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 1.

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 1.

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 m, n > 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 m, n > 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. ♦

warning It 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 2.

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 2.

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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Commutative Algebra 2

Radical of an Ideal

Properties of Radical

Exercise.

Hint (highlight to read)

Division of Ideals

Note

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Commutative Algebra 2

Radical of an Ideal

Properties of Radical

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Division of Ideals

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