A Mind for Madness

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Witt Vectors Form a Ring

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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

I am a mathematics graduate student fascinated in how all my interests fit together.

One thought on “Witt Vectors Form a Ring

  1. Pingback: Fourth Linkfest

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