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Part II Lecture Notes: Algebraic Topology, James Lingard Homotopy Given two maps $f_{1}, f_{2}: X \rightarrow Y$, a homotopy from $f_{1}$ to $f_{2}$ is a map $F:[0,1] \times X \rightarrow Y$ such that $F(0, x)=f_{1}(x)$ and $F(1, x)=f_{2}(x)$. If there is a homotopy from $f_{1}$ to $f_{2}$ then we say that $f_{1}$ and $f_{2}$ are homotopic and we write $f_{1} \sim f_{2}$. Homotopy is an equivalence relation, although to prove transitivity we need the following lemma: The 'Gluing Lemma' If a space $X$ is a union of two closed subsets $A$ and $B$ then for any two continuous maps $f: A \rightarrow Y$ and $g: B \rightarrow Y$ which agree on the intersection $A \cap B$ we get a continuous map $f \cup g: X \rightarrow Y$. This is easy to prove. The set of homotopy classes of maps $X \rightarrow Y$ is written $[X, Y]$. Relative homotopy Given two maps $f_{1}, f_{2}: X \rightarrow Y$ which agree on a subset $A \subset X$, we say that they are homotopic relative to $A$ if there is a map $F:[0,1] \times X \rightarrow Y$ such that $F(0, x)=f_{1}(x), F(1, x)=f_{2}(x)$ and $F(x, t)=f_{0}(x)$ for all $x \in A$. The fundamental group The fundamental group of $X$ based at $x$, written $\pi_{1}(X, x)$, is the group of homotopy classes relative to $\{0,1\}$ of loops in $X$ based at $x$, with the natural multiplication given by composition of loops. It is easy to show that the multiplication is well-defined and satisfies the group axioms. Note that for loops $p, q$, we write $p q$ to mean the loop which goes round $p$ first and then $q$. Induced homomorphisms of fundamental groups If we have a map $f: X \rightarrow Y$ then we get an induced homomorphism $f_{*}: \pi_{1}(X, x) \rightarrow \pi_{1}(Y, f(x))$ between the fundamental groups of $X$ and $Y$, by mapping a loop $p$ in $X$ to the loop $f p$ in $Y$. It is easy to check that this is a group homomorphism. Also, if we have the identity map $1_{X}: X \rightarrow X$ then $\left(1_{X}\right)_{*}$ is the identity homomorphism, and for any two maps $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ then we have $(g \circ f)_{*}=g_{*} \circ f_{*}$. Thus if $f$ is a homeomorphism $X \rightarrow Y$ then the induced homomorphism $f_{*}$ is an isomorphism, and hence the fundamental group is a topological invariant. If two maps $f, g: X \rightarrow Y$ are homotopic relative to $\{x\}$ then the induced homomorphisms $f_{*}$ and $g_{*}$ are the same, since the homotopy $F$ from $f$ to $g$ determines a homotopy $F(\alpha(s), t)$ from $f(\alpha)$ to $g(\alpha)$ for any loop $\alpha$ in $X$ based at $x$. This is a special case of a result below. Change of base point If $x, y \in X$, then $\pi_{1}(X, x)$ and $\pi_{1}(X, y)$ are unrelated if $x$ and $y$ are in different path-components, but isomorphic if they are in the same path component. For any path $\alpha$ from $x$ to $y$ we get a group isomorphism $\alpha_{*}: \pi_{1}(X, x) \rightarrow \pi_{1}(X, y)$. Note that in general different paths $\alpha$ may lead to different isomorphisms. Contractible spaces A non-empty space $X$ is said to be simply connected if it is path-connected and has trivial fundamental group. $X$ is said to be contractible if the identity map $X \rightarrow X$ is homotopic to a constant map. For example, any convex subset of $\mathbb{R}^{n}$ is contractible. A contractible space is simply connected. Note that the converse is not true $-S^{2}$ is simply connected but not contractible. To prove this, we need to show that any loop in $X$ is homotopic, relative to $\{0,1\}$, to the constant loop. Observe that the homotopy $F$ from the identity on $X$ to the