Commutative Algebra 2

Radical of an Ideal

In this installation, we will study more on ideals of a ring A.

Definition.

If \mathfrak a\subseteq A is an ideal, its radical is defined by

r(\mathfrak a) := \{ x \in A: x^n \in \mathfrak a \text{ for some } n>0\}.

To fix ideas, again consider the case A = \mathbb Z again. For the ideal (m) where m = p_1^{e_1}\ldots p_k^{e_k}, each e_i > 0, its radical is simply (m’) where m' = p_1 \ldots p_k with the repeated exponents removed. Thus one thinks of the radical as “retaining only the prime factors”.

Our first result is:

Lemma.

The radical of an ideal \mathfrak a is also an ideal.

Proof

Let \mathfrak b = r(\mathfrak a). Suppose x, y \in \mathfrak b; pick m, n > 0 such that x^m, y^n \in \mathfrak a.

As before, (x+y)^{m+n} is a sum of terms Cx^i y^j with i+j = m+n and so each term is a multiple of x^m or y^n. Thus (x+y)^{m+n} \in \mathfrak a so x+y \in \mathfrak b.

Also for any z\in A we have (xz)^m = x^m z^m \in \mathfrak a. Hence xz \in \mathfrak b. Finally since 0\in \mathfrak b, we see that \mathfrak b is an ideal of A. ♦

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Properties of Radical

The following properties relate the radical of an ideal to earlier constructions.

Proposition.

Let \mathfrak a, \mathfrak b be ideals of A, and (\mathfrak a_i) be any collection of ideals of A.

  • r(r(\mathfrak a)) = r(\mathfrak a).
  • r(\sum_i \mathfrak a_i) = r(\sum_i r(\mathfrak a_i)).
  • r(\mathfrak a \cap \mathfrak b) = r(\mathfrak {ab}) = r(\mathfrak a) \cap r(\mathfrak b).

Proof

Since \mathfrak a \subseteq r(\mathfrak a) we have proven ⊇ of the first claim. Conversely, if x \in r(r(\mathfrak a)) then x^n \in \mathfrak a for some n > 0 and so (x^n)^m = x^{mn} \in \mathfrak a for some mn > 0. Thus x\in r(\mathfrak a).

For second claim, since \mathfrak a_i \subseteq r(\mathfrak a_i), ⊆ is obvious. Conversely, if x lies in the RHS then x^n \in \sum_i r(\mathfrak a_i) for some n > 0, and so x^n = y_1 + \ldots + y_k with y_j \in r(\mathfrak a_{i_j}). Without loss of generality, there is an m > 0 such that y_j^m \in \mathfrak a_{i_j} for each j = 1, …, k (take m large enough). Then

(x^n)^{mk} = (y_1 + \ldots + y_k)^{mk} = sum of terms of the form y_1^{e_1} \ldots y_k^{e_k} with e_1 +\ldots + e_k = mk.

In each term, we have e_j \ge m for some j, hence the term is a multiple of y_j^m \in \mathfrak a_{i_j}. Thus (x^n)^{mk} \in \sum_i \mathfrak a_i and x lies in the LHS.

Finally for the last claim.

  • Since \mathfrak{ab} \subseteq \mathfrak a \cap \mathfrak b the second term is contained in the first.
  • Since \mathfrak a \cap \mathfrak b \subseteq \mathfrak a we have r(\mathfrak a \cap \mathfrak b) \subseteq r(\mathfrak a) and similarly r(\mathfrak a \cap \mathfrak b)\subseteq r(\mathfrak b) so the first term is contained in the third.
  • Finally if x\in r(\mathfrak a) \cap r(\mathfrak b) there exist mn > 0 such that x^m \in \mathfrak a, x^n \in\mathfrak b. Then x^{m+n} \in \mathfrak{ab} so the third term is contained in the second. ♦

warningIt is not true that r(\cap \mathfrak a_i) = \cap_i r(\mathfrak a_i) for any class of ideals \mathfrak a_i\subseteq A. For example, take A = \mathbb Z and \mathfrak a_n = (2^n) for n = 1, 2, \ldots. Then r(\mathfrak a_n) = (2) so

r(\cap_{n\ge 1} \mathfrak a_n) = r((0)) = (0), \quad \cap_n r(\mathfrak a_n) = \cap_n (2) = (2).

Definition.

An ideal \mathfrak a \subseteq A is called a radical ideal if r(\mathfrak a) = \mathfrak a.

Note that for any ideal \mathfrak a, r(\mathfrak a) is a radical ideal.

Exercise.

