# Explaining the product of two ideals

My textbook says that the product of two ideals $I$ and $J$ is the set of all finite sums of elements of the form $ab$ with $a \in I$ and $b \in J$. What does this mean exactly? Can you give examples?

• The particular case of $I$ and $J$ being principal is easy to understand and could be helpful, see here. – Watson Nov 10 '16 at 20:37
• So, according to the definition, if $x,y\in AB$ then $x=a_1b_1 + \ldots +a_nb_n$ and $y=a'_1b'_1 + \ldots +a'_mb'_m$? where m and n may be different? – yasir May 20 '18 at 10:57
• – Watson Nov 30 '18 at 8:19

One would like the product ideal to be $$IJ=\{ij\mid i\in I,j\in J\}$$ but we can easily see that there is a problem. It must be closed under addition, so $ij+i'j'$ must be in $IJ$. Can you find $i''\in I$, $j''\in J$ such that $ij+i'j'=i''j''$ so that it's in $IJ$ as defined above? Not in general, no. The natural way to allow for additive closure is to define $IJ$ as you did, including arbitrary finite sums of products.

• Can this also be written as $\sum_{i \in I, j \in J} (ij)$? (where $(ij)$ is the ideal generated by $ij$). – Al Jebr Jun 16 '19 at 17:47

For a complete answer let me add an example: $I=(2,X)$ and $J=(3,X)$ in $\mathbb Z[X]$. Then $IJ=(6,X)$ (why?), thus $X\in IJ$ and $X$ can't be written as $ij$ with $i\in I, j\in J$ (why?). (Note that if one of the ideals is principal one can't get such an example.)

• This is a good example, 1+! Almost the same, of course, works with $R[X,Y]$ and the two ideals $(X,Y)$ and $(1-X,Y)$. Geometrically, $Y$ vanishes on $\{(0,0),(1,0)\} \subseteq \mathbb{A}^2_R$, but cannot be written as a product of two polynomials which vanish on $(0,0)$ resp. $(1,0)$. Are there $1$-dimensional examples? – Martin Brandenburg Jan 30 '13 at 10:44
• A big +1 for actually giving an example of when $IJ$ is not simply the set of all products $ij$. – Pete L. Clark May 9 '13 at 17:10
• Can anyone tell me why $IJ=(6,X)$? – User Apr 23 '16 at 9:54
• @User $2 \cdot 3 = 6$, $3x - 2x = x$ and it cannot be larger than $(6,x)$ – user128245 Apr 2 '18 at 16:12

Another way to phrase this: The product ideal $IJ$ is the smallest ideal containing all the products of elements of $I$ with elements of $J$.

As for examples: In $\mathbb{Z}$, we have $$\langle a\rangle\langle b\rangle=\langle ab\rangle$$

• I like your version better, the other one I can't understand if it means that there can be multiple sums added together, like would $ax+by \in IJ$ if $a,b \in I$ and $x,y \in J$? – user39794 Jan 30 '13 at 1:45
• @AllisonCameron yes, by (your) definition of the product of two ideals, that sum would be in the product. – ferson2020 Jan 30 '13 at 1:48
• Notice that my definition is equivalent, because if an ideal contains some elements, then it contains all their finite sums. :) – pre-kidney Jan 30 '13 at 1:49

Since it was also asked for examples, let me mention how to compute the product of two ideals (beyond the already mentioned principal ideals).

If $I$ is generated by elements $\{a_i\}$ and $J$ is generated by elements $\{b_j\}$, then $I \cdot J$ is generated by the elements $\{a_i \cdot b_j\}$. You can verify this either using the element definition of $I \cdot J$, or using the more elegant definition of $I \cdot J$ as the smallest ideal containing all products.

For example, in $\mathbb{Q}[x,y]$, one computes $(x,y) \cdot (x^2,y^2)=(x^3,x y^2,x^2 y,y^3)$.

In general, one observes that $I \cdot J \subseteq I \cap J$. This is not an equality in general; in the above example the intersection is just $(x,y)$. However, one has (in the commutative case) $\sqrt{I \cdot J} = \sqrt{I \cap J}$.

• I wished that abstract algebra books were written with this clarity. Clarity requires mastery of the subject. – nilo de roock Sep 17 '14 at 21:06
• Wait, how is the intersection $(x,y)$? Since $(x^2,y^2)\subset (x,y)$, shouldn't $(x,y)\cap (x^2,y^2)=(x^2,y^2)$? – user2154420 Feb 15 '18 at 15:20

The main thing to notice is that it is not always, as a student might first guess, just $\{ab\mid a\in I, b\in J\}$. That works for groups, but in a ring you have two operations going on. Certainly in addition to having all the pairwise products, it would also have to have all possible sums of those products. Otherwise, given $ab$ and $a'b'$, you would be at a loss to write $ab+a'b'$ in the form $a''b''$ (the $a$'s are from $I$, the $b$'s are from $J$).

Just try it out: show that $\{\sum a_ib_i\mid a_i\in I, b_i\in J\}$ (finite sums) forms an ideal. Then show it's the smallest ideal containing the pairwise products.

• Why is the ideal $IJ$ not $\left\{\sum r_ia_ib_i \mid r_i \in R, a_i \in I, b_i \in J\right\}$? – Al Jebr Jun 16 '19 at 17:53
• @AlJebr what is the point of writing $r_ia_i$ when it is already in $I$? – rschwieb Jun 16 '19 at 18:04

This means that the product $IJ$ is the set of all sums $a_1b_1 + a_2b_2 + ... + a_nb_n$ where $a_1, a_2, ..., a_n \in I$, $b_1, b_2, ..., b_n \in J$.