Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Example 28.1.4 (Derivative of \( {x}^{3} \) is \( 3{x}^{2} \) ) | Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = {x}^{3} \) . For any point \( p \), and nonzero \( h \) we can compute\n\n\[ \n\frac{f\left( {p + h}\right) - f\left( p\right) }{h} = \frac{{\left( p + h\right) }^{3} - {p}^{3}}{h} \n\]\n\n\[ \n= \frac{3{p}^{2}h + {3p}{h}^{2} + {h}^{3}}{h} \n\]\n\... | Yes |
Theorem 28.2.1 (Your friendly high school calculus rules)\n\nIn what follows \( f \) and \( g \) are differentiable functions, and \( U, V \) are open subsets of \( \mathbb{R} \) .\n\n- (Sum rule) If \( f, g : U \rightarrow \mathbb{R} \) then then \( {\left( f + g\right) }^{\prime }\left( x\right) = {f}^{\prime }\left(... | Proof. - Sum rule: trivial, do it yourself if you care.\n\n- Product rule: for every nonzero \( h \) and point \( p \in U \) we may write\n\n\[ \frac{f\left( {p + h}\right) g\left( {p + h}\right) - f\left( p\right) g\left( p\right) }{h} = \frac{f\left( {p + h}\right) - f\left( p\right) }{h} \cdot g\left( {p + h}\right)... | No |
Let \( r \) be a real number. The function \( {\mathbb{R}}_{ > 0} \rightarrow \mathbb{R} \) by \( x \mapsto {x}^{r} \) has derivative \( {\left( {x}^{r}\right) }^{\prime } = r{x}^{r - 1} \) . | Proof. We knew this for integers \( r \) already, but now we can prove it for any positive real number \( r \) . Write\n\n\[ f\left( x\right) = {x}^{r} = {e}^{r\log x} \]\n\nconsidered as a function \( f : {\mathbb{R}}_{ > 0} \rightarrow \mathbb{R} \) . The chain rule (together with the fact that \( \left. {{\left( {e}... | Yes |
Corollary 28.2.7 (Derivative of \( \log \) is \( 1/x \) ) | The function \( \log : {\mathbb{R}}_{ > 0} \rightarrow \mathbb{R} \) has derivative \( {\left( \log x\right) }^{\prime } = 1/x \) . Proof. We have that \( x = {e}^{\log x} \) . Differentiate both sides, and again use the chain rule \( {}^{3} \n\n\[ \n1 = {e}^{\log x} \cdot {\left( \log x\right) }^{\prime }. \n\] \n\nTh... | Yes |
Theorem 28.3.3 (Fermat's theorem on stationary points)\n\nSuppose \( f : U \rightarrow \mathbb{R} \) is differentiable and \( p \in U \) is a local extrema. Then \( {f}^{\prime }\left( p\right) = 0 \) . | Proof. Assume for contradiction \( {f}^{\prime }\left( p\right) > 0 \) . Choose any \( \varepsilon > 0 \) with \( \varepsilon < {f}^{\prime }\left( p\right) \) . Then for sufficiently small \( \left| h\right| \) we should have\n\n\[ \frac{f\left( {p + h}\right) - f\left( p\right) }{h} > \varepsilon \]\n\nIn particular ... | Yes |
Proposition 28.3.5 \( \left( {{e}^{x} \geq 1 + x}\right) \)\n\nFor all real numbers \( x \) we have \( {e}^{x} \geq 1 + x \) . | Proof. Define the differentiable function\n\n\[ f\left( x\right) = {e}^{x} - \left( {x + 1}\right) \;f : \mathbb{R} \rightarrow \mathbb{R}. \]\n\nConsider the compact interval \( Z = \left\lbrack {-1,{100}}\right\rbrack \) . If \( x \leq - 1 \) then obviously \( f\left( x\right) > 0 \) . Similarly if \( x \geq {100} \)... | Yes |
Theorem 28.4.1 (Rolle's theorem)\n\nSuppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a continuous function, which is differentiable on the open interval \( \left( {a, b}\right) \), such that \( f\left( a\right) = f\left( b\right) \) . Then there is a point \( c \in \left( {a, b}\right) \) su... | Proof. Assume \( f \) is nonconstant (otherwise any \( c \) works). By compactness, there exists both a global maximum and minimum. As \( f\left( a\right) = f\left( b\right) \), either the global maximum or the global minimum must lie inside the open interval \( \left( {a, b}\right) \), and then Fermat’s theorem on sta... | Yes |
Theorem 28.4.2 (Mean value theorem)\n\nSuppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a continuous function, which is differentiable on the open interval \( \left( {a, b}\right) \) . Then there is a point \( c \in \left( {a, b}\right) \) such that\n\n\[{f}^{\prime }\left( c\right) = \frac{... | Proof of mean value theorem. Let \( s = \frac{f\left( b\right) - f\left( a\right) }{b - a} \) be the slope of the secant line, and define\n\n\[g\left( x\right) = f\left( x\right) - {sx}\]\nwhich intuitively shears \( f \) downwards so that the secant becomes vertical. In fact \( g\left( a\right) = g\left( b\right) \) n... | Yes |
Proposition 28.4.4 (Racetrack principle)\n\nLet \( f, g : \mathbb{R} \rightarrow \mathbb{R} \) be two differentiable functions with \( f\left( 0\right) = g\left( 0\right) \).\n\n(a) If \( {f}^{\prime }\left( x\right) \geq {g}^{\prime }\left( x\right) \) for every \( x > 0 \), then \( f\left( x\right) \geq g\left( x\rig... | Proof of racetrack principle. We prove (a). Let \( h = f - g \), so \( h\left( 0\right) = 0 \) . Assume for contradiction \( h\left( p\right) < 0 \) for some \( p > 0 \) . Then the secant joining \( \left( {0, h\left( 0\right) }\right) \) to \( \left( {p, h\left( p\right) }\right) \) has negative slope; in other words ... | Yes |
Theorem 28.4.5 (Ratio mean value theorem)\n\nLet \( f, g : \\left\\lbrack {a, b}\\right\\rbrack \\rightarrow \\mathbb{R} \) be two continuous functions which are differentiable on \( \\left( {a, b}\\right) \) , and such that \( g\\left( a\\right) \\neq g\\left( b\\right) \) . Then there is a \( c \\in \\left( {a, b}\\r... | Proof. Use Rolle's theorem on the function\n\n\[ \nh\\left( x\\right) = \\left\\lbrack {f\\left( x\\right) - f\\left( a\\right) }\\right\\rbrack \\left\\lbrack {g\\left( b\\right) - g\\left( a\\right) }\\right\\rbrack - \\left\\lbrack {g\\left( x\\right) - g\\left( a\\right) }\\right\\rbrack \\left\\lbrack {f\\left( b\... | Yes |
