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The product of two ideals is defined as the set of all finite sums $\sum f_i g_i$, with $f_i$ an element of $I$, and $g_i$ an element of $J$. I'm trying to think of an example in which $IJ$ does not equal $I \cap J$.

I'm thinking of letting $I = 2\mathbb{Z}$, and $J = \mathbb{Z}$, and $I\cap J = 2\mathbb{Z}$? Can someone point out anything faulty about this logic of working with an even ideal and then an odd ideal?

Thanks in advance.

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    $\begingroup$ But then $IJ = 2\mathbf Z$, so this doesn't quite work. In fact, $I \cap J = IJ$ will hold whenever $I + J$ is the whole ring, so you should try to avoid that. $\endgroup$ – Dylan Moreland May 29 '12 at 21:31
  • $\begingroup$ The only thing faulty is that it doesn't do what you want it to do. $IJ=I\mathbb{Z}=I=I\cap J$. $\endgroup$ – rschwieb May 29 '12 at 21:31
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    $\begingroup$ Since you seem to have difficulty with the topic, it would be a good exercise to explicitly calculate $IJ$ and $I \cap J$, and if you have trouble to ask that here. $\endgroup$ – user14972 May 29 '12 at 21:31
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    $\begingroup$ Also, I don't know what even ideal and odd ideal mean. The set of odd numbers $1 + 2\mathbf Z$ is a coset, but not an ideal (it definitely isn't closed under multiplication by even numbers). $\endgroup$ – Dylan Moreland May 29 '12 at 21:38
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But in your example $IJ = I = I \cap J$. How about taking $I = J = 2 \mathbb{Z}$. Then $I \cap J = 2\mathbb{Z}$ while $IJ = 4\mathbb{Z}$. In general $IJ \subset I \cap J$, and the equality holds if $I+J = R$ where $R$ is the ring you are working with.

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  • $\begingroup$ @Dylan Moreland: Thanks for your comment! I just edited my answer. $\endgroup$ – algebra_fan May 29 '12 at 21:41
  • $\begingroup$ @algebra_fanCan you say something about what conditions need to be satisfied by $I$ and $J$ so that, as you say, $I + J = R$. Thanks $\endgroup$ – user12802 May 29 '12 at 22:02
  • $\begingroup$ @Andrew But $I+J = R$ is a condition on $I$ and $J$! In general, $I+J$ is an ideal of $R$ which may or may not be equal to $R$.An equivalent way to state this condition is that $1 = i + j$ for some $i \in I$ and some $j \in J$. That is, $I+J=R$ if and only if you can write $1$ as a sum of an element of $I$ and an element of $J$. $\endgroup$ – algebra_fan May 30 '12 at 6:22
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    $\begingroup$ @Andrew As an example, $2\mathbb{Z} + 3\mathbb{Z} = \mathbb{Z}$ because we can write $1 = 4 + (-3)$ and the first summand belongs to $2 \mathbb{Z}$ and the second one belongs to $3 \mathbb{Z}$. On the other hand, $4\mathbb{Z} + 6 \mathbb{Z} \neq \mathbb{Z}$. Any element of $4 \mathbb{Z} + 6\mathbb{Z}$ is of the form $4x+6y$ for some integers $x,y$; in particular, it is always even so it can never be equal to $1$. I hope this helps. $\endgroup$ – algebra_fan May 30 '12 at 6:26
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    $\begingroup$ It should be noted that this argument works when $R$ is commutative: Let $x\in I\cap J$. Then since $\exists i\in I,\exists j\in J: 1=i+j,$ we have $x=x1=x(i+j)=xi+xj=ix+xj\in IJ$. For further reference see Lang's Algebra (3e), p. 87. $\endgroup$ – Alp Uzman Mar 5 '17 at 21:48
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In a PID, we have $(a) \cap (b) = (\mathrm{lcm}(m,n))$, whereas we have $(a) \cdot (b) = (a \cdot b)$. So this ideal-theoretic question becomes a number-theoretic one and we see: $(a) \cap (b) = (a \cdot b)$ iff $a \cdot b$ is a lcm of $a$ and $b$ iff $a,b$ are coprime.

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Maybe it is helpful for you to realise what really happens for ideals in the integers. You probably know that any ideal in $\mathbb Z$ is of the form $(a)$ for $a\in \mathbb Z$, i.e. is generated by one element. The elements in $(a)$ are all integers which are divisible by $a$.

If we are given two ideals $(a)$ and $(b)$, their intersection consists of those numbers which are divisible by $a$ and divisible by $b$. Their product consists of all numbers which are divisible by the product $ab$.

If $a$ and $b$ are coprime they are the same. E.g. all numbers which are divisible by $2$ and $3$ are also divisible by $6$, and vice versa. If they are not coprime the situation changes. If a number is divisible by $4$ and $2$, then it is not necessarily divisible by $8$.

Another way of saying that two integers $a$, $b$ are coprime is that there exist $x,y$, such that $xa+by=1$ (cf. Euclidean algorithm). In the language of ideals this translates to $(a)+(b)=\mathbb Z$ and the circle closes.

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  • $\begingroup$ Just in case the notation is new to you. What I denote by $(a)$ is $a\mathbb Z$ in your notation. $\endgroup$ – Simon Markett May 29 '12 at 22:44
  • $\begingroup$ Thanks for the clarification. The notation is new $\endgroup$ – James R. May 29 '12 at 22:46
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Pick any ring with a nonzero ideal $I$ such that $I^2=0$. Then $I=I\cap I\neq I*I=\{0\}$.

A concrete example of this is $I=2\mathbb{Z}/4\mathbb{Z}$ in the ring $R=\mathbb{Z}/4\mathbb{Z}$.

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