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I know that in general if $R[u]$ is the ring obtained by adjoining an element $u$ to a ring $R$, then $R[u]\cong R[x]/I$ for some ideal $I$ such that $I\cap R=\{0\}$.

In a particular instance, I'm working with $u=\sqrt{2}+\sqrt{3}$, so I'm wondering if there is a concrete description of an ideal $I$, say generated by some polynomials of $\mathbb{Q}[x]$ or something like that, such that $\mathbb{Q}[x]/I\cong\mathbb{Q}[u]$.

It's clear there is a surjective homomorphism from $\mathbb{Q}[x]\to\mathbb{Q}[u]$, so I would take the kernel, which is just the set of all polynomials with $u$ a root. Is this ideal generated by anything easy to write down? Or is that the best description I can give?

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Just to note, for any Noetherian ring $R$ the ring $R[x_1,\ldots,x_n]$ is Noetherian so the ideal $I$ you mention is always finitely generated. –  Alex Becker Jul 7 '12 at 7:58
    
you would just calculate the minimal polynomial, no? –  Holdsworth88 Jul 7 '12 at 8:43

2 Answers 2

up vote 6 down vote accepted

Let $p(x)$ be the minimimal polynomial of $\sqrt{2}+\sqrt{3}$. Then $I=(p(x))$. According to Wolfram|Alpha, $p(x)=x^4-10x^2+1$.

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We don't need WA for this: $$x=\sqrt 2+\sqrt 3\Longrightarrow x^2=5+2\sqrt 6\Longrightarrow (x^2-5)^2=(2\sqrt 6)^2\Longrightarrow$$ $$\Longrightarrow x^4-10x^2+25=24\Longrightarrow x^4-10x^2+1=0$$

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You still have to prove that there is no lower-degree polynomial equation that is satisfied by $\sqrt{2}+\sqrt{3}$. In other words, why isn't $\sqrt{2}+\sqrt{3}$ rational or a quadratic irrational? –  lhf Jul 7 '12 at 13:27
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Piece of cake: just write the pol. as a product of two real non-rational polynomials using the quadratic equation formula and the fact that the above pol. is biquadratic:$$x^4-10x^2+1=\left[x^2-(5+2\sqrt 6)\right]\left[x^2-(5-2\sqrt 6)\right]$$ –  DonAntonio Jul 7 '12 at 14:10

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