constant map at $x \in X$ gives us a homotopy $G(s, t)=F(\alpha(s), t)$ from any loop $\alpha$ based at $x$ to the constant loop at $x$ - but this homotopy is not necessarily relative to $\{0,1\}$. To get around this difficulty we use the fact that the square $[0,1] \times[0,1]$ is convex, so there is a straight line homotopy $H$ between any two paths from $a$ to $b$ in $[0,1] \times[0,1]$, and so any two such paths are homotopic relative to $\{0,1\}$. Using this, we see that the path along the left edge of the square is homotopic, relative to $\{0,1\}$, to the path along the bottom edge, up the right edge and back along the top edge of the square. Composing the homotopies $H$ and $G$ gives us the result. Homotopy equivalence Two spaces $X$ and $Y$ are homotopy equivalent if there is a map $f: X \rightarrow Y$ and a map $g: Y \rightarrow X$ such that $f \circ g \sim 1_{Y}$ and $g \circ f \sim 1_{X}$. Such an $f$ is called a homotopy equivalence. We also say that $X$ and $Y$ have the same homotopy type, and we write $X \simeq Y$. The relation $X \simeq Y$ is an equivalence relation. Any homeomorphism is clearly a homotopy equivalence. It is also very easy to prove that a space is contractible if and only if it is homotopy equivalent to a point. Two homotopy equivalent spaces have the same number of path-components. Furthermore, two homotopy equivalent spaces have isomorphic fundamental groups. To prove this, we first show that if $f$ and $g$ are homotopic maps from $X$ to $Y$ and $x \in X$, then $g_{*}=\alpha_{*} \circ f_{*}$ as maps from $\pi_{1}(X, x)$ to $\pi_{1}(Y, g(x))$, where $\alpha$ is the path in $Y$ from $f(x)$ to $g(x)$ given by the homotopy $F$ between $f$ and $g$. This follows by using the homotopy $F$ to show that for any loop $\alpha$ in $X$ based at $x$, the loops $g_{*}(\alpha)$ and $\left(\alpha_{*} \circ f_{*}\right)(\alpha)$ in $Y$ are homotopic. It then follows that the map $f_{*}$ is an isomorphism of fundamental groups if and only if $g_{*}$ is. Therefore, if $f$ is a homotopy equivalence with homotopy inverse $h$, then $f \circ h \simeq 1_{X}$ and so $(f \circ h)_{*}=f_{*} \circ h_{*}$ is an isomorphism. Thus $f_{*}$ is injective. Working the other way we see that $f_{*}$ is surjective. Hence $f_{*}$ gives an isomorphism between the fundamental groups of $X$ and $Y$. The fundamental group of the $n$-sphere If $X$ is a space which can be written as the union of two simply connected open subsets $A$ and $B$ in such a way that $A \cap B$ is path-connected, then $X$ is simply connected. To prove this we show that every loop in $X$ starting at a point $p$ in $A \cap B$ is homotopic to a (finite) product of loops, each of which is either in $A$ or in $B$. Using this result, we see that the fundamental group of $S^{n}$ is trivial for $n \geq 2$.

3 Theorem. Let $p ∈[1, ∞]$. Then there exists no $h ∈ L^p((0,2))$ such that \[ ∫_{(0,2)} h ⋅ φ ⋅ d λ^1=φ(1)   ∀ φ ∈ C_c^{∞}((0,2)) \] Proof. Assume there is an $h ∈ L^p((0,2))$ with the property above. Let $q$ be the conjugate of $p$. Further, let $φ_ϵ$ as in the lemma. Then we have \[ \begin{aligned} 1 & =\left|φ_ϵ(1)\right| \\ & =\left|∫ h φ_ϵ d λ^1\right| ⩽ ∫|h|\left|φ_ϵ\right| d λ^1=∫\left|h 1_{B_ϵ(1)}\right|\left|φ_ϵ\right| d λ^1 \\ & ⩽\| h 1_{B_ϵ(1)}\|_p ⋅\| φ_ϵ\|_q ⩽\| h 1_{B_ϵ(1)} \|_p \\ & =\left(∫\left|h 1_{B_ϵ(1)}\right|^p d λ^1\right)^{\frac{1}{p}} \xrightarrow{ϵ → 0+} 0 \end{aligned} \] (note the use of Hölder's inequality; the limit follows from the dominated convergence theorem which is applicable as \[ 0 \stackrel{\text { a.e. }}{⟵}\left|h 1_{B_ϵ(1)}\right|^p ⩽|h|^p \] and $|h|^p$ is integrable since $h ∈ L^p((0,2))$.) Contradiction!

