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I am wondering how to prove the following statement:

Let $R$ be a PID, $a,b$ are relatively prime. Then $\langle a\rangle \cap \langle b\rangle = \langle ab\rangle$


Progress: I think it holds when $a$ or $b$ is $0$. Or $a$ or $b$ is a unit. And of course $\langle ab\rangle\subseteq \langle a\rangle \cap \langle b\rangle $. Then I got stuck.

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  • $\begingroup$ What have you tried? Do you see how to prove inclusion in one of the directions? $\endgroup$ – Tobias Kildetoft Apr 29 '15 at 12:15
  • $\begingroup$ Sorry...I correct it now. $\endgroup$ – Allitee Apr 29 '15 at 12:15
  • $\begingroup$ I think it's hold when a or b is 0. Or a or b is a unit. And of course <ab> is in <a> intersect <b>. Then I got stuck. $\endgroup$ – Allitee Apr 29 '15 at 12:18
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Since $a$ and $b$ are relatively prime, there exist $x$, $y \in R$ such that $xa+yb=1$. Now let $c \in \langle a \rangle \cap \langle b \rangle$. Then there exist $z_1, z_2 \in R$ such that $c=z_1a=z_2b$. We now have $$c=1c=(xa+yb)c=xac+ybc=xz_2ba+yz_1ab=(xz_2+yz_1)ab \in \langle ab \rangle.$$

Note that we did not use here that $R$ is a principal ideal domain, only that it is commutative.

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    $\begingroup$ I'm not convinced. $\endgroup$ – ThorbenK Apr 29 '15 at 13:29
  • $\begingroup$ Do you require R to be Euclidean Domain to say there exist x, y such that xa + yb = 1? $\endgroup$ – Allitee Apr 30 '15 at 12:14
  • $\begingroup$ No. In fact, this follows from the definition of $x$ and $y$ being relatively prime. $\endgroup$ – Orlando Marigliano May 5 '15 at 10:52
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Before I give you a solution, its important you know what a generated ideal really means.

Let $R$ be a ring. I am sure you know the definition of an ideal, but what do we mean when we say a "generated" ideal?

Let $J$ = {$j_{1},j_{2}, \dots j_{s}$} be a subset of $R$, then the set defined as follows:

$$\sum_{i=1}^{s} h_{i}.j_{i}$$ where $h_{i} \in R$, $j_{i} \in J$ is infact an ideal in $R$, (It is a good exercise to show that this set really is an ideal) and we denote it as $I = \langle J \rangle$ and say that this ideal is "generated" by $J$. So a principal ideal is really just a special case of a generated ideal where the $J$ (our generating set) has just one element. Don't worry too much if you don't fully grasp the above definition. What's important is that a PID is an ideal generated by a single element. Now in response to your question:

We want to show that $\langle a \rangle \bigcap \langle b \rangle = \langle ab \rangle$, where $a,b$ are co-prime. These ideals are essentially sets and whenever you want to show two sets are equal you have to show each one is a subset of the other. So you have already shown that $\langle ab \rangle \subset \langle a \rangle \bigcap \langle b \rangle$. It only remains to show the other inclusion.

Before you proceed, think about what the intersection really means. The PID generated by $a$ is the set of all elements which are multiples of $a$ and the PID generated by $b$ is the set of all elements which are multiples of $b$ and their intersection means that you have to find a set of elements which are both multiples of $a$ and $b$, but you have said that $a,b$ are co-prime which means that this is only possible when we consider multiples of $ab$. (If this is hard for you to visualise, try a few examples out, take $a = 3, b = 5$ in the ring of $\mathbb{Z}$)

So what we have proved is that any element in $\langle a \rangle \bigcap \langle b \rangle$ is a multiple of $ab$ and therefore also belongs to $\langle ab \rangle$, hence $\langle a \rangle \bigcap \langle b \rangle \subset \langle ab \rangle$ and we have equality.

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