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Prove that $$\langle 4,2x,x^{2} \rangle=\langle 4,x \rangle\cap \langle 2,x^{2} \rangle $$ $$\langle 9,3x+3 \rangle=\langle 3 \rangle\cap \langle 9,x+1\rangle$$ are two primary decomposition in $\mathbb{Z}\left [ x \right ]$.

I put my answer below, that will be great if you check it and if you see mistakes please make me aware, and if you think it is right so tell me, I think my answer is right but I have doubts.

First I show that we have the equality: $$f(x)\in \left \langle 4,2x,x^{2} \right \rangle \Rightarrow f(x)=4f_{1}(x)+2xf_{2}(x)+x^{2}f_{3}(x)\Rightarrow f(x)=4(f_{1}(x))+x(2f_{2}(x)+xf_{3}(x)), f(x)=2(2f_{1}(x)+xf_{2}(x))+x^{2}f_{3}(x)\Rightarrow f(x)\in \left \langle 4,x \right \rangle\cap \left \langle 2,x^{2} \right \rangle$$for the inverse $$f(x)=2f_{1}(x)+x^{2}f_{2}(x)=4f_{3}(x)+xf_{4}(x)\Rightarrow f_{1}(x)=2f_{3}(x)\Rightarrow f(x)=4f_{3}(x)+x^{2}f_{2}(x)$$now I show that $\left \langle 4,x \right \rangle, \left \langle 2,x^{2} \right \rangle $are primary:$$1\notin \left \langle 4,x \right \rangle\Rightarrow \left \langle 4,x \right \rangle \neq \mathbb{Z}\left [ x \right ]$$$$f(x)g(x)\in \left \langle 4,x \right \rangle ,f(x)\notin \left \langle 4,x \right \rangle\Rightarrow f(x)g(x)=4f_{1}(x)+xf_{2}(x)\Rightarrow f(0)g(0)=4f_{1}(0)\Rightarrow 4\mid f(0)g(0),f(x)\notin \left \langle 4,x \right \rangle\Rightarrow 4\mid g(0)\Rightarrow g(x)=4k(x)+xh(x)$$ the other ones are the same.

kpax
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2 Answers2

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According to the post and comment, it suffices to show $\supseteq$.

Let $z \in (2,x^2) \cap (4,x)$. We may write $z = a2 + bx^2$ for some $a,b \in \mathbb{Z}[x]$. Since $a2 + bx \in (4,x)$ implies $a2 \in (4,x)$, there exists $c, d \in \mathbb{Z}[x]$ such that $a2 = c4 + dx$. Since $a2 \in (2)$, $dx \in (2)$. We have $d = 2d'$ for some $d' \in \mathbb{Z}[x]$ because $2$ is a prime element which does not divide $x$. Since $\mathbb{Z}[x]$ is a domain, $a2 = c4 + 2d'x$ implies $a = 2c + d'x$. Therefore, one has $z = a2 + bx^2 = 4c + 2d'x + bx^2 \in \hbox{LHS}$.

Youngsu
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  • By the way, one can also show $\sqrt(2,x^2) = (2,x)$ to say $(2,x^2)$ is $(2,x)$-primary since $(2,x)$ is maximal. – Youngsu Dec 23 '13 at 09:48
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I would avoid using so many variables. I don't see what is gained from it, and it seems to make it hard to spot mistakes.

All of these ideals can be explained more simply by describing intrinsic properties of the polynomials that belong to them. For example:

$$f\in \left \langle 4,x \right \rangle \iff f\in\mathbb{Z}[x]\text{ has constant term divisible by $4$.}$$

$$f\in \left \langle 2, x^2 \right \rangle \iff f\in\mathbb{Z}[x]\text{ has constant and linear terms divisible by $2$.}$$

Doesn't it seem more straightforward to compute their intersection now?

Andrew Dudzik
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