$\DeclareMathOperator{\tr}{tr}\DeclareMathOperator{\disc}{disc}$Suppose that $β$ is a root of $X^3+p X+q=0$, where $X^3+p X+q$ is an irreducible polynomial in $𝐙[X]$, and let $K=𝐐(β)$. Compute $\tr_{K/𝐐}(β^i)$ for $i=0,1, …, 4$. Deduce that $\disc_{K/𝐐}(1, β, β^2)=-4 p^3-27 q^2$. Hence, give an example of a cubic number field $K$ such that $𝒪_K$ has a power integral basis. $\{1,β,β^2\}$ is a 𝐐-basis of $K$ $β^3=-q-pβ$ $\tr_{K/𝐐}(1)=\tr\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=3$ $\tr_{K/𝐐}(β)=\tr\begin{pmatrix}0&1&0\\0&0&1\\-q&-p&0\end{pmatrix}=0$ $\tr_{K/𝐐}(β^2)=\tr\begin{pmatrix}0&0&1\\-q&-p&0\\0&-q&-p\end{pmatrix}=-2p$ $\tr_{K/𝐐}(β^3)=\tr\begin{pmatrix}-q&-p&0\\0&-q&-p\\pq&p^2&-q\end{pmatrix}=-3q$ $\tr_{K/𝐐}(β^4)=\tr\begin{pmatrix}0&-q&-p\\pq&p^2&-q\\q^2&2pq&p^2\end{pmatrix}=2p^2$ \begin{align*}\disc_{K/𝐐}(1, β, β^2)&=\begin{vmatrix}\tr_{K/𝐐}(1)&\tr_{K/𝐐}(β)&\tr_{K/𝐐}(β^2)\\\tr_{K/𝐐}(β)&\tr_{K/𝐐}(β^2)&\tr_{K/𝐐}(β^3)\\\tr_{K/𝐐}(β^2)&\tr_{K/𝐐}(β^3)&\tr_{K/𝐐}(β^4)\\\end{vmatrix}\\&=\begin{vmatrix}3&0&-2p\\ 0&-2p&-3q\\-2p&-3q&2p^2\end{vmatrix} \\&=-4 p^3 - 27 q^2\end{align*} When $p=q=1$, $\disc_{K/𝐐}(1, β, β^2)=-31$ is squarefree, by 2.18 $K=𝐐(β)$ is a cubic number field such that $𝒪_K$ has a power integral basis $\{1, β, β^2\}$.

Let $T\colon X → Y$ be a bounded linear operator between Hilbert spaces. Show the following equivalences: (a) $T$ is isometric $⇔⟨Tx,Ty⟩=⟨x,y⟩∀x,y∈X⇔T^* T=I_X$. (b) $T$ is unitary $⇔T^* T=I_X$ and $T T^*=I_Y$ (a) $3\!⟹\!1$: Suppose $T^* T=I_X$. Then $∀x∈X$ \[‖T x‖^2=⟨Tx, Tx⟩=⟨T^*Tx,x⟩=⟨I_Xx,x⟩=‖x‖^2\] Hence $T$ is an isometry. $1\!⟹\!2$: Suppose $T$ is an isometry. By polarisation $∀x,y∈X$ \[⟨Tx,Ty⟩=\frac{‖T(x+y)‖^2-‖T(x-y)‖^2}4=\frac{‖x+y‖^2-‖x-y‖^2}4=⟨x,y⟩\] [similarly for complex Hilbert space $⟨x,y⟩=\frac 14\left(\|x+y\|^{2}-\|x-y\|^{2}-i\|x+iy\|^{2}+i\|x-iy\|^{2}\right).$] $2\!⟹\!3$: Suppose $⟨Tx,Ty⟩=⟨x,y⟩∀x,y∈X$, then $⟨T^*Tx,y⟩=⟨x,y⟩$, then $T^* Tx=x$. Hence $T^* T=I_X$. (b) $1\!⟹\!2$: Suppose $T^*T=I_X$ and $TT^*=I_Y$. By (a) $T^*T=I_X$ implies that $T$ is isometric. Also $TT^*=I_Y$ implies that $T$ is onto, because $∀y∈Y,y=TT^*y∈T(X)$. $2\!⟹\!1$: Suppose $T$ is unitary. By (a) $T$ is isometric implies $T^*T=I_X$. It remains to show $TT^* = I_Y$. $∀y∈Y$, by surjectivity $∃x∈X,Tx=y$ \[TT^*y=TT^*Tx\xlongequal{T^*T=I_X}Tx=y\]Therefore $TT^*=I_Y$.