Theorem 29.2.4 (Cauchy-Hadamard theorem)\n\nLet \( \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{a}_{n}{z}^{n} \) be a power series with radius of convergence \( R \) . Let \( h \) be a real number, and consider the infinite series\n\n\[ \n\\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{a}_{n}{h}^{n} \n\]\n\nof real numb... | Proof. This is not actually hard, but it won't be essential, so not included. | No |
Example 29.2.6 \( \left( {\sum {z}^{n}}\right. \) has radius 1 \( ) \) | Consider the geometric series \( \mathop{\sum }\limits_{n}{z}^{n} = 1 + z + {z}^{2} + \ldots \) Since \( {a}_{n} = 1 \) for every \( n \) , we get \( R = 1 \), which is what we expected. | Yes |
Theorem 29.3.1 (Differentiation works term by term)\n\nLet \( \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{z}^{n} \) be a power series with radius of convergence \( R > 0 \), and consider the corresponding function\n\n\[ f : \left( {-R, R}\right) \rightarrow \mathbb{R}\;\text{ by }\;f\left( x\right) = \mathop{\sum }\limit... | Proof. Also omitted. The right way to prove it is to define the notion \ | No |
Corollary 29.3.2 (A description of power series coefficients)\n\nLet \( \\mathop{\\sum }\\limits_{{n \\geq 0}}{a}_{n}{z}^{n} \) be a power series with radius of convergence \( R > 0 \), and consider the corresponding function \( f\\left( x\\right) \) as above. Then\n\n\[ \n{a}_{n} = \\frac{{f}^{\\left( n\\right) }\\lef... | Proof. Take the \( n \) th derivative and plug in \( x = 0 \) . | No |
Example 29.4.4 (Examples of analytic functions)\n\n(b) The smooth function from before defined by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} \exp \left( {-1/x}\right) & x > 0 \\ 0 & x \leq 0 \end{array}\right. \] | is not analytic. Indeed, suppose for contradiction it was. As all the derivatives are zero, its Taylor series would be \( 0 + {0x} + 0{x}^{2} + \ldots \) . This Taylor series does converge, but not to the right value - as \( f\left( \varepsilon \right) > 0 \) for any \( \varepsilon > 0 \), contradiction. | Yes |
It now follows that \( f\left( x\right) = \sin \left( x\right) \) is analytic. | To see that, we can compute\n\n\[ f\left( 0\right) = \sin 0 = 0 \]\n\n\[ {f}^{\prime }\left( 0\right) = \cos 0 = 1 \]\n\n\[ {f}^{\prime \prime }\left( 0\right) = - \sin 0 = 0 \]\n\n\[ {f}^{\left( 3\right) }\left( 0\right) = - \cos 0 = - 1 \]\n\n\[ {f}^{\left( 4\right) }\left( 0\right) = \sin 0 = 0 \]\n\n\[ {f}^{\left( ... | Yes |
Proposition 29.5.3 (exp is multiplicative)\n\nUnder this definition,\n\n\[ \exp \left( {x + y}\right) = \exp \left( x\right) \exp \left( y\right) \] | Idea of proof. There is some subtlety here with switching the order of summation that we won't address. Modulo that:\n\n\[ \exp \left( x\right) \exp \left( y\right) = \mathop{\sum }\limits_{{n \geq 0}}\frac{{x}^{n}}{n!}\mathop{\sum }\limits_{{m \geq 0}}\frac{{y}^{m}}{m!} = \mathop{\sum }\limits_{{n \geq 0}}\mathop{\sum... | No |
Corollary 29.5.4 (exp is positive)\n\n(a) We have \( \exp \left( x\right) > 0 \) for any real number \( x \) .\n\n(b) The function exp is strictly increasing. | Proof. First\n\n\[ \exp \left( x\right) = \exp {\left( x/2\right) }^{2} \geq 0 \]\n\nwhich shows exp is nonnegative. Also, \( 1 = \exp \left( 0\right) = \exp \left( x\right) \exp \left( {-x}\right) \) implies \( \exp \left( x\right) \neq 0 \) for any \( x \), proving (a).\n\n(b) is just since \( {\exp }^{\prime } \) is... | Yes |
Theorem 30.1.3 (Uniform continuity free for compact spaces)\n\nLet \( M \) be a compact metric space. Then any continuous map \( f : M \rightarrow N \) is also uniformly continuous. | Proof. Assume for contradiction there is some bad \( \varepsilon > 0 \) . Then taking \( \delta = 1/n \), we find that for each integer \( n \) there exists points \( {p}_{n} \) and \( {q}_{n} \) which are within \( 1/n \) of each other, but are mapped more than \( \varepsilon \) away from each other by \( f \) . In sy... | Yes |
Theorem 30.2.4 (Extending uniformly continuous functions)\n\nLet \( M \) be a metric space, \( N \) a complete metric space, and \( S \) a dense subspace of \( M \) . Suppose \( \psi : S \rightarrow N \) is a uniformly continuous function. Then there exists a unique continuous function \( \widetilde{\psi } : M \rightar... | Outline of proof. As mentioned in the discussion, each \( x \in M \) can be approximated by a sequence \( {x}_{1},{x}_{2},\ldots \) in \( S \) with \( {x}_{i} \rightarrow x \) . The two main hypotheses, completeness and uniform continuity, are now used:\n\nExercise 30.2.5. Prove that \( \psi \left( {x}_{1}\right) ,\psi... | No |
Theorem 30.3.3 (The Riemann integral)\n\nThere exists a unique continuous map\n\n\\[ \n{\\int }_{a}^{b} : M\\left( \\left\\lbrack {a, b}\\right\\rbrack \\right) \\rightarrow \\mathbb{R} \n\\]\n\nsuch that the diagram  \\) is a dense subset of \\( M\\left( \\left\\lbrack {a, b}\\right\\rbrack \\right) \\) . In other words, for any continuous \\( f : \\left\\lbrack {a, b}\\right... | Yes |
Theorem 30.4.2 (Riemann integral)\n\nLet \( f : \\left\\lbrack {a, b}\\right\\rbrack \\rightarrow \\mathbb{R} \) be continuous. Then\n\n\[ \n{\\int }_{a}^{b}f\\left( x\\right) {dx} = \\mathop{\\lim }\\limits_{\\substack{{P\\text{ tagged partition }} \\\\ {\\text{ mesh }P \\rightarrow 0} }}\\left( {\\mathop{\\sum }\\lim... | Proof. The right-hand side corresponds to the areas of some rectangle functions \( {g}_{1},{g}_{2} \) , ...with increasingly narrow rectangles. As in the proof Theorem 30.3.3, as the meshes of those rectangles approaches zero, by uniform continuity, we have \( d\\left( {f,{g}_{n}}\\right) \\rightarrow 0 \) as well. Thu... | No |
