Some properties of ideal arithmetic

Let R be a ring with 1, and let I,J \subseteq R be ideals of R.

  1. Prove that I+J is the \subseteq-smallest ideal of R containing both I and J.
  2. Prove that IJ is an ideal contained in I \cap J.
  3. Give an example where IJ \neq I \cap J.
  4. Prove that if R is commutative and if I+J = R, then IJ = I \cap J.

  1. We wish to prove the following: (1) I+J is an ideal of R and (2) If K \subseteq R is an ideal and I,J \subseteq K, then I+J \subseteq K.
    1. First we show that I+J is an ideal of R. Let a_1+b_1, a_2+b_2 \in I+J for some a_i \in I and b_i \in J. Then (a_1+b_1)-(a_2+b_2) = (a_1-a_2) + (b_1-b_2) \in I +J since I and J are closed under subtraction. Now let r \in R. We have r(a+b) = ra + rb \in I+J since I and J absorb R on the right; similarly, (a+b)r \in I + J. Thus I+J is an ideal of R.
    2. Now suppose K is an ideal of R with I,J \subseteq K. If a+b \in I+J, then since K is closed under addition, we have a+b \in K. Hence I+J \subseteq K.
  2. We wish to prove that IJ is an ideal of R and that IJ \subseteq I \cap J.

    First, let \alpha,\beta \in IJ, with \alpha = \sum a_ib_i and \beta = \sum c_id_i, where a_i,c_i \in I and b_i,d_i \in J. Clearly \alpha + \beta \in IJ. Now let r \in R. Since I is an ideal of R, r\alpha = \sum (ra_i)b_i \in IJ. Similarly, \alpha r \in IJ. Thus IJ is an ideal of R.

    Now consider again \alpha = \sum a_ib_i. Since a_i \in I and I is an ideal, a_ib_i \in I, and thus \alpha \in I. Similarly, \alpha \in J. Thus IJ \subseteq I \cap J.

  3. Let R = \mathbb{Z} and let I = 2\mathbb{Z} and J = 4\mathbb{Z}. Evidently, (2\mathbb{Z})(4\mathbb{Z}) = 8\mathbb{Z}, while 2\mathbb{Z} \cap 4\mathbb{Z} = 4\mathbb{Z}. Thus it is not generally the case that IJ = I \cap J.
  4. Suppose I+J = R and that R is commutative. We have IJ \subseteq I \cap J by part (3) above.

    Note that (I \cap J)(I + J) = (I \cap J)R = I \cap J. Thus, if z \in I \cap J, we have z = w(x+y) for some x \in I and y \in J and w \in I \cap J. Then z = xw + wy \in IJ. Thus IJ = I \cap J.

Post a comment or leave a trackback: Trackback URL.


  • ysqure3John Magnum  On November 4, 2011 at 3:55 am

    Observation: Dummit-Foote includes a note at the beginning of this section of problems that R is assumed to be a ring with identity. This fact is necessary for part 4, because otherwise (I \cap J)R =/= I \cap J

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: