# Frobenius map of Witt vectors

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 $$\pi_{k} \in \mathbb{Z}[\mathbf{x},\mathbf{y}]$$

• 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. Mar 17 at 8:11
• 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 $$\Phi(\pi(x_0,x_1,\ldots,y_0,y_1,\ldots))=\pi(\Phi(x_0),\Phi(x_1),\ldots,\Phi(y_0),\Phi(y_1),\ldots).$$ Mar 17 at 8:15
• 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. Mar 18 at 18:46
• 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. Mar 18 at 18:51

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, $$F(\mathbf{a}\cdot \mathbf{b}) = F(\mathbf{a})\cdot F(\mathbf{b}) = (\pi_{0}(\phi(\mathbf{a}),\phi(\mathbf{b})),\pi_{1}(\phi(\mathbf{a}),\phi(\mathbf{b})),\dots )$$