Theorem 31.4.2 (Cauchy-Goursat theorem)\n\nLet \( \gamma \) be a loop, and \( f : \Omega \rightarrow \mathbb{C} \) a holomorphic function where \( \Omega \) is open in \( \mathbb{C} \) and simply connected. Then\n\n\[{\oint }_{\gamma }f\left( z\right) {dz} = 0\] | The typical proof of Cauchy's Theorem assumes additionally that the partial derivatives of \( f \) are continuous and then applies the so-called Green’s theorem. But it was Goursat who successfully proved the fully general theorem we've stated above, which assumed only that \( f \) was holomorphic. I’ll only outline th... | No |
Theorem 31.5.1 (Cauchy's integral formula)\n\nLet \( \gamma : \left\lbrack {0,{2\pi }}\right\rbrack \rightarrow \mathbb{C} \) be a circle in the plane given by \( t \mapsto R{e}^{it} \), which bounds a disk \( D \) . Suppose \( f : U \rightarrow \mathbb{C} \) is holomorphic such that \( U \) contains the circle and its... | Proof. There are several proofs out there, but I want to give the one that really draws out the power of Cauchy’s theorem. Here’s the picture we have: there’s a point \( a \) sitting inside a circle \( \gamma \), and we want to get our hands on the value \( f\left( a\right) \) .\n\n![1dfd4520-d1c6-4b94-9568-57d535640b4... | No |
Lemma 31.5.3 ( \( {ML} \) estimation lemma)\n\nLet \( f \) be a holomorphic function and \( \alpha \) a path. Suppose \( M = \mathop{\max }\limits_{{z\text{on }\alpha }}\left| {f\left( z\right) }\right| \), and let \( L \) be the length of \( \alpha \) . Then\n\n\[ \left| {{\oint }_{\alpha }f\left( z\right) {dz}}\right... | (This is straightforward to prove if you know the definition of length: \( L = {\int }_{a}^{b}\left| {{\alpha }^{\prime }\left( t\right) }\right| {dt} \) , where \( \alpha : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) .) | No |
Theorem 31.6.1 (Cauchy's differentiation formula)\n\nLet \( f : U \rightarrow \mathbb{C} \) be a holomorphic function and let \( D \) be a disk centered at point \( p \) bounded by a circle \( \gamma \) . Suppose \( D \) is contained inside \( U \) . Then \( f \) is given everywhere in \( D \) by a Taylor series\n\n\[ ... | Most importantly,\n\nOver any disk, a holomorphic function is given exactly by a Taylor series.\n\nThis establishes a result we stated at the beginning of the chapter: that a function being complex differentiable once means it is not only infinitely differentiable, but in fact equal to its Taylor series. | Yes |
Consider the function\n\\[ \n\frac{z + 1}{\sin z}\text{.}\n\\]\n\nIt is meromorphic, because it is holomorphic everywhere except at the zeros of \\( \\sin z \\) . | At each of these points we can put a Laurent series: for example at \\( z = 0 \\) we have\n\n\\[ \n\frac{z + 1}{\sin z} = \\left( {z + 1}\\right) \\cdot \\frac{1}{z - \\frac{{z}^{3}}{3!} + \\frac{{z}^{5}}{5!} - \\ldots }\n\\]\n\n\\[ \n= \\frac{1}{z} \\cdot \\frac{z + 1}{1 - \\left( {\\frac{{z}^{2}}{3!} - \\frac{{z}^{4}... | Yes |
Theorem 32.3.3 (Cauchy's residue theorem)\n\nLet \( f : \Omega \rightarrow \mathbb{C} \) be meromorphic, where \( \Omega \) is simply connected. Then for any loop \( \gamma \) not passing through any of its poles, we have\n\n\[ \n\frac{1}{2\pi i}{\oint }_{\gamma }f\left( z\right) {dz} = \mathop{\sum }\limits_{{\text{po... | Proof. Let the poles with nonzero winding number be \( {p}_{1},\ldots ,{p}_{k} \) (the others do not affect the sum). \( {}^{1} \) Then we can write \( f \) in the form\n\n\[ \nf\left( z\right) = g\left( z\right) + \mathop{\sum }\limits_{{i = 1}}^{k}{P}_{i}\left( \frac{1}{z - {p}_{i}}\right)\n\]\n\n\( {}^{1} \) To show... | Yes |
Proposition 32.4.1 (The logarithmic derivative)\n\nLet \( f : U \rightarrow \mathbb{C} \) be a meromorphic function. Then the logarithmic derivative \( {f}^{\prime }/f \) is meromorphic as a function from \( U \) to \( \mathbb{C} \) ; its zeros and poles are:\n\n(i) A pole at each zero of \( f \) whose residue is the m... | Proof. Dead easy with Taylor series. Let \( a \) be a zero/pole of \( f \), and WLOG set \( a = 0 \) for convenience. We take the Taylor series at zero to get\n\n\[ f\left( z\right) = {c}_{k}{z}^{k} + {c}_{k + 1}{z}^{k + 1} + \ldots \]\n\nwhere \( k < 0 \) if 0 is a pole and \( k > 0 \) if 0 is a zero. Taking the deriv... | Yes |
Theorem 32.4.2 (Argument principle)\n\nLet \( \gamma \) be a regular curve. Suppose \( f : U \rightarrow \mathbb{C} \) is meromorphic inside and on \( \gamma \), and none of its zeros or poles lie on \( \gamma \) . Then\n\n\[ \n\frac{1}{2\pi i}{\oint }_{\gamma }\frac{{f}^{\prime }}{f}{dz} = Z - P \n\] \n\nwhere \( Z \)... | Proof. Immediate by applying Cauchy's residue theorem alongside the preceding proposition. In fact you can generalize to any curve \( \gamma \) via the winding number: the integral is\n\n\[ \n\frac{1}{2\pi i}{\oint }_{\gamma }\frac{{f}^{\prime }}{f}{dz} = \mathop{\sum }\limits_{{\text{zero }z}}\mathbf{I}\left( {\gamma,... | Yes |
Example 34.5.1 (A geometric example of why \( \mathcal{A} = {2}^{\Omega } \) is unsuitable)\n\nLet \( \Omega \) denote the unit circle in \( {\mathbb{R}}^{2} \) and \( \mathcal{A} = {2}^{\Omega } \) . We will show that any measure \( \mu \) on \( \Omega \) with \( \mu \left( \Omega \right) = 1 \) will have undesirable ... | Let \( \sim \) denote an equivalence relation on \( \Omega \) defined as follows: two points are equivalent if they differ by a rotation around the origin by a rational multiple of \( \pi \) . We may pick a representative from each equivalence class, letting \( X \) denote the set of representatives. Then\n\n\[ \Omega ... | Yes |
Proposition 34.6.2 (Continuous implies Borel measurable)\n\nSuppose \( X \) and \( Y \) are topological spaces and we pick the Borel measures on both. A function \( f : X \rightarrow Y \) which is continuous as a map of topological spaces is also measurable. | Proof. Follows from the fact that pre-images of open sets are open, and the Borel measure is generated by open sets. | Yes |
