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4 votes

Isomorphic finite fields of a skew field

The following answer is nearly identical to @DustinCartwright's, which I saw after posting had been posted a few minutes earlier. I only construct a specific division algebra with a non-central ...
LSpice's user avatar
  • 11k
8 votes

Isomorphic finite fields of a skew field

No, this is not true. Suppose that $D$ is a skew field containing a finite field $F$ and $F$ is not contained in the center of $D$. Take $d \in D$ which does not commute with $F$. If $dFd^{-1}$ is ...
Dustin Cartwright's user avatar
3 votes

Matrices over a finite field: matrices for which some unipotent $U$ satisfies Trace$(ZU)=0$ for all $Z$ in the commutant

No, there are counterexamples with $p=5$. Let me restate the question. For a square matrix $A$ over any field, write $Y(A)=\{M:\forall Z\in C(A):\operatorname{Tr}(MZ)=0\}.$. Your question is whether $...
YCor's user avatar
  • 60k
3 votes

Loss of degree for polynomials

Yes. Let $P_0 = T$, $P_1 = -T$, $Q_0 = T^2+1$, $Q_1=T^2$. The degree of $(T^2+1)^{q^k}-(T^2)^{q^k}$ is at most $2q^k-2$ (the terms of degree $2q^k$ cancel), so the degree of $T(T^2+1)^{q^k}-T(T^2)^{q^...
paste bee's user avatar
  • 1,171
2 votes

Constructing a family of $3$-wise independence functions from $\mathbb{Z}_p^n \rightarrow \mathbb{Z}_p$

Let $\mathrm{tr}:GF(p^n)\rightarrow GF(p)$ be the trace function which is equidistributed. Choose a basis so that you consider $GF(p^n)$ as the vector space $GF(p^n)$. For any $c \in GF(p^n)^\ast,$ ...
kodlu's user avatar
  • 10k
2 votes

When is Chevalley Warning's bound best possible?

I would like to add to Will Sawin's answer and then mention two other results that are not as well known as they should be. Regarding Sawin's method, this is worked out in great detail in Ireland and ...
David Leep's user avatar
2 votes

Does the Plotkin bound mean that one can not achieve the Singleton bound asymptotically?

There do exist codes that achieve the Singleton bound, most famously Reed-Solomon codes. The asymptotic version of the Plotkin bound states that $R \le 1 - \frac q{q-1}\delta + o(1)$, where $R$ is the ...
Jonathan Mosheiff's user avatar

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