Field Characteristic & Prime Subfield


Def. $\text{ch}(F)$ = the smallest (necc. prime) pos integer $p$ such that $p\cdot 1_F = 0$, or $0$ otherwise.  Also makes sense for any ID $R$ & $\text{ch}(R) = \text{ch}(\text{Frac}(R))$.  Examples:
  •  $\text{ch}(\Bbb{R}) = \text{ch}(\Bbb{Q}) = \text{ch}(\Bbb{Z}) = 0$
  •  $\Bbb{F}_p = \Bbb{Z}/(p) \implies \text{ch}(\Bbb{F}_p) = p$
  • $\text{ch}(\Bbb{F}_p[X])  = p$
There is a ring hom $\Bbb{Z} \xrightarrow{\varphi} F, n \mapsto \begin{cases} n \cdot 1_F, \text{ if } n \geq 0 \\ -(n\cdot 1_F), \text{ if } n \lt 0\end{cases}$, such that $\ker \varphi = \text{ch}(F) \Bbb{Z}$.

 So $\Bbb{Z}/\text{ch}(F)\Bbb{Z} \rightarrowtail F$ is an injection & $F$ either contains an isomorphic copy of $\Bbb{Q}$ or $\Bbb{F}_p$.

 Def. $\text{Prime}(F) =$ the smallest subfield of $F$ containing $1_F =$ subfield generated by $1_F$.  $\text{Prime}(F) = \Bbb{Q}$ or $\Bbb{F}_p$.  Examples:
  • $\text{Prime}(\Bbb{R}) = \Bbb{Q}$
  • $\text{Prime}(\Bbb{F}_p(X)) \simeq \Bbb{F}_p$ given by constant polynomials in $\Bbb{F}_p[X]$

Comments

Post a Comment

Popular posts from this blog

Every Map $X \xrightarrow{f} \Omega$ is the Characteristic Map of a Monomorphism

  Proposition  Let $\mathcal{E}$ be a topos with subobject classifier $\Omega$.  If $f: X \to \Omega$, then $f = \chi_m$ for some mono  $m: Y \rightarrowtail X$. Proof A topos is closed under taking pullbacks.  We already have the diagram, Taking its pullback, we get a diagram, But a pullback of $\text{true}$ along any morphism $f$ is always monic, By the definition of subobject classifier, namely the part about uniqueness of $\chi_m$, we must have that $\chi_m = f$. $\blacksquare$
Read more

Elementary counting formula for the number of $2k$-separated prime pair averages in an interval.

Proof of 2k-separated prime pairs in an interval, counting formula Let all intervals be of integers. Fix a natural number k ≥ 1 k \geq 1 k ≥ 1 . Primorial p n # p_n\# p n ​ # is defined as usual to be p 1 p 2 ⋯ p n p_1 p_2 \cdots p_n p 1 ​ p 2 ​ ⋯ p n ​ . Let an undeclared p p p or q q q just mean any prime number. Lemma 1. The number of x ∈ [ 0 , y ] x \in [0,y] x ∈ [ 0 , y ] such that x 2 = k 2 ( m o d p i ) x^2 = k^2 \pmod {p_i} x 2 = k 2 ( mod p i ​ ) for no i = 1.. n i = 1..n i = 1.. n is given by: f ( y ) = ∑ d ∣ p n # ,   d ≤ y ( − 1 ) ω ( d ) ∑ c ∣ d ,   gcd ⁡ ( c , 2 k ) = 1 ⌊ y − ( z c , d ) ( d ) d ⌋ (0) f(y) = \sum_{d \mid p_n\#, \ \\ d \leq y } (-1)^{\omega(d)}\sum_{c \mid d, \ \\ \gcd(c,2k) = 1} \lfloor \dfrac{y - (z_{c,d})_{(d)}}{d}\rfloor \tag{0} f ( y ) = d ∣ p n ​ # ,   d ≤ y ∑ ​ ( − 1 ) ω ( d ) c ∣ d ,   g c d ( c , 2 k ) = 1 ∑ ​ ⌊ d y − ( z c , d ​ ) ( d ) ​ ​ ⌋ ( 0 ) Where ( z c , d ) ( d ) (z_{c,d})_{(d)} ( z c , d ​ ) ( d ) ​
Read more

90% visual proof of Contravariant Yoneda Lemma

By the fact that $\alpha_X : \text{Hom}_C(X,X) \to AX$ we have that $\alpha_X(\text{id}_X) =: u \in AX$ and we're done with mapping any natural map $\alpha : \text{Hom}_C(\cdot, Y) \to A$ to an element $u \in AX$. The sides of the triangular prism in the bottom diagram need to commute for each $f:Y\to X, g : Z \to Y$ in $C$. For one, the triangular endcaps must commute because $A, \text{Hom}_C(\cdot, X)$ are both contravariant functors. Next, the three square sides commute by naturality. We must then have that any time $\alpha_Y(f) = A(f)\circ u$, the diagram commutes.
Read more