Theorem 35.2.2 (Constructing outer measures from pre-measures)\n\nLet \( {\mu }_{0} \) be a pre-measure on an algebra \( {\mathcal{A}}_{0} \) on a set \( \Omega \) .\n\n(a) The map \( {\mu }^{ * } : {2}^{\Omega } \rightarrow \left\lbrack {0, + \infty }\right\rbrack \) defined by\n\n\[ \n{\mu }^{ * }\left( E\right) = \i... | Proof of Theorem 35.2.2. As alluded to already, part (a) is a special case of Problem \( {35}{\mathrm{\;A}}^{ \dagger } \) (and proving it in this generality is actually easier, because you won't be distracted by unnecessary properties).\n\nWe now check (b), that \( {\mu }^{ * }\left( A\right) = {\mu }_{0}\left( A\righ... | No |
Proposition 35.3.2 (Carathéodory measurability is compatible with the initial \( {\mathcal{A}}_{0} \) ) Suppose \( {\mu }^{ * } \) was obtained from a pre-measure \( {\mu }_{0} \) on an algebra \( {\mathcal{A}}_{0} \), as in Theorem 35.2.2. Then every set in \( {\mathcal{A}}_{0} \) is \( {\mu }^{ * } \) -measurable. | Proof. Let \( A \in {\mathcal{A}}_{0} \) and \( E \in {2}^{\Omega } \) ; we wish to prove \( {\mu }^{ * }\left( E\right) = {\mu }^{ * }\left( {E \cap A}\right) + {\mu }^{ * }\left( {E \smallsetminus A}\right) \) . The definition of outer measure already requires \( {\mu }^{ * }\left( E\right) \leq {\mu }^{ * }\left( {E... | Yes |
Theorem 35.4.7 (Lebesgue is completion of Borel)\n\nFor \( {\mathbb{R}}^{n} \), the Lebesgue measure is the completion of the Borel measure. | Proof. This actually follows from results in the next section, namely Exercise 35.5.1 and part (c) of Carathéodory for pre-measures (Theorem 35.5.5). | No |
Theorem 35.5.5 (Carathéodory extension theorem for pre-measures)\n\nLet \( {\mu }_{0} \) be a pre-measure on an algebra \( {\mathcal{A}}_{0} \) of \( \Omega \), and let \( \mathcal{A} \) denote the \( \sigma \) -algebra generated by \( {\mathcal{A}}_{0} \) . Let \( {\mathcal{A}}^{\mathrm{{cm}}},{\mu }^{\mathrm{{cm}}} \... | Proof of Theorem 35.5.5. For (a): this is just Theorem 35.2.2 and Theorem 35.3.4 put together, combined with the observation that \( {\mathcal{A}}^{ * } \supset {\mathcal{A}}_{0} \) and hence \( {\mathcal{A}}^{ * } \supset \mathcal{A} \) . Parts (b) and (c) are more technical, and omitted. | No |
Example 36.2.2 (Integrating \( {x}^{2} \) over \( \left\lbrack {1,4}\right\rbrack \) ) | \[ {\int }_{\left\lbrack 1,4\right\rbrack }{x}^{2}{d\mu } = {\int }_{1}^{4}{x}^{2}{dx} = \frac{1}{3} \cdot {4}^{3} - \frac{1}{3} \cdot {1}^{3} = {21}. \] | Yes |
Theorem 36.2.3 (Improper integrals are nice Lebesgue ones)\n\nLet \( f \geq 0 \) be a nonnegative continuous function defined on \( \left( {a, b}\right) \subseteq \mathbb{R} \), possibly allowing \( a = - \infty \) or \( b = \infty \) . Then\n\n\[ \n{\int }_{\left( a, b\right) }{fd\mu } = \mathop{\lim }\limits_{\substa... | The right-hand side makes sense since \( \left\lbrack {{a}^{\prime },{b}^{\prime }}\right\rbrack \varsubsetneq \left( {a, b}\right) \) is a compact interval on which \( f \) is continuous. This means that improper Riemann integrals of nonnegative functions can just be regarded as Lebesgue ones over the corresponding op... | Yes |
Example 37.1.1 ( \( {\mathbf{1}}_{\mathbb{Q}} \) is a limit of finitely supported functions) | We can define the sequence of functions \( {g}_{1},{g}_{2},\ldots \) by\n\n\[ \n{g}_{n}\left( x\right) = \left\{ \begin{array}{ll} 1 & \left( {n!}\right) x\text{ is an integer } \\ 0 & \text{ else. } \end{array}\right.\n\]\n\nThen each \( {g}_{n} \) is piecewise continuous and hence Riemann integrable on \( \left\lbrac... | Yes |
Corollary 37.4.1 (Monotone convergence theorem)\n\nLet \( f \) and \( {f}_{1},{f}_{2},\ldots : \Omega \rightarrow \left\lbrack {0, + \infty }\right\rbrack \) be a sequence of nonnegative measurable functions such that \( \mathop{\lim }\limits_{n}{f}_{n} = f \) and \( {f}_{n}\left( \omega \right) \leq f\left( \omega \ri... | Proof. We have\n\n\[ {\int }_{\Omega }{fd\mu } = {\int }_{\Omega }\left( {\mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}}\right) {d\mu } \]\n\n\[ \leq \mathop{\liminf }\limits_{{n \rightarrow \infty }}{\int }_{\Omega }{f}_{n}{d\mu } \]\n\n\[ \leq \mathop{\limsup }\limits_{{n \rightarrow \infty }}{\int }_{\Om... | Yes |
Example 37.4.3 (Monotone convergence gives \( {\mathbf{1}}_{\mathbb{Q}} \) ) | This already implies Example 37.1.1. Letting \( {g}_{n} \) be the indicator function for \( \frac{1}{n!}\mathbb{Z} \) as described in that example, we have \( {g}_{n} \leq {\mathbf{1}}_{\mathbb{Q}} \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{g}_{n}\left( x\right) = {\mathbf{1}}_{\mathbb{Q}}\left( x\right)... | Yes |
Corollary 37.4.4 (Fatou-Lebesgue theorem)\n\nLet \( f \) and \( {f}_{1},{f}_{2},\ldots : \Omega \rightarrow \mathbb{R} \) be a sequence of measurable functions. Assume that \( g : \Omega \rightarrow \mathbb{R} \) is an absolutely integrable function for which \( \left| {{f}_{n}\left( \omega \right) }\right| \leq \left|... | Proof. There are three inequalities:\n\n- The first inequality follows by Fatou on \( g + {f}_{n} \) which is nonnegative.\n\n- The second inequality is just \( \liminf \leq \operatorname{limsup.\ (This\ makes\ the\ theorem\ statement} \) easy to remember!)\n\n- The third inequality follows by Fatou on \( g - {f}_{n} \... | Yes |
Corollary 37.4.6 (Dominated convergence theorem)\n\nLet \( {f}_{1},{f}_{2},\ldots : \Omega \rightarrow \mathbb{R} \) be a sequence of measurable functions such that \( f = \) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n} \) exists. Assume that \( g : \Omega \rightarrow \mathbb{R} \) is an absolutely integra... | Proof. If \( f\left( \omega \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) \), then \( f\left( \omega \right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{f}_{n}\left( \omega \right) \) . So a... | Yes |
