ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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Primitive polynomials depend on the characteristic

Example

In the polynomial Hopf algebra $k [x]$ with $Δ (x) = x \otimes 1 + 1 \otimes x$ and $ε (x) = 0$ , call $z$ primitive if $Δ (z) = z \otimes 1 + 1 \otimes z$ and $ε (z) = 0$ . The generator $x$ is primitive over every field. If $char k = p > 0$ , then $x^{p}$ is primitive too. In characteristic zero, $x^{2}$ is not primitive.

Facts & Assumptions

Given: A field $k$ and the polynomial Hopf structure in the Example.

[F1]

For every field, $k [x]$ has $Δ (x) = x \otimes 1 + 1 \otimes x$ and $ε (x) = 0$ (Group algebras, finite-group function algebras, and polynomial Hopf algebras).

[F2]

If ${x^{i}}$ is the monomial basis, then ${x^{i} \otimes x^{j}}$ is a basis of $k [x] \otimes k [x]$ (The elementary tensors of two bases form the product basis of the tensor product).

[F3]

The binomial coproduct expansion for all $n$ is proved by induction in step 2.3 of Group algebras, finite-group function algebras, and polynomial Hopf algebras.

[F4]

Step 3.1 of Group algebras, finite-group function algebras, and polynomial Hopf algebras shows that $p$ is prime in characteristic $p$ and that the intermediate coefficients $(\binom{p}{j})$ vanish in $k$ .

Verification

technique · direct

1.1givenF1

The formulas in [F1] give $Δ (x) = x \otimes 1 + 1 \otimes x$ and $ε (x) = 0$ , so $x$ is primitive by the definition in the Example.

1.2givenF1F3F4algebra

Let $char k = p > 0$ and put $X = x \otimes 1$ , $Y = 1 \otimes x$ . By the polynomial bialgebra in [F1], $Δ (x^{p}) = (X + Y)^{p}$ ; [F3] gives its binomial expansion and [F4] shows that the intermediate coefficients vanish, so $Δ (x^{p}) = x^{p} \otimes 1 + 1 \otimes x^{p}$ . Since its counit is the algebra map with $ε (x) = 0$ from [F1], $ε (x^{p}) = ε (x)^{p} = 0$ . Thus $x^{p}$ is primitive.

1.3givenF1F2algebra

If $char k = 0$ , then $Δ (x^{2}) = x^{2} \otimes 1 + 2 x \otimes x + 1 \otimes x^{2}$ . By [F2], $x \otimes x$ is a basis vector and is nonzero; $2 \neq 0$ in characteristic zero, so the middle term is nonzero. Therefore $Δ (x^{2}) \neq x^{2} \otimes 1 + 1 \otimes x^{2}$ , and $x^{2}$ is not primitive.

2.1step 1.1step 1.2step 1.3F1∎

Step 1.1 verifies the characteristic-independent primitive element, while steps 1.2–1.3 give the characteristic-dependent distinction. The Hopf algebra itself exists in every characteristic by [F1].

Depends on

Used by

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Dependency tree · two levels

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Sources

Hans-Jürgen Schneider, Lectures on Hopf algebras, notes by Sonia Natale (standard reference, not scraped)
Pavel Etingof and Mykola Semenyakin, A brief introduction to quantum groups (standard reference, not scraped)