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?

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    $\begingroup$ The particular case of $I$ and $J$ being principal is easy to understand and could be helpful, see here. $\endgroup$
    – Watson
    Nov 10, 2016 at 20:37
  • $\begingroup$ 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? $\endgroup$
    – miyagi_do
    May 20, 2018 at 10:57
  • $\begingroup$ Related: math.stackexchange.com/questions/21440 $\endgroup$
    – Watson
    Nov 30, 2018 at 8:19

6 Answers 6


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.

  • $\begingroup$ Can this also be written as $\sum_{i \in I, j \in J} (ij)$? (where $(ij)$ is the ideal generated by $ij$). $\endgroup$ Jun 16, 2019 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.)

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    $\begingroup$ 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? $\endgroup$ Jan 30, 2013 at 10:44
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    $\begingroup$ A big +1 for actually giving an example of when $IJ$ is not simply the set of all products $ij$. $\endgroup$ May 9, 2013 at 17:10
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    $\begingroup$ Can anyone tell me why $IJ=(6,X)$? $\endgroup$
    – User
    Apr 23, 2016 at 9:54
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    $\begingroup$ @User $2 \cdot 3 = 6$, $3x - 2x = x$ and it cannot be larger than $(6,x)$ $\endgroup$
    – user128245
    Apr 2, 2018 at 16:12
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    $\begingroup$ @user782709 The notation $(n,X)$ denotes the ideal generated by the elements $n, X \in \mathbb{Z}[X]$, i.e. the intersection of all ideals in $\mathbb{Z}[X]$ that contain $n$ and $X$. $\endgroup$
    – Danny
    Jan 24, 2022 at 22:15

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 $$\langle a\rangle\langle b\rangle=\langle ab\rangle$$

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    $\begingroup$ 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 $x,y \in J$? $\endgroup$
    – user39794
    Jan 30, 2013 at 1:45
  • $\begingroup$ @AllisonCameron yes, by (your) definition of the product of two ideals, that sum would be in the product. $\endgroup$
    – ferson2020
    Jan 30, 2013 at 1:48
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    $\begingroup$ Notice that my definition is equivalent, because if an ideal contains some elements, then it contains all their finite sums. :) $\endgroup$
    – pre-kidney
    Jan 30, 2013 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) $\sqrt{I \cdot J} = \sqrt{I \cap J}$.

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    $\begingroup$ I wished that abstract algebra books were written with this clarity. Clarity requires mastery of the subject. $\endgroup$ Sep 17, 2014 at 21:06
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    $\begingroup$ Wait, how is the intersection $(x,y)$? Since $(x^2,y^2)\subset (x,y)$, shouldn't $(x,y)\cap (x^2,y^2)=(x^2,y^2)$? $\endgroup$ Feb 15, 2018 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.

  • $\begingroup$ Why is the ideal $IJ$ not $\left\{\sum r_ia_ib_i \mid r_i \in R, a_i \in I, b_i \in J\right\}$? $\endgroup$ Jun 16, 2019 at 17:53
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    $\begingroup$ @AlJebr what is the point of writing $r_ia_i$ when it is already in $I$? $\endgroup$
    – rschwieb
    Jun 16, 2019 at 18:04
  • $\begingroup$ Why only finite sums? If one of the ideals is infinite, then one can form infinitely many sums of sort you mentioned. $\endgroup$
    – Dhiraj Rao
    Apr 13 at 12:24
  • $\begingroup$ @DhirajRao Each sum only involves finitely many nonzero terms. I am not talking about how many such sums exist. In the polynomial ring $\mathbb Q[x]$, each element is a finite sum of nonzero monomials, but of course there are infinitely many polynomials. $\endgroup$
    – rschwieb
    Apr 13 at 13:56

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


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