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I know that $\operatorname{Spec} \mathbb{C}[x]$ can be identified with the set $\mathbb{C}\cup *$, where $*$ is a generic point via the correspondence $$ \prod_{i}(x-a_i) \leftrightarrow \{a_i\}_i , \ \ \ \ (0)\leftrightarrow *. $$ This correspondence holds for any $\operatorname{Spec} k[x]$ with algebraic closed field $k$.

When $k$ is not algebraic closed, how should one understand $\operatorname{Spec} k[x]$?

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This is a good exercise. What have you tried? –  Qiaochu Yuan Oct 19 '12 at 1:26
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Since polynomial does not decompose into linear factors over non-algebraic field $k$, there must exist correspondence between irreducible polynomial and points on $Spec k[x]$. –  M. K. Oct 19 '12 at 1:30
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The only prime ideals in $\mathbb{C}[x]$ are either the zero ideal or generated by a polynomial of the form $x-a$, as is true for $k[x]$ for any algebraically closed $k$. The correspondence above should read $(x-a) \leftrightarrow a, (0)\leftrightarrow *$. –  Brad Oct 19 '12 at 1:56
    
Oh you are right, Brad. –  M. K. Oct 19 '12 at 3:22

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up vote 7 down vote accepted

You note correctly in a comment that the points of $\operatorname{Spec} k[x]$ correspond to (monic) irreducible polynomials, and you are correct that this can be identified with the set $k\cup\{*\}$ when $k$ is algebraically closed.

Suppose $\bar k$ is an algebraic closure of $k$. Then the inclusion $k[x]\to\bar k[x]$ induces a morphism $\pi:\operatorname{Spec} \bar k[x]\to\operatorname{Spec} k[x]$ which is surjective as a map of sets. If we take a point of $\operatorname{Spec} k[x]$ and think of it as a monic irreducible polynomial $p(x)$, the fibre of $\pi$ over that point is exactly the set of roots of $p$. For simplicity, let's assume that $k$ is a perfect field (e.g. it has characteristic zero or it's a finite field), so that the extension $\bar k/k$ is Galois. The Galois group $G:=\operatorname{Gal}(\bar k/k)$ acts on $\operatorname{Spec} \bar k[x]\simeq \bar k\cup\{*\}$ in the obvious way (fixing the point $*$), and the fibres of $\pi$ are exactly the orbits of this action, so we can think of $\operatorname{Spec} k[x]$ as the quotient $(\bar k\cup\{*\})/G$.

For example, if $k=\mathbb{R}$, we can think of $\operatorname{Spec}\mathbb{R}[x]$ as something like what we get by taking the complex plane $\mathbb{C}$ and identifying complex conjugates, i.e. folding it over on itself across the real line. (That, along with the generic point $*$.)

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Let $\bar k$ be an algebraic closure of $k$. Let $G = Aut(\bar k/k)$. $\bar k/G$ be the set of $G$-orbits. We will show that there exists a canonical bijection $Spec$ $k[x] \rightarrow \bar k/G\cup \{*\}$.

Let $Y =$ $Spec$ $k[x] - \{(0)\}$. It suffices to prove that there exists a canonical bijection $\psi\colon Y \rightarrow \bar k/G$. Let $p \in Y$. There exists a unique monic irreducible polynomial $f(x) \in k[x]$ such that $p = (f(x))$. Let $S$ be the set of roots of $f(x)$ in $\bar k$. It is clear that $S$ is $G$-stable. Let $\alpha, \beta \in S$. There exists a $k$-isomorphism $k(\alpha) \rightarrow k(\beta)$ transforming $\alpha$ to $\beta$. It can be extended to a $k$-automorphism of $\bar k$. Hence there exists $\sigma \in G$ such that $\sigma(\alpha) = \beta$. Hence $S \in \bar k/G$. Hence we get a map $\psi\colon Y \rightarrow \bar k/G$ such that $\psi(p) = S$. Clearly $\psi$ is injective.

Let $T \in \bar k/G$. Let $\alpha \in T$. Let $f(x)$ be the minimal polynomial of $\alpha$ over $k$. Clearly $\psi((f(x)) = T$. Hence $\psi$ is surjective.

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Can you say more about why $G$ acts transitively on $S$, when $\bar{k}$ is possibly non-separable over $k$? –  Zhen Lin Oct 19 '12 at 11:16
    
@ZhenLin I added an explanation for it. –  Makoto Kato Oct 19 '12 at 11:51

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