Witt Vectors Form a Ring

Today we’ll check that the ring of Witt vectors is actually a ring. Let ${A}$ be a ring, then ${W(A)}$ as a set is the collection of infinite sequences of ${A}$. Recall that our construction involves lots of various polynomials and a strange definition addition and multiplication. I won’t rewrite those, since it was the entirety of the last post.

Now there is a nice trick to prove that ${W(A)}$ is a ring when ${A}$ is a ${\mathbb{Q}}$-algebra. Just define ${\psi: W(A)\rightarrow A^\mathbb{N}}$ by ${(a_1, a_2, \ldots) \mapsto (w_1(a), w_2(a), \ldots)}$. This is a bijection and the addition and multiplication is taken to component-wise addition and multiplication, so since this is the standard ring structure we know ${W(A)}$ is a ring. Also, ${w(0,0,\ldots)=(0,0,\ldots)}$, so ${(0,0,\ldots)}$ is the additive identity, ${W(1,0,0,\ldots)=(1,1,1,\ldots)}$ which shows ${(1,0,0,\ldots)}$ is the multiplicative identity, and ${w(\iota_1(a), \iota_2(a), \ldots)=(-a_1, -a_2, \ldots)}$, so we see ${(\iota_1(a), \iota_2(a), \ldots)}$ is the additive inverse.

We can actually get this idea to work for any characteristic ${0}$ ring by considering the embedding ${A\rightarrow A\otimes\mathbb{Q}}$. We have an induced injective map ${W(A)\rightarrow W(A\otimes\mathbb{Q})}$. The addition and multiplication is defined by polynomials over ${\mathbb{Z}}$, so these operations are preserved upon tensoring with ${\mathbb{Q}}$. We just proved above that ${W(A\otimes\mathbb{Q})}$ is a ring, so since ${(0,0,\ldots)\mapsto (0,0,\ldots)}$ and ${(1,0,0,\ldots)\mapsto (1,0,0,\ldots)}$ and the map preserves inverses we get that the image of the embedding ${W(A)\rightarrow W(A\otimes \mathbb{Q})}$ is a subring and hence ${W(A)}$ is a ring.

Lastly, we need to prove this for positive characteristic rings. Choose a characteristic ${0}$ ring that surjects onto ${A}$, say ${B\rightarrow A}$. Then since the induced map again preserves everything and ${W(B)\rightarrow W(A)}$ is surjective, the image is a ring and hence ${W(A)}$ is a ring.

So where does all this formal group stuff we started with come into play? Well, notice that what we were really implicitly using is that ${W:\mathbf{Ring}\rightarrow\mathbf{Ring}}$ is a functor. It takes a ring ${A}$ and gives a new ring ${W(A)}$. If ${\phi: A\rightarrow B}$ is a ring map, then ${W(\phi): W(A)\rightarrow W(B)}$ by ${(a_1, a_2, \ldots)\mapsto (\phi(a_1), \phi(a_2), \ldots)}$ is still a ring map. We also have ${w_n:W(A)\rightarrow A}$ by ${a\mapsto w_n(a)}$ are ring maps for all ${n}$.

Some people think it is cleaner to define the ring of Witt vectors as the unique functor ${W}$ that satisfies these three properties. From a functorial point of view it turns out that ${W}$ is representable. The representing ring via the ring axioms gives a Hopf algebra structure, and hence we get an affine group scheme out of it. Then as in the formal group discussion, we can complete this to get a formal group. This will be the discussion of next time.

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Author: hilbertthm90

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