Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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, $p(x)=x^4-10x^2+1$.

share|cite|improve this answer

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

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.