Explaining the product of two ideals

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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

Answer 4

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) $\sqrt{I \cdot J} = \sqrt{I \cap J}$.

Answer 5

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.

Answer 6

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

Answer 7

Here's an example that goes all the way back to Dedekind's favorite pedagogical example of failure of unique factorization in number rings: $(2+\sqrt{-5})(2-\sqrt{-5}) = 9 = (3)(3)$ in $ℤ[\sqrt{-5}]$. The ideals $I=(3,2+\sqrt{-5})$ and $J=(3,2-\sqrt{-5})$ are non-principal ideals with product $(3)$, but if we just multiply elements of $I$ by elements of $J$ we get a proper subset of $(3)$ that's missing 3 itself.

To see why 3 can't be written as an element of $I$ times an element of $J$, it's enough to look at elements of $I$ and $J$ of small norm (where the norm of $a+b\sqrt{-5}$ is $a^2+5b^2$). The nonzero elements of $I$ of smallest norm are $1+\sqrt{-5}$ (with norm 6), $2-\sqrt{-5}$ (with norm 9), and 3 (with norm 9). Likewise, a nonzero element of $J$ must have norm 6 or norm at least 9. So there's no way to multiply nonzero elements of $I$ and $J$ to get something of norm 9, specifically 3.