Show that the set of products need not be an ideal

Question

Let $R$ be a commutative ring, and $I$, $J$ be two ideals in $R$.

Show, by giving an explicit example, that the set of products $\{xy\ |\ x\in I, y\in J\}\subset R$ need not be an ideal.

The definition of "ideal" we're using is:

An ideal is a subset $I$ of a commutative ring $R$ that satisfies the following properties:

1) $0\in I$.

2) If $x\in I$ and $y\in I$, then $x+y\in I$.

3) If $x\in I$ and $a\in R$, then $ax\in R$.

The set of products will contain $0$, so property 2 or 3 has to fail. I can't come up with an example. Any suggestions?

I think $\mathbb{Z}$ already contains an example. Can you guess it? Hint: use the closed under addition property — Alex Vong, Mar 11 '17 at 01:55
@HenningMakholm Yes I was wrong. ${ax \mid x \in \mathbb{Z}}$ and ${bx \mid x \in \mathbb{Z}}$ gives you ${abx \mid x \in \mathbb{Z}}$, which is an ideal. — Alex Vong, Mar 11 '17 at 02:00

score 2 · Answer 1 · answered Mar 11 '17 at 01:57

2

Consider $k[x,y,z,t]$ where $k$ is a field, $I$ is the ideal generated by $x,y$ and $J$ is the ideal generated by $z,t$. $xz,yt\in IJ$ but $xz+yt$ is not in $I.J$.

answered Mar 11 '17 at 01:57

Tsemo Aristide

89,587

If you just have x and y, and I=(x) and J=(y) you can do the same, no? – Mariano Suárez-Álvarez Mar 11 '17 at 02:04
@MarianoSuárez-Álvarez: Not with principal ideals (in a commutative ring with unity); there you always have $(a)(b)=(ab)$. – hmakholm left over Monica Mar 11 '17 at 02:15
Why xz + yt isn't in I.J ? Thanks in advance – Pritam Roy Dec 17 '22 at 14:10

score 2 · Answer 2 · answered Mar 11 '17 at 02:15

Let $R,I,J$ be given by

\begin{align*} R&=\mathbb{Z}[s,t]\\[3pt] I&=(s,t)\\[3pt] J&=(s,t)\\[3pt] \end{align*}

Then, letting $A = \{xy\ |\ x\in I, y\in J\}$, we have

\begin{align*} &s^2 \in A\\[3pt] &t^2 \in A\\[3pt] &s^2 + t^2 \notin A \\[3pt] \end{align*}

hence $A$ is not an ideal of $R$.

Show that the set of products need not be an ideal

2 Answers2