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$I=(2,x)$ and $J=(3,x)$ be ideals.

$I\cdot J=(6,x)$.

Because $6=2\cdot 3$ and $x=3\cdot x-2\cdot x$, does this means that $I\cdot J$ contains $(6,x)$ or is it the inverse?

Bill Dubuque
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Enzo Creti
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    Well, the last sentence of the post means that $I \cdot J \supset (6,x)$. To deduce that $I \cdot J \subset (6,x)$ we can check that all products of generators of $I$ and $J$ are in $(6,x)$. – radekzak May 16 '22 at 19:46
  • @user26857 6 and x can be expressed as a linear combination of elements of J and I? Is it so? – Enzo Creti May 16 '22 at 19:47
  • I think you proved this. In fact, they are linear combinations of products of elements from $I$ and $J$. – user26857 May 16 '22 at 19:48
  • Also it is usually nice to mention what ring you are working with since the ideal generated in $\mathbb{Q}$ for example would be $(1)$ – MIO May 16 '22 at 22:10

2 Answers2

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$IJ=(2\cdot 3,2x,3x,x^2)$. But $x=3x-2x\in IJ$, so $IJ=(6,2x,3x,x^2,x)=(6,x)$.

user26857
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Expanding the product ideals below using ideal arithmetic (esp. distributive law)

yields $\,(2,x)(3,x) = (6,2x,3x,xx) = (6,x(\color{#c00}{2,3},x)) = (6,x)\,$ by $\,\color{#c00}{(2,3)=1}$

Generally, $ $ as for gcds, $ $ $(a,c)(b,c) = (ab,c(\color{#c00}{a,b,c})) = (ab,c)\,$ if $\,\color{#c00}{(a,b,c) = 1}$

Note on notation: when we view $\:\!\color{#c00}1\:\!$ an ideal (e.g. in above ideal equalities), we use the convention that this denotes the unit ideal $\:\!\color{#c00}{(1)}.\,$ This allows us to write proofs like above that work uniformly for both ideals and gcds, e.g. see these proofs of Euclid's Lemma (cf. comments below).

Bill Dubuque
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  • The proof shows more generally $,(a,b,c)=1\Rightarrow (a,c)(b,c) = (ab,c),\ $ [same for gcds] – Bill Dubuque May 16 '22 at 20:07
  • @Bill Dubuque so if the ideals were (2,x) and (4,x) because the gcd(2,4) is not one, it woul d have been different? – Enzo Creti May 16 '22 at 21:29
  • @pollaris See my prior comment. In particular, $(a,b,c) = (2,4,x)\neq (1)$ so the inference my first comment does not apply to your example. It should be clear from the proof that "$1$" denotes the unit ideal. – Bill Dubuque May 17 '22 at 01:49