Suppose one wishes to compute\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{\int }_{\left( 0,1\right) }\frac{n\sin \left( {{n}^{-1}x}\right) }{\sqrt{x}}}\right) {dx} \] | then one starts by observing that the inner term is bounded by the absolutely integrable function \( {x}^{-1/2} \) . Therefore it equals\n\n\[ {\int }_{\left( 0,1\right) }\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( \frac{n\sin \left( {{n}^{-1}x}\right) }{\sqrt{x}}\right) {dx} = {\int }_{\left( 0,1\right) }\fr... | Yes |
Example 38.1.3 (Examples of locally compact abelian groups) | - Any finite group \( Z \) with the discrete topology is LCA.\n- The circle group \( \mathbb{T} \) is LCA and also in fact compact.\n- The real numbers \( \mathbb{R} \) are an example of an LCA group which is not compact. | Yes |
Proposition 38.2.5 ( \( G \) compact \( \Leftrightarrow \widehat{G} \) discrete)\n\nLet \( G \) be an LCA group. Then \( G \) is compact if and only if \( \widehat{G} \) is discrete. | Proof. Problem 38B. | No |
Theorem 39.1.4 (Linearity of expectation)\n\nIf \( X \) and \( Y \) are random variables on \( \Omega \) then\n\n\[ \mathbb{E}\left\lbrack {X + Y}\right\rbrack = \mathbb{E}\left\lbrack X\right\rbrack + \mathbb{E}\left\lbrack Y\right\rbrack \] | Proof. \( \mathbb{E}\left\lbrack {X + Y}\right\rbrack = {\int }_{\Omega }X\left( \omega \right) + Y\left( \omega \right) {d\mu } = {\int }_{\Omega }X\left( \omega \right) {d\mu } + {\int }_{\Omega }Y\left( \omega \right) {d\mu } = \mathbb{E}\left\lbrack X\right\rbrack + \mathbb{E}\left\lbrack Y\right\rbrack \) . | Yes |
Example 41.1.1 (Being even greedier than 18th century France)\n\nIn the game above, we start by betting $ \$ 1 $ .\n\n- If we win, we leave having made $ \$ 1 $ .\n\n- If we lose, we then bet $ \$ {10} $ instead, and\n\n- If we win, then we leave having made $ \$ {10} - \$ 2 = \$ 8 $, and\n\n- If we lose then we bet $ ... | This chapter will show that under sane conditions such as \ | No |
Proposition 41.3.2 (Conditional expectation definition)\n\nLet \( X : \Omega \rightarrow \mathbb{R} \) be an absolutely integrable random variable (meaning \( \mathbb{E}\left\lbrack \left| X\right| \right\rbrack < \infty \) ) over a probability space \( \Omega \), and let \( \mathcal{F} \) be a sub- \( \sigma \) -algeb... | Proof. Omitted, but relevant buzzword used is \ | No |
Proposition 41.5.2 (Stopped supermartingales are still supermartingales)\n\nLet \( {X}_{0},{X}_{1},\ldots \) be a supermartingale. Then the sequence\n\n\[ \n{X}_{\tau \land 0},{X}_{\tau \land 1},\ldots \n\]\n\nis itself a supermartingale. | Proof. We have almost everywhere the inequalities\n\n\[ \n\mathbb{E}\left( {{X}_{\tau \land n} \mid {\mathcal{F}}_{n - 1}}\right) = \mathbb{E}\left( {{X}_{n - 1} + {\mathbf{1}}_{\tau \left( \omega \right) = n - 1}\left( {{X}_{n} - {X}_{n - 1}}\right) \mid {\mathcal{F}}_{n - 1}}\right)\n\]\n\n\[ \n= \mathbb{E}\left( {{X... | Yes |
Example 42.1.3 (Total derivative of \( f\left( {x, y}\right) = {x}^{2} + {y}^{2} \) ) | Let \( V = {\mathbb{R}}^{2} \) with standard basis \( {\mathbf{e}}_{1},{\mathbf{e}}_{2} \) and let \( W = \mathbb{R} \), and let \( f\left( {x{\mathbf{e}}_{1} + y{\mathbf{e}}_{2}}\right) = {x}^{2} + {y}^{2} \) . Let \( p = a{\mathbf{e}}_{1} + b{\mathbf{e}}_{2} \) . Then, we claim that\n\n\[ \n{\left( Df\right) }_{p} : ... | Yes |
Theorem 42.2.1 (Projection principle)\n\nLet \( U \) be an open subset of the vector space \( V \) . Let \( W \) be an \( n \) -dimensional real vector space with basis \( {w}_{1},\ldots ,{w}_{n} \) . Then there is a bijection between continuous functions \( f : U \rightarrow W \) and \( n \) -tuples of continuous \( {... | Proof. Obvious. | No |
Example 42.3.5 (Partial derivatives of \( f\left( {x, y}\right) = {x}^{2} + {y}^{2} \) ) | Let \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) by \( \left( {x, y}\right) \mapsto {x}^{2} + {y}^{2} \) . Then in our new language,\n\n\[ \n{Df} : \left( {x, y}\right) \mapsto {2x} \cdot {\mathbf{e}}_{1}^{ \vee } + {2y} \cdot {\mathbf{e}}_{2}^{ \vee }.\n\]\n\nThus the partials are\n\n\[ \n\frac{\partial f}{\parti... | Yes |
Theorem 42.3.6 (Continuous partials implies differentiable)\n\nLet \( U \subseteq V \) be open and pick any basis \( {e}_{1},\ldots ,{e}_{n} \). Let \( f : U \rightarrow \mathbb{R} \) and suppose that \( \frac{\partial f}{\partial {e}_{i}} \) is defined for each \( i \) and moreover is continuous. Then \( f \) is diffe... | Proof. Not going to write out the details, but...given \( v = {t}_{1}{e}_{1} + \cdots + {t}_{n}{e}_{n} \), the idea is to just walk from \( p \) to \( p + {t}_{1}{e}_{1}, p + {t}_{1}{e}_{1} + {t}_{2}{e}_{2},\ldots \), up to \( p + {t}_{1}{e}_{1} + {t}_{2}{e}_{2} + \cdots + {t}_{n}{e}_{n} = p + v \) , picking up the par... | No |
Example 42.3.8 (Actually computing a total derivative)\n\nLet \( f\left( {x, y}\right) = x\sin y + {x}^{2}{y}^{4} \) . Then | \n\[ \frac{\partial f}{\partial x}\left( {x, y}\right) = \sin y + {y}^{4} \cdot {2x} \]\n\n\[ \frac{\partial f}{\partial y}\left( {x, y}\right) = x\cos y + {x}^{2} \cdot 4{y}^{3}. \]\n\nSo Theorem 42.3.6 applies, and \( {Df} = \frac{\partial f}{\partial x}{\mathbf{e}}_{1}^{ \vee } + \frac{\partial f}{\partial y}{\mathb... | No |
Set \( V = {\mathbb{R}}^{3} \) . Suppose that at some point \( p \), the 2 -form \( \alpha \) returns\n\n\[ \n{\alpha }_{p} = 2{\mathbf{e}}_{1}^{ \vee } \land {\mathbf{e}}_{2}^{ \vee } + {\mathbf{e}}_{1}^{ \vee } \land {\mathbf{e}}_{3}^{ \vee }\n\]\n\nLet \( {v}_{1} = 3{\mathbf{e}}_{1} + {\mathbf{e}}_{2} + 4{\mathbf{e}... | \[ \n{\alpha }_{p}\left( {{v}_{1},{v}_{2}}\right) = 2\det \left\lbrack \begin{array}{ll} 3 & 8 \\ 1 & 9 \end{array}\right\rbrack + \det \left\lbrack \begin{array}{ll} 3 & 8 \\ 4 & 5 \end{array}\right\rbrack = {21}.\n\] | Yes |
