2
$\begingroup$

I have been trying to solve Neukirch's exercise II.4.5 on Witt vectors and need some help. The question (only the part I am having issue with) is as follows:

Let $k$ be a field of characteristic $p$. Then, $F: W(k) \rightarrow W(k)$ given by $(a_{0}, a_{1}, \dots )\mapsto (a_{0}^p, a_{1}^p, \dots )$ is a ring homomorphism.

I have checked closed under addition but I am not sure how to check multiplication. More concretely, why is $$F(\mathbf{a}\cdot \mathbf{b})=F(\mathbf{a})\cdot F(\mathbf{b})$$ where $\mathbf{a}\cdot \mathbf{b} = (\pi_{0}(\mathbf{a},\mathbf{b}),\pi_{1}(\mathbf{a},\mathbf{b}), \dots )$ and $\pi_{k} \in \mathbb{Z}[\mathbf{x},\mathbf{y}]$

$\endgroup$
4
  • $\begingroup$ The sum of two Witt vectors also involves multivariate polynomials with integer coefficients with inputs from the field $k$, so it seems to me that $F$ respecting Witt vector addition is no different from Witt vector products? Both require $a\mapsto a^p$ to be a homomorphism of rings from $k$ to itself. $\endgroup$
    – Jyrki Lahtonen
    Mar 17 at 8:11
  • $\begingroup$ If $\Phi:k\to k$ is any homomorphism of rings, then for all polynomials $\pi\in\Bbb{Z}[x_0,x_1,\ldots,y_0,y_2,\ldots]$ we have $$\Phi(\pi(x_0,x_1,\ldots,y_0,y_1,\ldots))=\pi(\Phi(x_0),\Phi(x_1),\ldots,\Phi(y_0),\Phi(y_1),\ldots).$$ $\endgroup$
    – Jyrki Lahtonen
    Mar 17 at 8:15
  • $\begingroup$ Right. Of course, it makes sense. I was being dumb. Since Frobenius $F$ is a ring homomorphism and the fact that the coefficients are in $\mathbb{Z}$ means that $z^p=z \; ,z\in \mathbb{Z}$ using small FLT. There is no difference whether I am considering Witt vector addition or multiplication. Thanks! @JyrkiLahtonen. $\endgroup$
    – Hillimath
    Mar 18 at 18:46
  • $\begingroup$ Glad to hear you figured it out. Feel free to write the argument as an answer also. I'm a bit reluctant to do it, because I may be using a different definition for the Witt vector arithmetic. An answer written by you would consistently use the notation from your source. $\endgroup$
    – Jyrki Lahtonen
    Mar 18 at 18:51

1 Answer 1

Reset to default
1
$\begingroup$

Say $\phi:k \rightarrow k$ is the Frobenius automorphism $x \mapsto x^p$. Then, $F(a_{0},a_{1},\dots ) = (\phi(a_{0}),\phi(a_{1}), \dots)$.

\begin{align*} F(\mathbf{a}\cdot \mathbf{b}) &= F(\pi_{0}(\mathbf{a},\mathbf{b}),\pi_{1}(\mathbf{a},\mathbf{b}),\dots)\\ &= (\phi(\pi_{0}(\mathbf{a},\mathbf{b})),\phi(\pi_{1}(\mathbf{a},\mathbf{b})),\dots) \end{align*} So, it boils down to checking what happens to $\phi(\pi(\mathbf{a},\mathbf{b}))$ with $\pi(\mathbf{x},\mathbf{y}) \in \mathbb{Z}[\mathbf{X},\mathbf{Y}]$.

For simplicity, just view $\pi$ as a polynomial in $x$, say it were $z_{0} + z_{1}x + z_{2}x^2 + \cdots + z_{n}x^n$. Since $\phi$ is a ring homomorphism of $k$ and $z_{i} \in \mathbb{Z}$, therefore

\begin{align*} \phi(z_{0} + z_{1}a + z_{2}a^2 + \cdots + z_{n}a^n) &= \phi(z_{0}) + \phi(z_{1})\phi(a) + \phi(z_{2})\phi(a^2) + \cdots + \phi(z_{n})\phi(a^n)\\ &= z_{0} + z_{1}\phi(a) + z_{2}\phi(a)^2 + \cdots + z_{n}\phi(a)^n \\ &= \pi(\phi(a)) \\ \therefore \phi(\pi_{i}(\mathbf{a},\mathbf{b})) &= \pi_{i}(\phi(\mathbf{a}),\phi(\mathbf{b})) \end{align*} The second equality comes from small FLT ($z_{i}^p = z_{i}$ since $z_{i} \in \mathbb{Z}$). To conclude, $F(\mathbf{a}\cdot \mathbf{b}) = F(\mathbf{a})\cdot F(\mathbf{b}) = (\pi_{0}(\phi(\mathbf{a}),\phi(\mathbf{b})),\pi_{1}(\phi(\mathbf{a}),\phi(\mathbf{b})),\dots )$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .