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In $\mathbb{Q}[x]$, is the ideal $\langle x^2-4, x^3-8 \rangle$ equal to $\langle (x-2)(x+2)(x^2+2x+4)\rangle$?

Since $x^2-4=(x-2)(x+2)$ and $x^3-8=(x-2)(x^2+2x+4)$, then can we take the ideal generated by $x^2-4$ and $x^3-8$ to be the ideal generated by $(x-2)(x+2)(x^2+2x+4)$?

In general, is the ideal $\langle x,y \rangle$ equal to $\langle lcm(x,y) \rangle$? Or this only works when $x$ and $y$ are coprime?

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    $x^2 - 4$ is in $\langle x^2 - 4, x^3 - 8 \rangle$, but not in $\langle (x - 2)(x + 2)(x^2 + 2x + 4) \rangle$... – 2017-02-03
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    It should be gcd instead of lcm in the last sentence, so $\langle x^2-4,x^3-8\rangle$ does actually equal $\langle x-2\rangle$. – 2017-02-03
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    When $x, y$ are coprime, $\operatorname{lcm}(x, y) = xy$. So if it only worked then, there would be little point in even mentioning $\operatorname{lcm}$. – 2017-02-03
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    No, I mean x,y live the initial ring which is the polynomials over Q in one variable (or any other PID). – 2017-02-03

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No. In a P.I.D., $\langle a,b\rangle=\langle\gcd(a,b)\rangle$.

So here $\langle x^2-4 ,x^3-8\rangle=\langle (x-2)(x+2) ,(x-2)(x^2+2x+4)\rangle=\langle x-2\rangle$.

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    It's clear to me that $\langle x^2-4, x^3-8 \rangle \subset \langle x-2\rangle$, but how does the reverse inclusion happen? – 2017-02-03
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    For this particular case, without using many general theorems, you can see that $\frac{1}{4}(x^3-8) - \frac{x}{4}\cdot (x^2-4) = (x-2)$. – 2017-02-03
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    It's a general theorem on P.I.Ds. Its proof relies on a Bézout's relation: $x-2$ can be written as $ u(x^2-4)+v(x^3-8)$ for some coefficients $u(x)$ and $v(x)$. – 2017-02-03
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No, $$4x-8\in (x^2-4,x^3-8)$$ but not in the other ideal.

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    $x^2 - 4$ is a more obvious example. – 2017-02-03
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Hint $\ $ In a PID we have that $\ (a,b) = (\gcd(a,b))\ $ since

$$\begin{align} &(c) \supseteq (a)\!+\!(b)\\[.2em] \iff\ &(c) \supseteq (a),(b)\\[.2em] \iff\ &\ c\ \ \mid\ \ \ a,\ \ b\\[.2em] \iff\ &\ c\mid \gcd(a,b))\\[.2em] \iff\ & (c) \supseteq (\gcd(a,b)) \end{align}$$