1. Prove that a prime ideal is radical.

2. Decide which of the following is true. Find counter-examples for the false claims.

  • If \mathfrak a, \mathfrak b are radical ideals, so is \mathfrak a + \mathfrak b.
  • If \mathfrak a, \mathfrak b are radical ideals, so is \mathfrak a \cap \mathfrak b.
  • If \mathfrak a, \mathfrak b are radical ideals, so is \mathfrak a\mathfrak b.
  • If (\mathfrak a_i) is a collection of radical ideals, so is \sum_i \mathfrak a_i.
  • If (\mathfrak a_i) is a collection of radical ideals, so is \cap_i \mathfrak a_i.

Hint (highlight to read)

[For a counter-example to the first claim, take the ring A = ℤ[X], the ring of polynomials with integer coefficients.]

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Division of Ideals

Finally, we wish to divide ideal \mathfrak a by \mathfrak b.

Definition.

Let \mathfrak a,\mathfrak b\subseteq A be ideals. Write

(\mathfrak a : \mathfrak b) := \{ x \in A: x\mathfrak b\subseteq \mathfrak a\}.

Here, the notation x\mathfrak b means \{xy : y\in \mathfrak b\}; note that this is an ideal of A. As a convenient mnemonic for the definition (whether it is x\mathfrak b \subseteq \mathfrak a or x \mathfrak a \subseteq \mathfrak b), just recall that in the ring ℤ we have (mnℤ : nℤ) = mℤ.

Lemma.

The set \mathfrak c := (\mathfrak a : \mathfrak b) is an ideal of A.

Proof

Clearly 0 \in \mathfrak c since 0\mathfrak b = (0).

Next suppose x, y \in \mathfrak c, so x\mathfrak b, y\mathfrak b \subseteq \mathfrak a. Then (x-y)\mathfrak b \subseteq x\mathfrak b + y\mathfrak b \subseteq \mathfrak a.

Finally if x\in \mathfrak c, so x\mathfrak b\subseteq \mathfrak a, then any z\in A gives us xz \mathfrak b \subseteq z\mathfrak a \subseteq \mathfrak a since \mathfrak a is an ideal of A. ♦

Finally, we go through some basic properties of ideal division.

Proposition

Let \mathfrak a, \mathfrak b, \mathfrak c be ideals of A, and (\mathfrak a_i), (\mathfrak b_i) be any collection of ideals of A.

  • (\mathfrak a : \mathfrak b)\mathfrak b \subseteq \mathfrak a.
  • ((\mathfrak a : \mathfrak b) : \mathfrak c) = (\mathfrak a : \mathfrak {bc}).
  • (\cap_i \mathfrak a_i : \mathfrak b) = \cap_i (\mathfrak a_i : \mathfrak b).
  • (\mathfrak a : \sum_i \mathfrak b_i) = \cap_i (\mathfrak a : \mathfrak b_i).

Proof

First claim: if x \in (\mathfrak a : \mathfrak b), y\in\mathfrak b then by definition xy\in \mathfrak a. Hence \mathfrak a also contains any finite sum of x_i y_i with x_i \in (\mathfrak a : \mathfrak b), y_i \in \mathfrak b.

Second claim: for x\in A,

x \in ((\mathfrak a : \mathfrak b) : \mathfrak c) \iff x\mathfrak c \subseteq (\mathfrak a : \mathfrak b) \iff (x\mathfrak c)\mathfrak b \subseteq \mathfrak a \iff x \in (\mathfrak a : \mathfrak {cb}).

Third claim: for x \in A,

x\in (\cap_i \mathfrak a_i : \mathfrak b) \iff x\mathfrak b \subseteq \cap_i \mathfrak a_i \iff (\forall i, x\mathfrak b\subseteq \mathfrak a_i) \iff (\forall i, x \in (\mathfrak a_i : \mathfrak b)).

Fourth claim: x \in (\mathfrak a : \sum_i \mathfrak b_i) \iff x(\sum_i \mathfrak b_i) \subseteq \mathfrak a \iff (\forall i, x\mathfrak b_i \subseteq \mathfrak a), where the second equivalence follows from: x\sum_i \mathfrak b_i = \sum_i (x\mathfrak b_i) and for any collection of ideals \mathfrak b_i, \sum_i \mathfrak b_i \subseteq \mathfrak c \iff (\forall i, \mathfrak b_i \subseteq \mathfrak c). ♦

Note

In the next article, we will be looking at some basic ideas in algebraic geometry to motivate many of our subsequent concepts. The concept of a radical ideal is of paramount importance there. We will not be seeing much of ideal division for a while, until we encounter invertible ideals.

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