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Can someone point me towards a resource that proves that the spectrum of $\mathbb{Z}[x]$ consists of ideals $(p,f)$ where $p$ prime or zero and $f$ irred mod $p$? In particular I remember this can be proved simply using localizations, but can't quite remember how to do it! I definitely don't want a link to a long involved argument about polynomials, I can find quite enough of those!

Many thanks in advance.

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  • $\begingroup$ @Edward: IMO the most difficult part of the argument is that if $I$ is a non-zero prime ideal in $\mathbb{Q}[x]$, then it contains a generator $g$ such that $(g)$ is a prime ideal of $\mathbb{Z}[x]$. This requires some non-trivial facts about polynomial rings: e.g. the fact that $\mathbb{Z}[x]$ is a unique factorization domain, and what the primes elements are. $\endgroup$ – Hurkyl Oct 25 '12 at 22:56
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    $\begingroup$ @Hurkyl: It is Gauss' Lemma. $\endgroup$ – Martin Brandenburg Oct 26 '12 at 7:41
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    $\begingroup$ There's a wonderful picture by Mumford on this: neverendingbooks.org/index.php/mumfords-treasure-map.html $\endgroup$ – only Oct 27 '12 at 14:23
  • $\begingroup$ Related: math.stackexchange.com/questions/199990/… $\endgroup$ – Watson Jan 7 '17 at 17:13
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Here is a geometric (schemes!) way to think about it.

The inclusion $\Bbb Z\to\Bbb Z[x]$ defines a morphism $\operatorname{Spec}(\Bbb Z[x])\to\operatorname{Spec}(\Bbb Z).$ Thus to figure out the primes of $\operatorname{Spec}(\Bbb Z[x]),$ we can simply determine all the fibres of this map. How do we compute the fibres of this map?

For $\langle p\rangle\subseteq\Bbb Z$ a prime ideal, we pull back the morphism given above over the map $\operatorname{Spec}(\kappa(p))\to\operatorname{Spec}(\Bbb Z)$ induced by $\Bbb Z\to\Bbb Z_p/\frak{m_p}$ $=\kappa(p)$, where $\kappa(p)$ is the residue field of $p.$ The residue field $\kappa(0)=\Bbb Q$ and for all other primes $p$ we have $\kappa(p)=\Bbb F_p.$

The fibre over $\langle 0\rangle$ is thus $\operatorname{Spec}(\Bbb Q\otimes_{\Bbb Z}\Bbb Z[x])=\operatorname{Spec}(\Bbb Q[x])$ which is all irreducible polynomials over $\Bbb Q$ and the zero ideal. Similarly, the fibre over $\langle p\rangle$ is $\operatorname{Spec}(\Bbb F_p[x]),$ which is just the irreducible polynomials over $\Bbb F_p$ along with its zero ideal. (The zero ideals correspond to those in $\Bbb Z.$)

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The prime ideals of $\mathbb Z[x]$ are of three kinds depending on their heights

  1. (height $0$): $\{ 0\}$;

  2. (height $1$): $F(x)\mathbb Z[x]$ with $F(x)$ an irreducible element in $\mathbb Z[x]$. Equivalently: $F(x)$ is a prime number $p$ or is primitive and irreducible in $\mathbb Q[x]$.

  3. (height $2$, maximal ideals): the $(p, f(x))$ as you describe.

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The intersection of a prime ideal with $\mathbb Z$ is again prime, thus we obtain a prime $p$ (or 0). By localizing at $p$, we make all non-multiples of $p$ invertible and are left with an ideal in the principle ideal ring $\mathbb Z_p[X]$, i.e. $(f)$ with $f\in\mathbb Z_p[X]$. If we had a nontrivial factorization $f\equiv gh\pmod p$, this could be lifted to a factorization in $\mathbb Z_p[X]$, which is impossible. Hence $f$ is irreducible $\bmod p$. This also holds if we replace $f$ with an approximation in $\mathbb Z[X]$. We also see that any $g$ in the ideal becomes a multiple of $f$ in $\mathbb Z_p[X]$, hence can be written as a multiple of $p$ plus a multiple of $f$ in $\mathbb Z[X]$.

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  • $\begingroup$ I can't quite see the argument for when the prime ideal is (0). Could you possibly clarify it? Many thanks! $\endgroup$ – Edward Hughes Oct 25 '12 at 22:40
  • $\begingroup$ How is $\mathbb Z_p[X]$ a principal ideal domain? How about the ideal $(p,X)$? I don't think this ideal is principal, since if $f(X)$ divides both $p$ and $X$, dividing $p$ implies having degree $0$ and dividing $X$ while being of degree $0$ means being a unit. $\endgroup$ – Patrick Da Silva Jan 9 '14 at 0:36
  • $\begingroup$ Any polynomial ring F[x] over a field is a PID $\endgroup$ – rschwieb Jan 9 '14 at 3:41
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There is a sketch here but I didn't proofread it. I am gambling that it is useful, so I apologize (and will delete this) if it turns out to be useless.

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  • $\begingroup$ It was useful for me. Thanks! $\endgroup$ – Patrick Da Silva Jan 8 '14 at 23:45
  • $\begingroup$ Although after thinking about it I think it is wrong ; Z[x] is not an Euclidean domain, the Euclidean algorithm doesn't work. How could you compute the gcd of $x^2 - 1$ and $2x$? $\endgroup$ – Patrick Da Silva Jan 9 '14 at 0:00
  • $\begingroup$ Did you post this under the wrong thread or something? Z[x] is definitely not Euclidean because it's not a PID $\endgroup$ – rschwieb Jan 9 '14 at 3:39
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    $\begingroup$ @PatrickDaSilva You can perform the extended Euclidean algorithm first in $\Bbb Q[x]$, and then multiply through by the gcd of any fractions that appear. If the two polynomials were coprime in $\Bbb Q[x]$, then they'll have linear combination in $\Bbb Z[x]$ that's an integer. $\endgroup$ – rschwieb Jan 10 '14 at 14:01
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    $\begingroup$ @user26857 Replaced with a wayback link. Thanks $\endgroup$ – rschwieb Mar 2 at 12:54
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There are two cases: if a prime ideal $P$ of $Z[x]$ does not contain any constant then considering the ideal generated by $P$ in $Q[x]$, it is a proper ideal of $Q[x]$; so that it is a principal ideal generated by a single polynomial $f(x)\in Q[x]$. But then an irreducible element $g(x)\in P\subset Z[x]$ can be written as $g(x)=f(x)h(x)$ with $h(x)\in Q[x]$. By considering the common denominators, $f(x)=f_1(x)/b$ and $h(x)=h_1(x)/c$ where $f_1(x), h_1(x) \in Z[x]$. Now $bcg(x)=f_1(x)h_1(x)$. Looking at the content of polynomials we can conclude that $P=(f_1(x))$ is a principle ideal of $Z[x]$.

The case where $P$ contains a constant is easy to solve by passing to the quotient ring.

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