituting one gives two, and substituting negative one gives four.
Any other polynomial with the same pair differs by an element of both linear ideals, hence by a multiple of $(x-1)(x+1)=x^2-1$ . A nonzero polynomial of degree at most one cannot be such a multiple, proving uniqueness of this representative.

The result

The class is represented uniquely in degree below two by $3-x$ , corresponding to $(2,4)\in\mathbb{Q}\times\mathbb{Q}$ .

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Compare two evaluation components

Choose a nonzero coefficient a in the lab's $ax^2$ family. Predict and record the values at position one and negative one; both equal a. Explain why this restricted family reaches only diagonal evaluation pairs, although general polynomial representatives can reach every pair. The lab does not perform CRT reconstruction or modular reduction.

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 forA complete ring map may be surjective even when a small displayed family covers only part of its image.

Pause & reason

Check your understanding

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

Read the checkpoint explanation

A certificate e + f = 1 with e in I and f in J constructs af + be for any target representatives a and b. The other conditions do not supply that independence.

Your turn

Try it yourself

Use a comaximal certificate to find the integer class modulo fifteen that is two modulo three and one modulo five.

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

Give me a hint

The identity 6 − 5 = 1 gives an element of (3) plus an element of (5).

Compare with the solution

Take e = 6 in the ideal (3) and f = −5 in the ideal (5). For target representatives a = 2 and b = 1, the reconstruction af + be is −10 + 6 = −4, which is the class eleven modulo fifteen. Eleven is two modulo three and one modulo five. Any other solution differs from eleven by a multiple of both three and five, hence by a multiple of fifteen.

Keep the idea

Take it with you

Comaximal ideals admit a certificate expressing one as a sum from the two ideals.
The CRT map always has intersection kernel; comaximality makes its image the full product.
A product of nonzero rings has zero divisors, even when each component is a field.

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