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We've been given the definition of intersection of ideals, that it is equal to:

$$\bigcap _{\alpha \in A}\:I_{\alpha }\:=\sum _{\alpha \in A}\:I_{\alpha }\:=\left\{a_{\alpha _1}+...+a_{\alpha _n}\::\:n\in \mathbb{N},\:a_{\alpha _i}\in I_{\alpha _i}\right\}$$

I see that those $a_{\alpha _i}$ are arbitrary elements of Ideals $I_{\alpha }$, but let's say that for every element $a_{\alpha _i}$ except $a_{\alpha _1}$ we have $a_{\alpha _i}$ = 0. Then $a_{\alpha _1}$ would be an element of $\bigcap _{\alpha \in A}\:I_{\alpha }$, but it doesn't have to be true that $\forall _{i\:}a_{\alpha _1}\in \:I_{\alpha _i}$.

So what's the deal here?

user
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Your doubts are completely justified. Take a very simple example: $\mathbb{Z}$ as the ring, $A=\{1,2\}$ and $I_1=(2),I_2=(3)$ the ideals generated by the primes $2$ and $3$ respectively. Clearly $I_1\cap I_2=(6)$, but $I_1+I_2=\{2x+3y:x,y\in\mathbb{Z}\}$. Obviously $I_1\cap I_2\subsetneq I_1+I_2$, since e.g. $2\in I_1+I_2$. So there must be something wrong with the "definition" that you got.

Peter Melech
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  • Asking out of curiosity: Is there a specific reason why we define a sum of Ideals in the way you did right now? Or did mathematicians just decide to define it in that way because it's the most intuitive way to describe a sum of Ideals? Does the same go for a product of ideals? – user Mar 13 '21 at 14:54
  • The sum as defined above forms an ideal, if you would try to define the product by $I_1I_2={xy:x\in I_1,y\in I_2}$ you would quickly realize this is not closed under addition, so you define it to be sums of products, see e.g. here – Peter Melech Mar 13 '21 at 15:00
  • I understand that, but what I don't understand is why didn't we just define the sum or product as something entirely different like $I_1I_2:=:\left{a^b:mod:\left(a+b\right):::a\in I_1,:b\in I_2\right}$ – user Mar 13 '21 at 15:04
  • If $a^b mod (a+b)$ makes sense in the given structure one might of course define the "product" or whatever in this way, but of course it would be counterintuitive to call it a product – Peter Melech Mar 13 '21 at 15:07
  • So we define it like we do because it's intuitive? – user Mar 13 '21 at 15:08
  • Yes! In general definitions should be made out of some intuition I think. – Peter Melech Mar 13 '21 at 15:09
  • Shouldn't it rather be that $\bigcap :{\alpha :\in :A}:I{\alpha :}:=\sum :{i=1}^n:\left(a{\alpha :{i,1}}\cdot ...\cdot a{\alpha :_{i,n}}\right)$? – user Mar 13 '21 at 15:27
  • The intersection of ideals is the intersection of the sets $\bigcap_{\alpha\in A}I_{\alpha}={x\in R:x\in I_{\alpha}\forall\alpha\in A}$ and turns out to be an ideal. The ideal containing all sums of products ( what the right hand side seems to mean) is the product of the ideals $\Pi_{\alpha\in A}I_{\alpha}$ – Peter Melech Mar 13 '21 at 15:35
  • So why is it true that $IJ:\subset :I:\cap :J$? We explained it because $xy:\in :I:\cap :J$ where x is element of $I$, and y is element of $J$. – user Mar 13 '21 at 15:37
  • Yes by the ideal property and because it is closed under addition – Peter Melech Mar 13 '21 at 15:39
  • But I don't understand why $xy:\in :I:\cap :J$? Let's say $I=\left{2,:4\right}$ and $J=\left{3,:4\right}$. $I:\cap ::J:=:\left{4\right}$, but 2*3 = 6 – user Mar 13 '21 at 15:47
  • If $I,J$ are two-sided ideals in a ring $R$ then $y\in J$ implies $ry\in J$ for all $r\in R$ and $x\in I$ implies $xr\in I$ for all $r\in R$ and thus $xy\in I\cap J$. The sets you consider do not form ideals in $\mathbb{Z}$, if we stick to this example for a moment – Peter Melech Mar 13 '21 at 15:55
  • So basically if ry is element of J and should be element of the intersection of I and J, and the same goes if xr is element of I and should be element of the intersection of I and J, a good example would be if we just take r=x (or r=y), right? But we have no guarantee that x (y) exists in the intersection of I and J – user Mar 13 '21 at 16:06
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    Yes we have a guarantee because $xy\in J$ because $y\in J$ and $xy\in I$ because $x\in I$, thus $xy\in I\cap J$. – Peter Melech Mar 13 '21 at 16:09