## Commutative Algebra 1

Recall that we defined three operations on ideals: intersection, sum and product. We can take intersection and sum of any collection of ideals (even infinitely many of them), but we can only define the product of finitely many ideals.

As we mentioned earlier, ideals are a generalization of elements of the ring. In the case of $A = \mathbb Z$ with ideals $\mathfrak a = (m), \mathfrak b = (n)$, we have

$\mathfrak a \mathfrak b = (mn), \quad \mathfrak a + \mathfrak b = (\gcd(m,n)), \quad \mathfrak a \cap \mathfrak b = (\mathrm{lcm}(m,n)), \quad \mathfrak a\subseteq \mathfrak b \iff n | m.$

Thus, philosophically, we have the following correspondences in mind

• product of ideals ↔ product,
• sum of ideals ↔ gcd,
• intersection of ideals ↔ lcm,
• reverse containment ↔ divides.

Armed with the above picture, the following properties become quite natural.

Proposition 1.

Given any ideals $\mathfrak a, \mathfrak b$ and collection of ideals $\mathfrak b_i$ of the ring $A$, we have:

• $\mathfrak a(\sum_i \mathfrak b_i) = \sum_i (\mathfrak{ab}_i)$.
• $\mathfrak a\mathfrak b \subseteq \mathfrak a \cap \mathfrak b$.
• $(\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) \subseteq \mathfrak{ab}$.

Proof

For the first claim, ⊆ follows from: if $x\in \mathfrak a$ and $y \in \sum_i \mathfrak b_i$, then

$y = z_1 + \ldots + z_k$ for some $z_1 \in \mathfrak b_{i_1}, \ldots, z_k \in \mathfrak b_{i_k}$.

Then $xy = \sum_{j=1}^k xz_j$ where each $xz_j \in \mathfrak {ab}_{i_j}$. Hence $xy \in \sum_i (\mathfrak{ab}_i)$.

For the other containment ⊇, note that since $\mathfrak b_i \subseteq \sum_i \mathfrak b_i$ we have $\mathfrak {ab}_i \subseteq \mathfrak a\sum_i \mathfrak b_i$ and hence $\sum_i \mathfrak{ab}_i \subseteq \mathfrak a\sum_i \mathfrak b_i$.

Second claim: if $x\in \mathfrak a, y\in \mathfrak b$, then $xy \in \mathfrak a$ since it is a multiple of $x$, and similarly it lies in $\mathfrak b$ since it is a multiple of $y$. Thus $xy\in \mathfrak a\cap \mathfrak b$ and so $\mathfrak a\cap \mathfrak b$ must contain all finite sums $x_1 y_1 + \ldots + x_k y_k$ with $x_i \in \mathfrak a$ and $y_i \in \mathfrak b$.

Last claim: $(\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) = (\mathfrak a \cap \mathfrak b)\mathfrak a + (\mathfrak a \cap \mathfrak b)\mathfrak b \subseteq \mathfrak b\mathfrak a + \mathfrak a \mathfrak b = \mathfrak {ab}$. ♦

# Coprime Ideals

Suppose $m,n\in \mathbb Z$ are coprime. The following properties of $\mathbb Z$ are familiar to us.

• (Bezout’s theorem) There exist integers a, b such that $am+bn = 1$.
• Any common multiple of m and n is a multiple of mn.
• (Chinese Remainder Theorem) For any c mod m and d mod n, there is a unique a mod mn such that $a\equiv c \pmod m$ and $a\equiv d \pmod n$.
• Any powers $m^k$, $n^j$ are also coprime.

Let us generalize this to the setting of ideals.

Definition.

Let $A$ be a ring. We say that ideals $\mathfrak{a}, \mathfrak{b} \subseteq A$ are coprime if $\mathfrak{a} + \mathfrak{b} = (1) = A$.

Note

It is clear that ideals $\mathfrak a, \mathfrak b\subseteq A$ are coprime if and only if there exist $x\in \mathfrak a, y\in\mathfrak b$ such that $x+y = 1$.

Also, if $\mathfrak m$ is a maximal ideal of $A$, then it is coprime to any ideal $\mathfrak a$ not contained in it, because $\mathfrak m + \mathfrak a$ is an ideal which strictly contains $\mathfrak m$, so by the maximality of $\mathfrak m$, $\mathfrak m + \mathfrak a = A$.

The following two results give us a recipe for producing more coprime pairs of ideals given existing ones. Philosophically the results make sense if we imagine coprime to mean “not sharing any factor”.

Proposition 2.

If ideals $\mathfrak a, \mathfrak b\subseteq A$ are coprime, then so are any powers $\mathfrak a^m, \mathfrak b^n$.

Proof

Pick $x\in \mathfrak a, y\in\mathfrak b$ such that $x+y = 1$. Taking this to the (m+n)-th power, the LHS is a sum of m+n+1 terms, each of the form $Cx^i y^j$ where C is an integer, $i, j\ge 0$ are integers such that $i+j=m+n$. Since either $i\ge m$ or $j \ge n$, this term is a multiple of $x^m$ or $y^n$ so it lies in $\mathfrak a^m$ or $\mathfrak b^n$. Hence the whole sum lies in $\mathfrak a^m + \mathfrak b^n$ and $1 \in \mathfrak a^m + \mathfrak b^n$. ♦

Proposition 3.

If $\mathfrak a, \mathfrak b, \mathfrak c\subseteq A$ are ideals such that $(\mathfrak a, \mathfrak b)$ is coprime and $(\mathfrak a,\mathfrak c)$ is coprime, then $(\mathfrak a, \mathfrak {bc})$ is coprime.