Consider the following four functions on $ℝ$: \[f_1(x)=𝖾^{-x^2+2 x},   f_2(x)=𝖾^{-x} H(x),   f_3(x)=𝖾^{-|x|},   f_4(x)=\frac{1}{x^2+1},\] where $H$ is Heaviside’s function. (a) Verify that these functions all belong to $𝖫^1(ℝ)$. Which of them belong to $𝖫^2(ℝ)$ and which to $𝒮(ℝ)$? \[\begin{array}{ll}\int_ℝ{|f_1|}=\int_ℝ𝖾^{-(x-1)^2+1}𝖽x=𝖾\sqrtπ,&f_1∈𝖫^1(ℝ)\\\int_ℝ{|f_1|}^2=\int_ℝ𝖾^{-(\sqrt2x-\sqrt2)^2+2}𝖽x=𝖾^2\sqrt{π/2},&f_1∈𝖫^2(ℝ)\\\int_ℝ{|f_2|}=\int_0^∞𝖾^{-x}𝖽x=1,&f_2∈𝖫^1(ℝ)\\\int_ℝ{|f_2|}^2=\int_0^∞𝖾^{-2x}𝖽x=1/2,&f_2∈𝖫^2(ℝ)\\\int_ℝ{|f_3|}=2\int_0^∞𝖾^{-x}𝖽x=2,&f_3∈𝖫^1(ℝ)\\\int_ℝ{|f_3|}^2=2\int_0^∞𝖾^{-2x}𝖽x=1,&f_3∈𝖫^2(ℝ)\\\int_ℝ{|f_4|}=\int_ℝ\frac1{x^2+1}𝖽x=\left[\tan^{-1}x\right]_{-∞}^∞=π,&f_4∈𝖫^1(ℝ)\\\int_ℝ{|f_4|}^2=\int_ℝ\frac1{(x^2+1)^2}𝖽x=\frac12\left[\frac{x}{1+x^2} +\tan^{-1}x\right]_{-∞}^∞=\frac{π}2,&f_4∈𝖫^2(ℝ)\end{array}\] $f_1∈𝖢^∞$ and $(𝖾^{-x^2})^{(n)}=p_n(x)𝖾^{-x^2}$ for some polynomial $p_n(x)$. Since $𝖾^{x^2}=\sum_{n=0}^{∞}\frac{x^{2n}}{n!}$, for $n≥0$, for $x∈ℝ$, $|x^n𝖾^{-x^2}|≤\left|\frac{x^n}{1+\frac{x^{2n}}{n!}}\right|$, so $x^n𝖾^{-x^2}$ is bounded, thus $𝖾^{-x^2}$ and all of its derivatives are rapidly decreasing, so $f_1∈𝒮(ℝ)$. $\lim_{x→0+}f_2(x)=1,\lim_{x→0-}f_2(x)=0$, so $f_2∉𝖢^0$, so $f_2∉𝒮(ℝ)$. $\lim_{x→0+}f_3'(x)=-1,\lim_{x→0-}f_3'(x)=1$, so $f_3∉𝖢^1$, so $f_3∉𝒮(ℝ)$. $f_4∈𝖢^∞$ but $\lim_{x→+∞}x^3f_4(x)=+∞$, so $f_4∉𝒮(ℝ)$. (b) Calculate the Fourier transforms of these functions. Deduce Laplace’s integral identity \[∫_0^∞ \frac{\cos (x ξ)}{x^2+1}𝖽x=\frac{π}{2}𝖾^{-|ξ|}\] valid for all $ξ ∈ ℝ$. \begin{align*}\widehat{f_1}(ξ)&=∫_{-∞}^{∞}𝖾^{-x^2+2x}𝖾^{-𝗂ξx}𝖽x\\&=∫_{-∞}^{∞}𝖾^{(1-\frac{𝗂ξ}2)^2-(x-1+\frac{𝗂ξ}2)^2}𝖽x\\&=\sqrtπ𝖾^{(1-\frac{𝗂ξ}2)^2}\\&=\sqrtπ𝖾^{-\frac14(ξ+2𝗂)^2}\end{align*} \begin{align*}\widehat{f_2}(ξ)&=∫_{0}^{∞}𝖾^{-x}𝖾^{-𝗂ξx}𝖽x\\&=-\left[\frac{𝖾^{-(1+𝗂ξ)x}}{1+𝗂ξ} \right]_{0}^{∞}\\&=\frac{1}{1+𝗂ξ}\end{align*} \begin{align*}\widehat{f_3}(ξ)&=∫_{-∞}^{∞}𝖾^{-|x|}𝖾^{-𝗂ξx}𝖽x\\&=∫_{-∞}^{0}𝖾^x𝖾^{-𝗂ξx}𝖽x+∫_{0}^{∞}𝖾^{-x}𝖾^{-𝗂ξx}𝖽x\\&= \left[ \frac{𝖾^{(1-𝗂ξ)x}}{1-𝗂ξ} \right]_{-∞}^0-\left[\frac{𝖾^{-(1+𝗂ξ)x}}{1+𝗂ξ} \right]_{0}^{∞}\\&=\frac{1}{1-𝗂ξ}+\frac{1}{1+𝗂ξ}=\frac2{1+ξ^2}\end{align*} By Fourier inversion formula, \[𝖾^{-|ξ|}=\frac1{2π}\int_{-∞}^{∞}\frac{2}{1+x^2}𝖾^{𝗂ξx}𝖽x\] changing $ξ$ to $-ξ$ we get \[π𝖾^{-|ξ|}=\int_{-∞}^{∞}\frac1{1+x^2}𝖾^{-𝗂ξx}𝖽x\] so $\widehat{f_4}(ξ)=π𝖾^{-|ξ|}$, we deduce \begin{align*}\int^\infty_0 f_4(x)\cosξx\,𝖽x&=\frac{1}{2}\int^\infty_{-\infty} f_4(x)\cosξx\,𝖽x\\ &=\frac{1}{2}\int^\infty_{-\infty} f_4(x)\frac{𝖾^{𝗂ξx}+𝖾^{-𝗂ξx}}{2}\,𝖽x\\ &=\frac{1}{4}(\widehat{f_4}(-ξ)+\widehat{f_4}(ξ))=\frac12\widehat{f_4}(ξ)=\frac{π}2𝖾^{-|ξ|}.\end{align*} (c) For each of the Fourier transforms $\widehat{f_j}$, determine whether it is a function in $𝒮(ℝ)$, in $𝖫^1(ℝ)$, or in $𝖫^2(ℝ)$. $|\widehat{f_1}(ξ)|=\sqrtπ𝖾^{1-\frac14ξ^2}$, so $\widehat{f_1}∈𝒮(ℝ)$ by (a). $|\widehat{f_2}(ξ)|=\frac{1}{\sqrt{1+ξ^2}}$, $\int_{0}^{t}|\widehat{f_2}(ξ)|=\sinh^{-1}t→∞$ as $t→∞$, so $\widehat{f_2}(ξ)∉𝖫^1,\widehat{f_2}(ξ)∈𝖫^2,\widehat{f_2}(ξ)∉𝒮(ℝ)$. $|\widehat{f_3}(ξ)|=2f_4(ξ)$, so $\widehat{f_3}(ξ)∈𝖫^1,\widehat{f_3}(ξ)∈𝖫^2,\widehat{f_3}(ξ)∉𝒮(ℝ)$. $|\widehat{f_4}(ξ)|=πf_3(ξ)$, so $\widehat{f_4}(ξ)∈𝖫^1,\widehat{f_4}(ξ)∈𝖫^2,\widehat{f_4}(ξ)∉𝒮(ℝ)$.

