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I am trying to find Spec $\mathbb Q[x]$. Since $\mathbb Q$ is a field, the prime ideal is just generated by irreducible polynomial with coefficients in $\mathbb Q$.

I know the case of $\mathbb R$, where the irreducible polynomial has degree less than 3. Is this still true for $\mathbb Q$? What is the spectrum exactly?

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    I don't know what kind of answer you expect when asking 'what the spectrum is' but in any case: it's the prime ideals. As $\mathbb{Q}$ is a field, $\mathbb{Q}[X]$ is a principal ideal domain, hence, every non-zero prime ideal is maximal and generated by an irreducible polynomial. However, unlike $\mathbb{R}[X]$, $\mathbb{Q}[X]$ contains irreducible polynomials of arbitrary degree, as Jendrik Stelzner pointed out. – Bib-lost Jan 05 '16 at 22:23
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    $\mathbb{Q}[X]$ contains irreducible polynomials of every degree $d \geq 1$, for example $X^d - 2$. – Jendrik Stelzner Jan 05 '16 at 22:24
  • @Bib-lost I am confused too actually, maybe what Jendrik pointed out is the answer...... –  Jan 05 '16 at 22:29
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    @Learning I wrote something about the points in $Spec \mathbb{Q}[x]$ forming groupoids. But it got locked. The main idea is that for every irreducible polynomial there's a transitive action of the galois group of that polynomial on the roots. So in a sense points in $Spec \mathbb{Q}[x]$ are actually "galois orbits". Notice that this is true for non-algebraically closed points in any scheme. – Saal Hardali Jan 05 '16 at 23:13

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