Spectrum and induced homomorphisms

Question

Let $K$ be an algebraically closed field of characteristic two. Consider the the $K$-algebra homomorphism, $\varphi: K[x] \to K[x]$ defined by $\varphi(x)= x^2$. How can I describe the induced map $\varphi^*: \operatorname{Spec} K[x] \to \operatorname{Spec} K[x]$?

I'm not quite sure how this works. What comprises the domain $Spec K[x]$. In other words how does the homomorphism effect the domain $Spec K[x]$. Intuitively I would say we have $x^2-1$ in the fiber of $(x-1)$.

I apologize for rambling, someone set me straight.

Answer 1

The spectrum of $K[x]$ is called the affine line over $K$ and is denoted by $ \mathbb A^1_K$.
Its closed points correspond to the elements of $K$ and apart from these $ \mathbb A^1_K$ has another point $\eta$ corresponding to the zero ideal of $K[x]$, which is dense and called the generic point of $ \mathbb A^1_K$ .

Corresponding to the $K$-algebra morphism $\phi$ there is the morphism of affine schemes $F=\phi^*: \mathbb A^1_K\to \mathbb A^1_K $ called the Frobenius morphism.
It is, strangely, a bijective map (and even a homeomorphism) but not an isomorphism of schemes.

This Frobenius morphism is a perfectly admissible morphism of schemes and has or does not have the properties studied in scheme theory: it is a faithfully flat, finite morphism of degree $2$, everywhere ramified, nowhere smooth, etc., etc.

Answer 2

I will try to give you a thorough answer. We first describe the set $Spec K[x].$ This is by definition the set of prime ideals of $K[X].$ This is an integral domain so we have $(0) \in Spec K[X].$ Moreover this is a PID, so all other prime ideals are given by $(f)$ (the ideal generated by an irreducible element, i.e. an irreducible polynomial). As $K$ is algebraically closed, it follows that the only irreducible polynomials in $K[X]$ are the polynomials of the form $x-a$ where $a \in K.$ Now we take a look at the map $\phi^{*}: Spec K[X] \to Spec K[X].$ This is given by $\phi^{*}((x-a)) = \phi^{-1}((x-a)) = \{f\in K[X] : \phi(f(x)) = f(x^2) \in (x-a)\}.$ As $(x-a)(x+a) = x^2 - a^2.$ We see that $x^2 - a^2 \in (x-a).$ Moreover we have $\phi(x-a^2) = x^2 - a^2.$ From this we see that $(x-a^2) \subset \phi^{*}((x-a)).$ Moreover we have equality as $(x-a^2)$ is a maximal ideal. Thus $\phi^{*}((x-a)) = (x - a^2)$ and clearly we have $\phi^{*}((0)) = (0).$ Thus we have determined what the set map does to points. Since $K$ is algebraically closed, we have that for any $a \in K$ we have $\sqrt{a} \in K.$ This shows surjectivity as we then have $\phi^{*}((x-\sqrt{a})) = (x - \sqrt{a}^2) = (x-a).$ For injectivity we have: If $\phi^{*}((x-a)) = (x - a^2) = (x - b^2) = \phi^{*}((x-b)).$ Then $a^2 = b^2.$ We have that $x^2 - a^2$ has solution(s) $x = a$ and $x= -a$, but since $K$ is of characteristic $2$ we have $a = -a.$ Thus there is only one solution to this equation and from this we deduce that $a = b.$ Thus $\phi^{*}$ is injective as well. The map $(\phi^{*},\phi)$ does not give an isomorphism of schemes however because the ring map $\phi$ is clearly not even surjective.