KEITH CONRAD Characteristic 0 fields have a very handy feature: every irreducible polynomial in characteristic 0 is separable. Fields in characteristic $p$ may or may not have this feature. Definition 1. A field $K$ is called perfect if every irreducible polynomial in $K[T]$ is separable. Every field of characteristic 0 is perfect. We will see that finite fields are perfect too. The simplest example of a nonperfect field is the rational function field $\mathbf{F}_{p}(u)$, since $T^{p}-u$ is irreducible in $\mathbf{F}_{p}(u)[T]$ but not separable. Recall that an irreducible polynomial $\pi(T)$ is inseparable if and only if $\pi^{\prime}(T)=0$. Here is the standard way to check a field is perfect. Theorem 2. A field $K$ is perfect if and only if it has characteristic 0 , or it has characteristic $p$ and $K^{p}=K$. Proof. When $K$ has characteristic 0 , each irreducible $\pi(T)$ in $K[T]$ is separable since $\pi^{\prime}(T) \neq$ 0 (after all, $\operatorname{deg} \pi^{\prime}=\operatorname{deg} \pi-1$ ). It remains to show when $K$ has characteristic $p$ that every irreducible in $K[T]$ is separable if and only if $K^{p}=K$. To do this we will show the negations are equivalent: an inseparable irreducible exists in $K[T]$ if and only if $K^{p} \neq K$. If $K^{p} \neq K$, pick $a \in K-K^{p}$. Then $T^{p}-a$ has only one root in a splitting field: if $\alpha^{p}=a$ then $T^{p}-a=T^{p}-\alpha^{p}=(T-\alpha)^{p}$ since we are working in characteristic $p$. The polynomial $T^{p}-a$ is irreducible in $K[T]$ too: each nontrivial proper monic factor of $T^{p}-a$ is $(T-\alpha)^{m}$ where $1 \leq m \leq p-1$. The coefficient of $T^{m-1}$ in $(T-\alpha)^{m}$ is $-m \alpha$, so if $T^{p}-a$ has a nontrivial proper factor in $K[T]$ then $-m \alpha \in K$ for some $m$ from 1 to $p-1$. Then $m \in \mathbf{F}_{p}^{\times} \subset K^{\times}$, so $\alpha \in K$, which means $a=\alpha^{p} \in K^{p}$, a contradiction. Thus $T^{p}-a$ is irreducible and inseparable in $K[T]$. Now suppose there is an inseparable irreducible $\pi(T) \in K[T]$. Then $\pi^{\prime}(T)=0$, so $\pi(T)$ is a polynomial in $T^{p}$, say $$ \pi(T)=a_{m} T^{p m}+a_{m-1} T^{p(m-1)}+\cdots+a_{1} T^{p}+a_{0} \in K\left[T^{p}\right] $$ If $K^{p}=K$ then we can write $a_{i}=b_{i}^{p}$ for some $b_{i} \in K$, so $$ \pi(T)=\left(b_{m} T^{m}+b_{m-1} T^{m-1}+\cdots+b_{1} T+b_{0}\right)^{p} $$ (It was crucial for this conclusion that the coefficients of $\pi(T)$ are $p$ th powers and not only that $\pi(T)$ is a polynomial in $T^{p}$.) Since $\pi(T)$ is irreducible we have a contradiction, which shows $K^{p} \neq K$. Corollary 3. Fields of characteristic 0 and finite fields are perfect. Proof. By Theorem 2, fields of characteristic 0 are perfect. It remains to show a finite field $K$ of characteristic $p$ satisfies $K^{p}=K$. The $p$ th power map $K \rightarrow K$ is injective and therefore surjective because $K$ is finite, so we are done. The two types of fields listed in Corollary 3 are the most basic examples of perfect fields. Other perfect fields can show up, especially in the middle of technical proofs about fields. Corollary 3 says irreducibles in $\mathbf{F}_{p}[T]$ have no repeated roots, but the proof was rather indirect. We now give a more elementary proof that each irreducible $\pi(T)$ in $\mathbf{F}_{p}[T]$ is separable. The idea is to show $\pi(T)$ is a factor of a polynomial which we can directly check has no repeated roots. We may assume $\pi(T) \neq T$, so $T \bmod \pi$ is a unit in $\mathbf{F}_{p}[T] /(\pi)$. Since $\pi(T)$ is irreducible, $\mathbf{F}_{p}[T] /(\pi)$ is a field with size $p^{d}$, where $d=\operatorname{deg} \pi$. The nonzero elements of $\mathbf{F}_{p}[T] /(\pi)$ are a group of size $p^{d}-1$, so $T^{p^{d}-1} \equiv 1 \bmod \pi$. Multiplying through by $T$, we get $T^{p^{d}} \equiv T \bmod \pi$, so $\pi(T) \mid\left(T^{p^{d}}-T\right)$ in $\mathbf{F}_{p}[T]$. The polynomial $T^{p^{d}}-T$ has no repeated roots (in a splitting field): if $\alpha^{p^{d}}-\alpha=0$ then $$ \begin{aligned} T^{p^{d}}-T & =T^{p^{d}}-T-\left(\alpha^{p^{d}}-\alpha\right) \\ & =\left(T^{p^{d}}-\alpha^{p^{d}}\right)-(T-\alpha) \\ & =(T-\alpha)^{p^{d}}-(T-\alpha) \\ & =(T-\alpha)\left((T-\alpha)^{p^{d}-1}-1\right) \end{aligned} $$ The second factor $(T-\alpha)^{p^{d}-1}-1$ has value -1 at $T=\alpha$, so $\alpha$ is a root with multiplicity 1. Since $\alpha$ was taken to be an arbitrary root of $T^{p^{d}}-T$, we see that $T^{p^{d}}-T$ is separable. Therefore its factor $\pi(T)$ is also separable. Theorem 4. A field $K$ is perfect if and only if every finite extension of $K$ is a separable extension. Proof. Suppose $K$ is perfect: every irreducible in $K[T]$ is separable. If $L / K$ is a finite extension then the minimal polyomial in $K[T]$ of every element of $L$ is irreducible and therefore separable, so $L / K$ is a separable extension. Now suppose every finite extension of $K$ is a separable extension. To show $K$ is perfect, let $\pi(T) \in K[T]$ be irreducible. Consider the field $L=K(\alpha)$, where $\pi(\alpha)=0$. This field is a finite extension of $K$, so a separable extension by hypothesis, so $\alpha$ is separable over $K$. Since $\pi(T)$ is the minimal polynomial of $\alpha$ in $K[T]$, it is a separable polynomial. A lot of results about field theory that are valid in characteristic 0 carry over to perfect fields in characteristic $p$ (but not everything), and the reader should be attentive to this point when reading texts which try to make life easy by always assuming fields have characteristic 0 . You should always check if the theorems (and even proofs) go through to general perfect fields.

