Notes - Rings and Modules HT24, Chinese remainder theorem

Created: April 18, 2024

Flashcards

What does it mean for two ideals $I$ and $J$ of some commutative ring $R$ to be coprime?
\[I + J = R\]

(there is no condition that $I \cap J = \{0\}$),

State a general version of the “Chinese Remainder Theorem” for rings, and quickly note how the standard version is a special case.

Suppose:

  • $R$ is a ring
  • $I, J$ ideals of $R$
  • $I + J = R$

Then:

\[\frac{R}{I \cap J} \cong R/I \oplus R/J\]

The standard Chinese Remainder Theorem states that for coprime $m, n$

\[\frac{\mathbb Z}{mn\mathbb Z} \cong \mathbb Z/m\mathbb Z \times \mathbb Z/n\mathbb Z\]

Quickly prove the general version of the Chinese Remainder Theorem for rings, i.e. that if

  • $R$ is a ring
  • $I, J$ ideals of $R$
  • $I + J = R$

then:

\[\frac{R}{I \cap J} \cong R/I \oplus R/J\]

By the first isomorphism theorem, we are done if we can find a surjective homomorphism $q : R \to R/I \oplus R/J$ such that $\ker q = I \cap J$.

Let $q _ 1 : R \to R/I$ be the quotient homomorphism for $R/I$ and $q _ 2 : R \to R/I$ be the quotient homormorphism for $R/J$. Define $q : R \to R/I \oplus R/J$ via $q(r) = (q _ 1(r), q _ 2(r))$.

Clearly $\ker(q) = I \cap J$, since both need to vanish to be $(0, 0)$.

To see surjectivity, suppose $(r + I, s + J) \in R/I \oplus R/J$. Then since $I + J = R$, we have $r = i _ 1 + j _ 1$, $s = i _ 2 + j _ 2$. Then note $q(i _ 2 + j _ 1) = (j _ 1 + I, i _ 2 + J) = (r + I, s + J)$.

Quickly prove the following generalisation of the Chinese Remainder Theorem, that if

  • $d _ 1, \cdots, d _ k \in R$ pairwise coprime elements of a PID

Then

\[\frac{R}{(d_1 d_2 \cdots d_k )R} \cong \bigoplus^k_{i=1} \frac{R}{d_i R}\]

You can assume the standard Chinese Remainder Theorem as a result. (This generalisation is useful in the primary decomposition form of the structure theorem for finitely generated modules over a Euclidean domain).

We induct on the number of pairwise coprime elements. The case where $k = 1$ is easy. For $k > 1$, say we have pairwise coprime elements $d _ 1, \cdots, d _ k$. Define for $i < k$

\[c_i = d_{i+1} \cdots d_{k}\]

Then for each $i$ we have $\text{hcf}(d _ i, c _ i) = 1$. Thus for each $i$ with $1 \le i \le r-1$, we have

\[d_i R + c_i R = R, \quad d_iR \cap c_iR = d_i c_i R\]

Thus by the Chinese Remainder Theorem and induction on $k$, we see that

\[\begin{aligned} \frac{R}{(d_1 \cdots d_k)R} &= \frac{R}{(d_1 c_1 R)} \\\\ &\cong \frac{R}{d_1 R} \oplus \frac{R}{c_1R} \\\\ &\cong \bigoplus^k_{i=1} \frac{R}{d_i R} \end{aligned}\]

As required.

(When we apply this to get the primary decomposition version of the structure theorem, we factorise some $c$ into $p _ 1^{q _ 1} \cdots p _ k^{q _ k}$).



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