Mg Problem 33

PRELIMS:

33.

a. Prove that if $ \neq [a] \in \mathbb{F}_{p}$, then $\{[a],[2a], ... ,[(p-1)a]\}= \mathbb{F}_{p}^{\times}$

Since $p$ is prime, no multiple of $a$ has $p$ as a factor until $pa$. Therefore, every multiple of $a$ from $a$ to $(p-1)a$ is not a member of the equivalence class of $p$.

b.

c.

d. Prove that $\mathbb{Q}[x]$ is isomorphic to the ring of polynomial functions $Func_{\mathbb{Q}[x]}$ by showing injectivity of the natural map from $\mathbb{Q}[x]$ into $Func_{\mathbb{Q}[x]}$.

FINALS

33(a) THM: If $ \neq [a] \in \mathbb{F}_{p}$, then $\{[a],[2a], ... ,[(p-1)a]\}= \mathbb{F}_{p}^{\times}$ for $p$-prime.

The set $\mathbb{F}_{p}^{\times}$ is defined as the set $\mathbb{F}_{p}\textbackslash{}$. Since $\mathbb{F}_{p}$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, we know that it contains $p$ elements, and as such $\mathbb{F}_{p}^{\times}$ contains $p-1$ elements. Therefore, to prove some set $S = \{[a],[2a], ... ,[(p-1)a]\}$ with $[a] \neq $ is in fact $\mathbb{F}_{p}^{\times}$, suffice it to show that it contains $p-1$ distinct nonzero elements.

Since every element of $S$ is an integer multiple of $[a]$, to prove that all elements of $S$ are nonzero, suffice it to prove that no multiple $[na]$ of $[a]$ for $1 \leq n \leq p-1$ is equivalent to zero. Suppose there exists $a$ some representative element of $[a]$ such that $a \in p\mathbb{Z}$. This would imply that $a = pn$ for some $n \in \mathbb{Z}$, thus indicating $p|a$. However, since there exists a representative element $a$ of $[a]$ such that $0\leq a<|p|$, and since $p$ is prime $GCD(a,p) = 1$, it is impossible that $p|a$. Therefore, given $[a] \neq $, any multiple $[na]$$can only be equivalent to$$if$[n] \neq $. All elements of$S$are of the form$[n_{i}a]$such that$n_i < p$for all$i$. Therefore, since$p$is prime,$GCD(n_{i},p) = 1$for all$i$as well, thereby proving that$[n_{i}] \neq $for all$i$. Thus, all elements of$S$are nonzero. Finally suffice it to prove that all elements of$S$are distinct. Consider$[n_{1}a]$and$[n_{2}a]$, two arbitrary elements of$S$. Consider$[n_{1}a]-[n_{2}a]$.$[n_{1}a] = [n_{2}a]$if and only if their difference is in the ideal$p\mathbb{Z}$. However, since without loss of generality,$n_{1},n_{2}$can always be chosen to be strictly less than$p$and greater than or equal to$1$. Therefore, their difference is always less than$p$and greater than$0$unless$n_{1}=n_{2}$. Thus, the difference between two distinct elements of$S$cannot be contained in the ideal$p\mathbb{Z}$, and therefore, all elements of$S$are distinct. QED (b) Fermat's Little Theorem: if$\neq [a]\in \mathbb{F}_p$, then$[a]^{p-1}=$Consider the set$\mathbb{F}_{p}^{\times} = \{[a],[2a],...,[(p-1)a]\}$. The product of all elements in$\mathbb{F}_{p}^{\times}$is$[(p-1)!]\cdot{[a]^{p-1}}$. Since the set$\mathbb{F}_{p}^{\times}$is equal up to ordering for any$a$, the product of all its elements,$[(p-1)!]\cdot{[a]^{p-1}}$is the same as the product of all the elements in the case that$[a]=$, which is to say that$[(p-1)!]\cdot{[a]^{p-1}} = [(p-1)!]$. By cancellation, we get$[a]^{p-1} = 1$. QED (c) Thm:$\mathbb{F}_{p}[x] \not\cong Func_{\mathbb{F}_{p}[x]}$Suffice it to show that a map$\mathbb{F}_{p}[x] \rightarrow Func_{\mathbb{F}_{p}[x]}$is not surjective by providing a counterexample in which two distinct elements of the domain$\mathbb{F}_{p}[x]$map to the same element of the codomain$Func_{\mathbb{F}_{p}[x]}$. From (b), we know that$x^{p-1} \mapsto 1$for$x \in \{[a] \mid a \not\sim 0\}$,$x^{p-1} \mapsto 0$for$x \in \{[a] \mid a \sim 0\}$. However,$(x^{p-1})^2 \mapsto 1$for$x \in \{[a] \mid a \not\sim 0\}$,$(x^{p-1})^2 \mapsto 0$for$x \in \{[a] \mid a \sim 0\}$. Therefore,$\mathbb{F}_{p}[x] \not\cong Func_{\mathbb{F}_{p}[x]}$. QED (d) Thm:$\mathbb{Q}[x] \cong Func_{\mathbb{Q}[x]}$(NOTE: Must only show injectivity, as remainder of proof shown in class.) Suppose the ring homomorphism is not injective. Then there would be some two distinct elements of$\mathbb{Q}[x]$that always map to the same$Func_{\mathbb{Q}[x]}$. It is obvious that distinct polynomials of the same degree have distinct evaluations, as otherwise they would be equivalent functions. Therefore, the only polynomials, polynomials that could map to the same evaluations would be polynomials of differing degrees in$\mathbb{Q}[x]$. However, since$\mathbb{Q}[x]$is an infinite ring,$x^n \not\sim x^m$unless$n=m$, unlike the ring from (c). Therefore, polynomials of different degrees also do not map to the same evaluations. Thus, the map$\mathbb{Q}[x] \rightarrow Func_{\mathbb{Q}[x]}$is injective, and therefore$\mathbb{Q}[x] \cong Func_{\mathbb{Q}[x]}\$.

page revision: 23, last edited: 25 Apr 2018 03:21
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License