Let \( V = {\mathbb{R}}^{3} \) with standard basis \( {\mathbf{e}}_{1},{\mathbf{e}}_{2},{\mathbf{e}}_{3} \) . Let \( f\left( {x, y, z}\right) = {x}^{4} + {y}^{3} + {2xz} \) . Then we compute\n\n\[ \n{df} = {Df} = \left( {4{x}^{3} + {2z}}\right) {dx} + 3{y}^{2}{dy} + {2xdz}.\n\] | Next, we can evaluate \( d\left( {df}\right) \) as prescribed: it is\n\n\[ \n{d}^{2}f = \left( {{12}{x}^{2}{dx} + {2dz}}\right) \land {dx} + \left( {6ydy}\right) \land {dy} + 2\left( {{dx} \land {dz}}\right)\n\]\n\n\[ \n= {12}{x}^{2}\left( {{dx} \land {dx}}\right) + 2\left( {{dz} \land {dx}}\right) + {6y}\left( {{dy} \... | Yes |
Let \( V = {\mathbb{R}}^{2} \) with basis \( {\mathbf{e}}_{1} \) and \( {\mathbf{e}}_{2} \), and suppose \( \phi : V \rightarrow {V}^{\prime } \) is given by sending\n\n\[ \phi \left( {a{\mathbf{e}}_{1} + b{\mathbf{e}}_{2}}\right) = \left( {{a}^{2} + {b}^{2}}\right) {w}_{1} + \log \left( {{a}^{2} + 1}\right) {w}_{2} + ... | \[ {\left( {\phi }^{ * }\alpha \right) }_{p} = f\left( {\phi \left( p\right) }\right) \cdot \left( {{2a}{\mathbf{e}}_{1}^{ \vee } + {2b}{\mathbf{e}}_{2}^{ \vee }}\right) \land \left( {3{b}^{2}{\mathbf{e}}_{2}^{ \vee }}\right) = f\left( {\phi \left( p\right) }\right) \cdot {6a}{b}^{2} \cdot {\mathbf{e}}_{1}^{ \vee } \la... | Yes |
Example 44.3.3 (Area of a circle)\n\nConsider \( V = {\mathbb{R}}^{2} \) and let \( c : \left( {r,\theta }\right) \mapsto \left( {r\cos \theta }\right) {\mathbf{e}}_{1} + \left( {r\sin \theta }\right) {\mathbf{e}}_{2} \) on \( \left\lbrack {0, R}\right\rbrack \times \left\lbrack {0,{2\pi }}\right\rbrack \) as before. T... | \[{c}^{ * }\alpha = \left( {\cos {\theta dr} - r\sin {\theta d\theta }}\right) \land \left( {\sin {\theta dr} + r\cos {\theta d\theta }}\right)\]\n\n\[= r\left( {{\cos }^{2}\theta + {\sin }^{2}\theta }\right) \left( {{dr} \land {d\theta }}\right)\]\n\n\[= {rdr} \land {d\theta }\]\n\nThus,\n\n\[{\int }_{c}\alpha = {\int... | Yes |
Theorem 44.3.5 (Changing variables doesn't affect integrals)\n\nLet \( c \) be a \( k \) -cell, \( \alpha \) a \( k \) -form, and \( \phi \) a reparametrization. Then\n\n\[ \n{\int }_{c \circ \phi }\alpha = \left\{ \begin{array}{ll} {\int }_{c}\alpha & \phi \text{ preserves orientation } \\ - {\int }_{c}\alpha & \phi \... | Proof. Use naturality of the pullback to reduce it to the corresponding theorem in normal calculus. | No |
Example 45.1.5 (An atlas on \( {S}^{1} \) ) | Here is a picture of an atlas for \( {S}^{1} \), with two open sets.\n\n | No |
Theorem 45.3.1 (Regular value theorem)\n\nLet \( V \) be an \( n \) -dimensional real normed vector space, let \( U \subseteq V \) be open and let \( {f}_{1},\ldots ,{f}_{m} : U \rightarrow \mathbb{R} \) be smooth functions. Let \( M \) be the set of points \( p \in U \) such that \( {f}_{1}\left( p\right) = \cdots = {... | For a proof, see [sj05, Theorem 6.3]. | Yes |
Example 45.3.3 (The circle \( {x}^{2} + {y}^{2} - c = 0 \) ) | Let \( f\left( {x, y}\right) = {x}^{2} + {y}^{2} - c, f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \), where \( c \) is a positive real number. Note that\n\n\[ \n{Df} = {2x} \cdot {dx} + {2y} \cdot {dy} \n\]\n\nwhich in particular is nonzero as long as \( \left( {x, y}\right) \neq \left( {0,0}\right) \), i.e. as long as... | Yes |
Theorem 45.6.3 (Stokes' theorem for manifolds)\n\nLet \( M \) be a smooth oriented \( n \) -manifold with boundary and let \( \alpha \) be a compactly supported \( n - 1 \) -form. Then\n\n\[ \n{\int }_{M}{d\alpha } = {\int }_{\partial M}\alpha \n\] | All the omitted details are developed in full in [Sj05]. | No |
Example 46.1.1 (Examples of minimal polynomials)\n\n(a) \( \sqrt{2} \) has minimal polynomial \( {x}^{2} - 2 \) .\n\n(b) The imaginary unit \( i = \sqrt{-1} \) has minimal polynomial \( {x}^{2} + 1 \) .\n\n(c) A primitive \( p \) th root of unity, \( {\zeta }_{p} = {e}^{\frac{2\pi i}{p}} \), has minimal polynomial \( {... | Note that \( {100}{x}^{2} - {200} \) is also a polynomial of the same degree which has \( \sqrt{2} \) as a root; that's why we want to require the polynomial to be monic. That's also why we choose to work in the rational numbers; that way, we can divide by leading coefficients without worrying if we get non-integers.\n... | Yes |
Example 46.2.5 (Examples of algebraic integers)\n\nThe numbers\n\n\[ \n4, i = \sqrt{-1},\sqrt[3]{2},\sqrt{2} + \sqrt{3} \]\n\nare all algebraic integers, since they are the roots of the monic polynomials \( x - 4 \) , \( {x}^{2} + 1,{x}^{3} - 2 \) and \( {\left( {x}^{2} - 5\right) }^{2} - {24} \) . | The number \( \frac{1}{2} \) has minimal polynomial \( x - \frac{1}{2} \), so it’s an algebraic number but not an algebraic integer. (In fact, the rational root theorem also directly implies that any monic integer polynomial does not have \( \frac{1}{2} \) as a root!) | No |
Proposition 46.2.6 (Rational algebraic integers are rational integers)\n\nAn algebraic integer is rational if and only if it is a rational integer. In symbols,\n\n\[ \overline{\mathbb{Z}} \cap \mathbb{Q} = \mathbb{Z} \] | Proof. Let \( \alpha \) be a rational number. If \( \alpha \) is an integer, it is the root of \( x - \alpha \), hence an algebraic integer too.\n\nConversely, if \( P \) is a monic polynomial with integer coefficients such that \( P\left( \alpha \right) = 0 \) then (by the rational root theorem, say) it follows \( \al... | Yes |
Proposition 46.2.7 ( \( \overline{\mathbb{Z}} \) is a ring and \( \overline{\mathbb{Q}} \) is a field) | We could prove this now if we wanted to, but the results in the next chapter will more or less do it for us, and so we take this on faith temporarily. | No |
