Let $I=(x, 2)$ and $J=(x, 3)$ be ideal in the ring $\Z[x]$.

(a) Prove that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

Hint.

If $I=(a_1,\dots, a_m)$ and $J=(b_1, \dots, b_n)$ are ideals in a commutative ring, then we have
\[IJ=(a_ib_j),\] where $1\leq i \leq m$ and $1\leq j \leq n$.

Proof.

(a) Prove that $IJ=(x, 6)$.

Note that the product ideal $IJ$ is generated by the products of generators of $I$ and $J$, that is, $x^2, 2x, 3x, 6$. That is, $IJ=(x^2, 2x, 3x, 6)$.

It follows that $IJ$ contains $x=3x-2x$ as well. As the first three generators can be generated by $x$, we deduce that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

Assume that $x=f(x)g(x)$ for some $f(x)\in I$ and $g(x) \in J$.
As $\Z[x]$ is a UFD, we have either
\[f(x)=\pm x, g(x)=\pm 1, \text{ or } f(x)=\pm 1, g(x)=\pm x.\]

In the former case, we have $1\in J$ and hence $J=\Z[x]$, which is a contradiction.
Similarly, in the latter case, we have $1\in I$ and hence $I=\Z[x]$, which is a contradiction.
Thus, in either case, we reached a contradiction.

Hence, $x$ cannot be written as the product of elements in $I$ and $J$.

Comment.

Let $I$ and $J$ be an ideal of a commutative ring $R$.
Then the product of ideals $I$ and $J$ is defined to be
\[IJ:=\{\sum_{i=1}^k a_i b_i \mid a_i\in I, b_i\in J, k\in \N\}.\]

The above problems shows that in general, there are elements in the product $IJ$ that cannot be expressed simply as $ab$ for $a\in I$ and $b\in J$.

The post Example of an Element in the Product of Ideals that Cannot be Written as the Product of Two Elements first appeared on Problems in Mathematics.

Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:

(a) The ideal $P$ is a prime ideal.

(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.

Proof.

(a) $\implies$ (b)

Suppose that $P$ is a prime ideal. Let $I$ and $J$ be ideals such that $IJ \subset P$. Assume that
\[I \not \subset P \text{ and } J \not \subset P.\] Then there exist
\[a \in I \setminus P \text{ and } b\in J \setminus P.\]

Then the element $ab$ is in both $I$ and $J$ since $I, J$ are ideals. Then we have
\[ab \in IJ \subset P\] and this implies either $a \in P$ or $b\in P$ since $P$ is a prime ideal.

However, this contradicts the choice of the elements $a, b$.
Therefore, we must have
\[I \subset P \text{ or } J \subset P.\]

(b) $\implies$ (a)

Now we assume statement (b) is true.
Suppose that $ab \in P$, where $a, b \in R$.
Let $I=(a)$, $J=(b)$ be ideals generated by $a$ and $b$, respectively.

Then we have
\[IJ=(ab)\subset P\] since $ab \in P$, and statement (b) implies that we have either $(a)=I\subset P $ or $(b)=J \subset P$.

Hence we have either $a \in P$ or $b\in P$.
Thus $P$ is a prime ideal.

The post Equivalent Conditions For a Prime Ideal in a Commutative Ring first appeared on Problems in Mathematics.