数论 5E: Number Theory/4H: Number Theory MATHS5074_1/MATHS4108_1 4
问题 1.
If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then the function $(f \circ \phi) \phi^{\prime}:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and $$ \int_{\phi(\alpha)}^{\phi(\beta)} f(x) d x=\int_{\alpha}^{\beta} f(\phi(t)) \phi^{\prime}(t) d t $$
证明 .
If $f:[a, b] \rightarrow \mathbb{R}$ is integrable and $\phi^{\prime}(t) \neq 0$ for every $t \in(\alpha, \beta)$, then the function $(f \circ \phi)\left|\phi^{\prime}\right|:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and $$ \int^{b} f(x) d x=\int^{\beta} f(\phi(t))\left|\phi^{\prime}(t)\right| d t $$
Now take $\lambda \in \operatorname{Sp}(T)$. Put $$ \lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) . $$ Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain $$ g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) . $$ Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus, $$ f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T)) $$ i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$
Proof. Assume, by way of contradiction, that there are only a finite number of prime numbers, say: $$ p_{1}, p_{2}, \ldots, p_{n} $$ Define $$ N=p_{1} p_{2} \cdots p_{n}+1 $$
证明 .
Since $p_{1} \geq 2$, clearly $N \geq 3$. So by Lemma $10.2 N$ has a prime divisor $p$. By assumption $p=p_{i}$ for some $i=1, \ldots, n$. Let $a=p_{1} \cdots p_{n}$. Note that $$ a=p_{i}\left(p_{1} p_{2} \cdots p_{i-1} p_{i+1} \cdots p_{n}\right), $$ so $p_{i} \mid a$. Now $N=a+1$ and by assumption $p_{i} \mid a+1$. So by Exercise $3.2$ $p_{i} \mid(a+1)-a$, that is $p_{i} \mid$
$$ a, a+1, a+2, \ldots, a+(n-1) $$ are all composite. Proof. Given $n \geq 1$ let $a=(n+1) !+2$. We claim that all the numbers $$ a+i, \quad 0 \leq i \leq n-1 $$ composite. Since $(n+1) \geq 2$ clearly $2 \mid(n+1)$ ! and $2 \mid 2$. Hence $2 \mid(n+1) !+2$. Since $(n+1) !+2>2,(n+1) !+2$ is composite. Consider $$ a+i=(n+1) !+i+2 $$
Hence $$ L(r f)=r L(f)=r U(f)=U(r f) . $$ On the other hand, if $r<0$, then for any partition $P$ of $[a, b]$, we see that $$ L(P, r f)=r U(P, f) \quad \text { and } \quad U(P, r f)=r L(P, f) $$
证明 .
and so $$ L(r f)=r U(f)=r L(f)=U(r f) . $$ In both the cases, we see that $r f$ is integrable and $$ \int_{a}^{b}(r f)(x) d x=r \int_{a}^{b} f(x) d x . $$
Hence $$ F(x)-F(a)=\sum_{i=1}^{n}\left[F\left(x_{i}\right)-F\left(x_{i-1}\right)\right]=\sum_{i=1}^{n} g\left(s_{i}\right)\left(x_{i}-x_{i-1}\right) $$ and so $$ L\left(P_{e}, g\right) \leq F(x)-F(a) \leq U\left(P_{e}, g\right) $$ Since we also have $$ L\left(P_{\epsilon}, g\right) \leq \int_{a}^{x} g(t) d t \leq U\left(P_{\epsilon}, g\right) $$
The mathematics of Classical Greece is known to us either through the reports of contemporary non-mathematicians or through mathematical works from the early Hellenistic period.In the case of number theory, this means, by and large, Plato and Euclid, respectively.
数论课后作业代写
$$ \mathrm{H}{\lambda, y}(\mathcal{F})=h{y *}^{\rightarrow}\left(h_{y}^{\leftarrow *}(\mathcal{F}) \otimes \mathrm{IC}{\lambda}\right) $$ The functors $\mathrm{H}{\lambda}$ are obtained by “gluing” together $\mathrm{H}{\lambda, y}$ for $y \in X$. Now the specialization of the Hecke property (9.9) to $y \in X$ amounts to the existence of a compatible collection of isomorphisms $$ v{\lambda}: \mathrm{H}{\lambda, y}\left(\Delta{x}\left(V_{\chi x}\right)\right) \stackrel{\sim}{\longrightarrow} V_{\lambda} \otimes c \Delta_{x}\left(V_{\chi_{x}}\right), \quad \lambda \in P_{+,} $$ where $V_{\lambda}$ is the irreducible representation of ${ }^{L} G$ of highest weight $\lambda$. We will now explain how Beilinson and Drinfeld derive $(9.9)$. Let us consider a “twopoint” realization of the localization functor, namely, we choose as our set of points $S \subset X$ the set ${x, y}$ where $x \neq y$. Applying the isomorphism (9.8) in this case, we find that $$ \Delta_{x}\left(V_{\chi_{x}}\right) \simeq \Delta_{x, y}\left(V_{\chi_{x}}, V_{-h^{\vee}}(g)_{y}\right) $$
