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Background

Definition: If the ideals $\mathfrak{a}=\langle m\rangle$ and $\mathfrak{b}=\langle n\rangle$ are both principal, i.e., generated by a single element, then their intersection $\mathfrak{a}\cap\mathfrak{b}=\langle \text{lcm}(m,n)\rangle$. For monomial ideals we get

$$\langle m_1,\ldots,m_r\rangle\cap \langle n_1,\ldots,n_s\rangle=\sum_{i=1}^{r}\sum_{j=1}^{s}\langle m\rangle\cap\langle n\rangle=\sum_{i=1}^{r}\sum_{j=1}^{s}\langle \text{lcm}(m_i,n_j) \rangle$$

Example (from the following youtube video at time index: 13:19 to 15:37:

$(x^2,y), (x,z)\subset \Bbb{C}[x,y,z]$

Intersection:

$(x^2,y)\cap (x,z)=\{p: p\in (x^2,y),p\in (x,z)\}$

$x^2\mid p\text{ or }y\mid p$

$\Rightarrow\; x\mid p \;\text{ either } xy\mid p \text{ or } yz\mid p$

$=(x^2,xy,yz)$

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Question

Should the example in the above screenshot from the cited youtube video be: $(x^2,y)\cap (x,z)=(x^2, x^2z, yx,yz)$ instead of $(x^2,xy,yz)$?

Thank you in advance

Seth
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  • $x^2 \in (x^2, y)$ and $x^2 \in (x, z)$, so it also must lie in the intersection, so $(x^3,x^2z,yx,yz)$ is clearly not correct. The answer from the video is right, what exactly is not clear for you in the their explanation? – Vladimir Lysikov Oct 06 '24 at 10:02
  • @VladimirLysikov how do I reconcile with the video lectures answer to the formula about the intersection of two ideals equal the LCM of the two of them? I must be not understanding something subtle here. I am not sure what though. – Seth Oct 06 '24 at 12:29
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    $\mathrm{lcm}(x^2, x) = x^2$, not $x^3$ – Vladimir Lysikov Oct 06 '24 at 12:35
  • @VladimirLysikov you are right about the $x^2$. I also notice in the video reasoning where he says "$x^2\mid p\text{ or }y\mid p$ $\Rightarrow; x\mid p ;\text{ either } xy\mid p \text{ or } yz\mid p$".

    Isn't $z\mid p$ missing in the line $x^2\mid p\text{ or }y\mid p$?

    According to his reasonings, if $c\mid (a,b)$ then $a\mid c$ or $b\mid p$, hence $x^2\mid p\text{ or }y\mid p \text{ or } z\mid p$. Then the rest of the argument proceed as it is.

     – Seth Oct 06 '24 at 17:27
  • @BillDubuque I am quoting a youtube video lecture, and I am not sure what the lecturer Zvi Rosen is discussing. That is why I am asking on here. Also is one thing to prove theorems about intersection of ideals, but it is another about computing it. – Seth Oct 06 '24 at 20:42
  • See the linked dupes for proofs of the claim about intersection of monomial ideals. – Bill Dubuque Oct 06 '24 at 20:44
  • @BillDubuque then if $p\in (x^2,y)$, if we can't conclude that $x^2\mid p$ or $y\mid p$, then what can one conclude about divisibility? – Seth Oct 06 '24 at 20:47
  • It is best to read the simple standard proof in the linked dupes rather than waste time trying to decipher some poorly presented argument. Is the author trying to give an example of using the claim or proving it? More context is needed. – Bill Dubuque Oct 06 '24 at 20:54
  • @BillDubuque he is trying to teach concepts from commutative algebra based on Atyiah and McDonald's text. The problem is, things in commutative algebra overlaps with abstract algebra, as far as some of the computational aspect of are concerned. – Seth Oct 06 '24 at 21:12
  • Likely it means this: It is easy to see that $,p,$ is in a monomial ideal iff each monomial of $,p,$ is, so wlog $,p,$ is a monomial. Let $I = (x^2,y)\cap (x,z).$ $,p\in (x^2,y),$ iff $,x^2\mid p,$ or $,y\mid p.,$ If $,\color{#c00}{x^2}\mid p,$ then $,p\in I,$ by $,x\mid x^2\mid p,$ so $,p\in (x,z).,$ Else $,y\mid p,$ so $,p\in I,$ iff $,p\in (x,z),$ iff $,x\mid p,$ or $,z\mid p,$ iff $,\color{#0a0}{xy\ ,{\rm or},\ yz}\mid p.,$ In summary: $,p\in I\iff p\in (\color{#c00}{x^2},\color{#0a0}{xy,yz})$. A general proof follows similarly - see the dupes. – Bill Dubuque Oct 06 '24 at 22:23

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