Consider the field\n\n\[ K = \mathbb{Q}\left( \sqrt{2}\right) = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q}\} . \]\n\nThis is a field extension of \( \mathbb{Q} \), and has degree 2 (the basis being 1 and \( \sqrt{2} \) ). | You might be confused that I wrote \( \mathbb{Q}\left( \sqrt{2}\right) \) (which should permit denominators) instead of \( \mathbb{Q}\left\lbrack \sqrt{2}\right\rbrack \), say. But if you read through Example 5.5.4, you should see that the denominators don’t really matter: \( \frac{1}{3 - \sqrt{2}} = \frac{1}{7}\left( ... | No |
Theorem 46.4.1 (Artin's primitive element theorem)\n\nEvery number field \( K \) is isomorphic to \( \mathbb{Q}\left( \alpha \right) \) for some algebraic number \( \alpha \) . | The proof is left as Problem 52E, since to prove it I need to talk about field extensions first. | No |
An irreducible polynomial in \( \mathbb{Q}\left\lbrack x\right\rbrack \) cannot have a complex double root. | Proof. Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be the irreducible polynomial and assume it has a double root \( \alpha \) . Take the derivative \( {f}^{\prime }\left( x\right) \) . This derivative has three interesting properties.\n\n- The degree of \( {f}^{\prime } \) is one less than the ... | No |
Example 47.1.5 (Norm of \( a + b\sqrt{2} \) ) | Let \( \alpha = a + b\sqrt{2} \in \mathbb{Q}\left( \sqrt{2}\right) = K \) . If \( b \neq 0 \), then \( \alpha \) and \( K \) have the degree 2 . Thus the only conjugates of \( \alpha \) are \( a \pm b\sqrt{2} \), which gives the norm\n\n\[ \left( {a + b\sqrt{2}}\right) \left( {a - b\sqrt{2}}\right) = {a}^{2} - 2{b}^{2}... | Yes |
Proposition 47.1.6 (Norms and traces are rational integers)\n\nIf \( \alpha \) is an algebraic integer, its norm and trace are rational integers. | Question 47.1.7. Prove it. (Vieta formula.) | No |
Theorem 47.1.8 (Morally correct definition of norm and trace)\n\nLet \( K \) be a number field of degree \( n \), and let \( \alpha \in K \) . Let \( {\mu }_{\alpha } : K \rightarrow K \) denote the map\n\n\[ x \mapsto {\alpha x} \]\n\nviewed as a linear map of \( \mathbb{Q} \) -vector spaces. Then,\n\n- the norm of \(... | Since the trace and determinant don't depend on the choice of basis, you can pick whatever basis you want and use whatever definition you got in high school. Fantastic, right? | No |
Example 47.1.9 (Explicit computation of matrices for \( a + b\sqrt{2} \) ) Let \( K = \mathbb{Q}\left( \sqrt{2}\right) \), and let \( 1,\sqrt{2} \) be the basis of \( K \) . Let \[ \alpha = a + b\sqrt{2} \] (possibly even \( b = 0 \) ), and notice that \[ \left( {a + b\sqrt{2}}\right) \left( {x + y\sqrt{2}}\right) = \l... | Proof. I'll prove the result for just the norm; the trace falls out similarly. Set \[ n = \deg \alpha ,\;{kn} = \deg K. \] The proof is split into two parts, depending on whether or not \( k = 1 \) . Proof if \( k = 1 \) . Set \( n = \deg \alpha = \deg K \) . Thus the norm actually is the product of the Galois conjugat... | Yes |
Example 47.2.3 (Ring of integers of \( K = \mathbb{Q}\left( \sqrt{3}\right) \) )\n\nLet \( K \) be as above. We claim that\n\n\[ \n{\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \sqrt{3}\right\rbrack = \{ m + n\sqrt{3} \mid m, n \in \mathbb{Z}\} .\n\] | We set \( \alpha = a + b\sqrt{3} \) . Then \( \alpha \in {\mathcal{O}}_{K} \) when the minimal polynomial has integer coefficients.\n\nIf \( b = 0 \), then the minimal polynomial is \( x - \alpha = x - a \), and thus \( \alpha \) works if and only if it’s an integer. If \( b \neq 0 \), then the minimal polynomial is\n\... | No |
We claim that in this case\n\n\[ \n{\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \frac{1 + \sqrt{5}}{2}\right\rbrack = \left\{ {m + n \cdot \frac{1 + \sqrt{5}}{2} \mid m, n \in \mathbb{Z}}\right\} .\n\] | The proof is exactly the same, except the exercise tells us instead that for \( b \neq 0 \), we have both the possibility that \( a, b \in \mathbb{Z} \) or that \( a, b \in \mathbb{Z} - \frac{1}{2} \) . This reflects the fact that \( \frac{1 + \sqrt{5}}{2} \) is the root of \( {x}^{2} - x - 1 = 0 \) ; no such thing is ... | No |
Theorem 47.2.6 \( \left( {K = \mathbb{Q} \cdot {\mathcal{O}}_{K}}\right) \)\n\nLet \( K \) be a number field, and let \( x \in K \) be any element. Then there exists an integer \( n \) such that \( {nx} \in {\mathcal{O}}_{K} \) ; in other words,\n\n\[ x = \frac{1}{n}\alpha \]\n\nfor some \( \alpha \in {\mathcal{O}}_{K}... | Exercise 47.2.7. Prove this yourself. (Start by using the fact that \( x \) has a minimal polynomial with rational coefficients. Alternatively, take the norm.) | No |
Lemma 47.2.8 \( \left( {\alpha \in \overline{\mathbb{Z}} \Leftrightarrow \mathbb{Z}\left\lbrack \alpha \right\rbrack \text{finitely generated}}\right) \) | Proof. Note that \( \alpha \) is an algebraic integer if and only if it’s the root of some nonzero, monic polynomial with integer coefficients. Suppose first that\n\n\[{\alpha }^{N} = {c}_{N - 1}{\alpha }^{N - 1} + {c}_{N - 2}{\alpha }^{N - 2} + \cdots + {c}_{0}.\n\]\n\nThen the set \( 1,\alpha ,\ldots ,{\alpha }^{N - ... | Yes |
Example 47.2.9 ( \( \frac{1}{2} \) isn’t an algebraic integer) | We already know \( \frac{1}{2} \) isn’t an algebraic integer. So we expect\n\n\[ \mathbb{Z}\left\lbrack \frac{1}{2}\right\rbrack = \left\{ {\left. {\frac{a}{{2}^{m}}\;}\right| \;a, m \in \mathbb{Z}\text{ and }m \geq 0}\right\} \]\n\nto not be finitely generated, and this is the case. | No |
Theorem 47.2.11 (Algebraic integers are closed under + and \( \times \) ) The set \( \overline{\mathbb{Z}} \) is closed under addition and multiplication; i.e. it is a ring. In particular, \( {\mathcal{O}}_{K} \) is also a ring for any number field \( K \) . | Proof. Let \( \alpha ,\beta \in \overline{\mathbb{Z}} \) . Then \( \mathbb{Z}\left\lbrack \alpha \right\rbrack \) and \( \mathbb{Z}\left\lbrack \beta \right\rbrack \) are finitely generated. Hence so is \( \mathbb{Z}\left\lbrack {\alpha ,\beta }\right\rbrack \) . (Details: if \( \mathbb{Z}\left\lbrack \alpha \right\rbr... | Yes |