Proposition. Let $l_1,l_2,l_3$ be lines in $P(ℝ^4)$ that do not intersect pairwise. Then there are an infinite number of lines in $l$ in $P(ℝ^4)$ that intersect $l_1,l_2,l_3$. Proof. Let $l_j=P(U_j),j=1,2,3$. Then $U_1,U_2,U_3$ are vector subspaces of $ℝ^4$, $\dim U_j=2$. Since $l_1\cap l_2=\emptyset$, $U_1\cap U_2=\{0\}$. Hence $ℝ^4=U_1\oplus U_2$, counting dimensions. Let $v\in U_3$ be non-zero, then $v=u_1+u_2,u_1\in U_1,u_2\in U_2$. If $u_1=0$ then $[v]\in l_2\cap l_3$❌ If $u_2=0$ then $[v]\in l_1\cap l_3$❌ Let $U_4=\langle u_1,u_2\rangle\leℝ^4$. Then $\dim U_4=2$. So $l_4=P(U_4)$ is a line in $P(ℝ^4)$. Also, $l_4$ contains $[u_1]\in l_1$ and $[u_2]\in l_2$ and $[u_1+u_2]=[v]\in l_3$. So $l_4$ intersects $l_1,l_2,l_3$. Unique such line intersecting $l_3$ in the point $[v]$. By construction, there exists such a line through each point $[v]$ in $l_3$, so there are infinitely many such lines $l$. □ Here is a more difficult proof. Desargues’ theorem Let $P,Q_1,Q_2,Q_3,R_1,R_2,R_3$ be distinct points in a projective plane $P(V)$, such that the lines $Q_1R_1,Q_2R_2,Q_3R_3$ are distinct and concurrent at $P$. Let $S_1$ be the intersection of $Q_2Q_3,R_2R_3$ Let $S_2$ be the intersection of $Q_3Q_1,R_3R_1$ Let $S_3$ be the intersection of $Q_1Q_2,R_1R_2$ Then $S_1,S_2,S_3$ are collinear. Proof. Choose representatives $p,q_i,r_i$ such that $p=q_i+r_i$. As $p=q_2+r_2=q_3+r_3$, put $s_1:=q_2-q_3=r_3-r_2$. Since $Q_2,Q_3$ are distinct, $s_1$ is nonzero. Hence $s_1$ represents a point on the line $Q_2Q_3$. Similarly, $s_1$ represents a point on the line $R_2R_3$. Hence $s_1$ represents $S_1$, the intersection of $Q_2Q_3,R_2R_3$. Similarly, put $s_2:=q_3-q_1=r_1-r_3,s_3:=q_1-q_2=r_2-r_1$, and then $s_2,s_3$ represent $S_2,S_3$, Now, $s_1+s_2+s_3=(q_2-q_3)+(q_3-q_1)+(q_1-q_2)=0$. Hence $S_1,S_2,S_3$ are collinear, as $S_1,S_2,S_3$ are linearly dependent. □

Theorem 1.2. For any $f ∈ L^1$ and any $ε>0$ there is a $g$ in $C_c(ℝ^n)$ such that $∫|f-g|dm<ε$. Since $L^1$ functions can be approximated in $L^1$ by integrable simple functions, it suffices to prove the theorem when $f$ is a simple function. Next, since an integrable simple function is a linear combination of functions of the form $χ_E$, where $E$ is a measurable set of finite measure, it suffices to prove the theorem when $f=χ_E$. Therefore, Theorem 1.2 will be proved by establishing Proposition 1.4. Let $E$ be a measurable subset of $ℝ^n$ with finite measure. Then for any $ε>0$ there is a $g ∈ C_c(ℝ^n)$ with $∫|χ_E-g| d m<ε$. Proof. By regularity properties of Lebesgue measure