Proof

Find $x\in \mathfrak a, y\in \mathfrak b$ such that $x+y = 1$. Also find $x'\in \mathfrak a, z\in \mathfrak c$ such that $x'+z = 1$. Then

$1 = (x+y)(x'+z) = \overbrace{x(x'+z) + x'y}^{\in \mathfrak a}+ yz \in \mathfrak a + \mathfrak {bc}$

and we are done. ♦

Corollary 1.

If $\mathfrak a, \mathfrak b_1, \ldots, \mathfrak b_n\subseteq A$ are ideals such that $\mathfrak a$ and $\mathfrak b_i$ are coprime for each $1 \le i \le n$, then $\mathfrak a$ and $\prod_{i=1}^n \mathfrak b_i$ are coprime.

Proof

Repeatedly apply proposition 3: since $(\mathfrak a, \mathfrak b_1)$ and $(\mathfrak a, \mathfrak b_2)$ are coprime pairs, so is $(\mathfrak a, \mathfrak b_1 \mathfrak b_2)$. Together with the fact that $(\mathfrak a, \mathfrak b_3)$ is coprime, we see that $(\mathfrak a, \mathfrak b_1 \mathfrak b_2 \mathfrak b_3)$ is coprime, etc. ♦

# Consequences

Finally we discuss the consequences given two coprime ideals of a ring A. Immediately we obtain Bezout’s theorem from the definition of coprimality: if $\mathfrak a,\mathfrak b$ are coprime, there exist $x\in \mathfrak a,y \in\mathfrak b$ such that $x+y = 1$. Next we have:

Proposition 4.

If $\mathfrak a,\mathfrak b\subseteq A$ are coprime ideals, then $\mathfrak{ab} = \mathfrak a \cap \mathfrak b$.

Proof

By proposition 1, $\mathfrak a \cap \mathfrak b \supseteq \mathfrak{ab} \supseteq (\mathfrak a \cap \mathfrak b)(\mathfrak a + \mathfrak b) = (\mathfrak a \cap \mathfrak b)(1) = \mathfrak a \cap \mathfrak b.$ so equality holds throughout. ♦

Corollary 2.

If $\mathfrak a_1, \ldots, \mathfrak a_n$ are pairwise coprime ideals of A, then $\mathfrak a_1 \cap \ldots \cap \mathfrak a_n = \mathfrak a_1 \ldots \mathfrak a_n$.

Note

In general, a collection of items is said to satisfy pairwise X if any two of them satisfy X.

Proof

When n = 1, clear. For $n\ge 2$, suppose it holds for n – 1 so that $\mathfrak a_1 \cap\ldots \cap \mathfrak a_{n-1}= \mathfrak a_1 \ldots \mathfrak a_{n-1}$. By corollary 1, $\mathfrak a_n$ and $\mathfrak a_1 \ldots \mathfrak a_{n-1}$ are coprime. By proposition 4 we have

$\mathfrak a_1 \ldots \mathfrak a_{n-1} \mathfrak a_n = (\mathfrak a_1 \cap \ldots \cap \mathfrak a_{n-1}) \mathfrak a_n = \mathfrak a_1 \cap \ldots \cap \mathfrak a_{n-1} \cap \mathfrak a_n.$

Chinese Remainder Theorem (CRT).

If $\mathfrak a, \mathfrak b\subseteq A$ are coprime ideals, then the natural map

$A/(\mathfrak{ab}) \longrightarrow (A/\mathfrak a) \times (A/\mathfrak b), \qquad a + \mathfrak{ab} \mapsto (a + \mathfrak a, a + \mathfrak b)$,

is an isomorphism.

Proof

The kernel of the map is clearly $\mathfrak a\cap \mathfrak b$ which is $\mathfrak {ab}$ by proposition 4. To prove that the map is surjective, pick $x\in \mathfrak a, y\in \mathfrak b$ such that $x+y = 1$. Now for any $(a + \mathfrak a, b + \mathfrak b)$ in the RHS, let $z := ay + bx \in A$. Since $x\equiv 1 \pmod {\mathfrak{b}}$ and $y \equiv 1 \pmod {\mathfrak{a}}$ we have

$z \equiv ay \equiv a \pmod{\mathfrak{a}}, \qquad z \equiv bx \equiv b \pmod{\mathfrak{b}}$

so $z + \mathfrak{ab}$ maps to $(a+\mathfrak a, b + \mathfrak b)$. ♦

As before, this immediately generalizes to the following.

Corollary 3.

If $\mathfrak a_1, \ldots, \mathfrak a_n \subseteq A$ are pairwise coprime ideals, then the natural map

$A/(\mathfrak a_1 \ldots \mathfrak a_n) \longrightarrow (A/\mathfrak a_1) \times \ldots \times (A/\mathfrak a_n)$

is an isomorphism.

Proof

For n = 1, OK. For n > 1, by corollary 1, $\mathfrak a_n$ and $\mathfrak a_1 \ldots \mathfrak a_{n-1}$ are coprime so CRT gives us

$A/(\mathfrak a_1 \ldots \mathfrak a_n) \cong A/(\mathfrak a_1\ldots \mathfrak a_{n-1}) \times A/\mathfrak a_n$

via the natural map. Now proceed inductively. ♦

The following is an important special case.

Example.

Let $\mathfrak m_1, \ldots, \mathfrak m_n$ be distinct maximal ideals of the ring A. Then their powers $\mathfrak m_1^{k_1}, \ldots, \mathfrak m_n^{k_n}$ are pairwise coprime for any $k_1, \ldots, k_n \ge 1$. Hence we have:

\begin{aligned}\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n} &= \mathfrak m_1^{k_1} \cap \ldots \cap \mathfrak m_n^{k_n}, \\ A/(\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}) &\cong A/\mathfrak m_1^{k_1} \times \ldots \times A/\mathfrak m_n^{k_n}.\end{aligned}

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