there is a compact set $K$ and an open set $U$ such that $K ⊂ E ⊂ U$ and $m(U∖K)<ε$. By Lemma 1.3, there is a $g ∈ C_c(ℝ^n)$ such that $0 ≤ g ≤ 1$ everywhere, $g=1$ on $K$, and $g$ vanishes outside of $U$. It follows that $|g-χ_E|$ vanishes outside of $U∖K$, and that $|g-χ_E| ≤ 1$ on $U∖K$. Therefore $∫|g-χ_E| d m=∫_{U∖K}|g-χ_E| d m ≤ m(U∖K)<ε$. ∎ In this section, we'll prove the version of Urysohn's Lemma that we used in the proof of Theorem 1.2. We'll work in a locally compact metric space $(X, d)$. We'll use the notation $B_r(x)$ for the open ball in $X$ with center $x$ and radius $r$. Lemma 2.1. Let $x_0 ∈ X$ and $r>0$. There is $a χ ∈ C_c(X)$ with 1. $0 ≤ χ(x) ≤ 1$ for every $x ∈ X$; 2. $χ(x)=1$ for every $x ∈ B_r(x_0)$; 3. The support of $χ$ is a subset of $B_{2 r}(x_0)$. Proof. Let $φ$ be the continuous function on $ℝ$ defined by $φ(t)=1$ for $0 ≤ t ≤ r$, $φ(t)=1-\frac{2}{r}(t-r)$ for $r\frac{3 r}{2}$. Let $χ(x)=φ(d(x, x_0))$. Proof of Lemma 1.3. Cover $K$ by finitely many balls $B_j=B_{r_j}(x_j)$ such that $\overline{B_{2 r_j}(x_j)}$ is a compact subset of $U$. Let $χ_j$ be the function obtained from Lemma 2.1 with $x_0=x_j$ and $r=r_j$. Let $φ=\sum χ_j$. Then 1. $φ$ is continuous on $X$; 2. The support of $φ$ is a compact subset of $U$; 3. $φ ≥ 0$ everywhere, and $φ ≥ 1$ on $K$. Let $χ=\min \{φ, 1\}$. Then $χ$ has all the required properties. ∎

The Acyclic Assembly Lemma gives conditions for the total complex of a double complex of modules over a ring to be acyclic. A full statement and proof can be found in An Introduction to Homological Algebra by Chuck Weibel on pages 59 and 60 (Lemma 2.7.3). $C$ is an upper half-plane double acyclic with exact columns. Thus $\operatorname{Tot}^Π(C)$ is exact. Proof: Enough to check that $H_0(\operatorname{Tot}^Π(C))=0$. $\operatorname{Tot}^Π(C)_0=Π_{p∈ℤ}C_{-p,p}∋c=(…,c_{-2,2},c_{-1,1},c_{0,0},0,…)$ be a 0-cycle, we will construct by induction elements $b_{-p,p+1}$ s.t. $d^v(b_{-p,p+1})+d^h(b_{-p+1,p})=c_{-p,p}$, this will give an element of $b∈Π_{p∈ℤ}C_{-p,p+1}$ s.t. $d(b)=c$ showing that $H_0(\operatorname{Tot}^Π(C))=0$ Let $b_{10}=0$ for $n=1$. $c_{01}=0$, so $d^v(c_{00})=0$ since the 0th column is exact, there is a $b_{01}∈C_{01}$ s.t. $d^v(b_{01})=c_{00}$. By induction $d^v(c_{-p,p}-d^h(b_{-p+1,p}))=d^v(c_{-p,p})+d^hd^v(b_{-p+1,p})=d^v(c_{-p,p})+d^h(c_{-p+1,p-1})-d^hd^h(b_{-p+2,p-1})=0$ since the $-p$th column is exact there is a $b_{-p,p+1}$ s.t. $d^v(b_{-p,p+1})=c_{-p,p}-d^h(b_{-p+1,p})$. Remark: spectral sequences gives you a way to calculate the homology of total complex