Proposition 48.2.2 (Ideal arithmetic via generators)\n\nSuppose \( \mathfrak{a} = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) and \( \mathfrak{b} = \left( {{b}_{1},\ldots ,{b}_{m}}\right) \) are ideals in a ring \( R \) . Then\n\n(a) \( \mathfrak{a} + \mathfrak{b} \) is the ideal generated by \( {a}_{1},\ldots ,... | Proof. Pretty straightforward; just convince yourself that this result is correct.\n\nIn other words, for sums you append the two sets of generators together, and for products you take products of the generators. Note that for principal ideals, this coincides with \ | No |
Proposition 48.3.4 ( \( {\mathcal{O}}_{K} \) integrally closed and Noetherian)\n\nFor any number field \( K \), the ring \( {\mathcal{O}}_{K} \) is integrally closed and Noetherian. | Proof. Boring, but here it is anyways for completeness.\n\nSince \( {\mathcal{O}}_{K} \cong {\mathbb{Z}}^{\oplus n} \), we get that it’s Noetherian.\n\nNow we show that \( {\mathcal{O}}_{K} \) is integrally closed. Suppose that \( \eta \in K \) is the root of some polynomial with coefficients in \( {\mathcal{O}}_{K} \)... | Yes |
Theorem 48.3.5 (Important: prime ideals divide rational primes)\n\nLet \( {\mathcal{O}}_{K} \) be a ring of integers and \( \mathfrak{p} \) a nonzero prime ideal inside it. Then \( \mathfrak{p} \) contains a rational prime \( p \) . Moreover, \( \mathfrak{p} \) is maximal. | Proof. Take any \( \alpha \neq 0 \) in \( \mathfrak{p} \) . Its Galois conjugates are algebraic integers so their product \( \mathrm{N}\left( \alpha \right) /\alpha \) is in \( {\mathcal{O}}_{K} \) (even though each individual conjugate need not be in \( K \) ). Consequently, \( \mathrm{N}\left( \alpha \right) \in \mat... | Yes |
Example 48.5.1 (Factoring 6 in the integers of \( \mathbb{Q}\left( \sqrt{-5}\right) \) ) | Let \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \sqrt{-5}\right\rbrack \) arise from \( K = \mathbb{Q}\left( \sqrt{-5}\right) \) . We’ve already seen that\n\n\[ \left( 6\right) = \left( 2\right) \cdot \left( 3\right) = \left( {1 + \sqrt{-5}}\right) \left( {1 - \sqrt{-5}}\right) \]\n\nand you can't get any further wit... | Yes |
Example 48.5.3 (Factoring 3 in the integers of \( \mathbb{Q}\left( \sqrt{-{17}}\right) \) ) | Let \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \sqrt{-{17}}\right\rbrack \) arise from \( K = \mathbb{Q}\left( \sqrt{-{17}}\right) \) . We know \( {\mathcal{O}}_{K} \cong \mathbb{Z}\left\lbrack x\right\rbrack /\left( {{x}^{2} + {17}}\right) \) . Now\n\n\[ \n{\mathcal{O}}_{K}/3{\mathcal{O}}_{K} \cong \mathbb{Z}\left\... | Yes |
Theorem 48.5.4 (Factoring algorithm / Dedekind-Kummer theorem)\n\nLet \( K \) be a number field. Let \( \theta \in {\mathcal{O}}_{K} \) with \( \left\lbrack {{\mathcal{O}}_{K} : \mathbb{Z}\left\lbrack \theta \right\rbrack }\right\rbrack = j < \infty \), and let \( p \) be a prime not dividing \( j \) . Then \( \left( p... | In particular, if \( K \) is monogenic with \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \theta \right\rbrack \) then \( j = 1 \) and the theorem applies for all primes \( p \) .\n\nIn almost all our applications in this book, \( K \) will be monogenic; i.e. \( j = 1 \) . Here \( \bar{\psi } \) denotes the image in \(... | No |
Example 48.6.2 \( \left( {\frac{5}{2}\mathbb{Z}\text{is a fractional ideal}}\right) \) | The set\n\[ \frac{5}{2}\mathbb{Z} = \left\{ {\frac{5}{2}n \mid n \in \mathbb{Z}}\right\} = \frac{1}{2}\left( 5\right) \]\nis a fractional ideal of \( \mathbb{Z} \) . | Yes |
Example 48.7.2 (Ideal norm of (5) in the Gaussian integers)\n\nLet \( K = \mathbb{Q}\left( i\right) ,{\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack i\right\rbrack \) . Consider the ideal (5) in \( {\mathcal{O}}_{K} \) . We have that | \[ {\mathcal{O}}_{K}/\left( 5\right) \cong \{ a + {bi} \mid a, b \in \mathbb{Z}/5\mathbb{Z}\} \]\n\nso (5) has ideal norm 25, corresponding to the fact that \( {\mathcal{O}}_{K}/\left( 5\right) \) has \( {5}^{2} = {25} \) elements. | Yes |
Example 48.7.3 (Ideal norm of \( \left( {2 + i}\right) \) in the Gaussian integers) | You'll notice that\n\n\[ \n{\mathcal{O}}_{K}/\left( {2 + i}\right) \cong {\mathbb{F}}_{5} \n\] \n\nsince \( {\;\operatorname{mod}\;2} + i \) we have both \( 5 \equiv 0 \) and \( i \equiv - 2 \) . (Indeed, since \( \left( {2 + i}\right) \) is prime we had better get a field!) Thus \( \mathrm{N}\left( \left( {2 + i}\righ... | No |
Theorem 49.2.1 (The \( n \) embeddings of a number field)\n\nLet \( K \) be a number field of degree \( n \) . Then there are exactly \( n \) field homomorphisms \( K \hookrightarrow \mathbb{C} \), say \( {\sigma }_{1},\ldots ,{\sigma }_{n} \), which fix \( \mathbb{Q} \) . | Proof. Deferred to Theorem 52.3.1, once we have the tools of Galois theory. | No |
Example 49.2.3 (Discriminant of \( K = \mathbb{Q}\left( \sqrt{2}\right) \) ) | We have \( {\mathcal{O}}_{K} = \mathbb{Z}\left\lbrack \sqrt{2}\right\rbrack \) and as discussed above the discriminant is\n\n\[ \n{\Delta }_{K} = {\left( -2\sqrt{2}\right) }^{2} = 8.\n\] | Yes |
Example 49.2.5 (Discriminant of \( \mathbb{Q}\left( \sqrt{5}\right) \) ) | Let \( K = \mathbb{Q}\left( \sqrt{5}\right) \) . This time, \( {\mathcal{O}}_{K} = \mathbb{Z} \oplus \frac{1 + \sqrt{5}}{2}\mathbb{Z} \), and so the discriminant is going to\n\nlook a little bit different. The embeddings are still \( a + b\sqrt{5} \mapsto a + b\sqrt{5}, a - b\sqrt{5} \) .\n\nApplying this to the \( \ma... | Yes |
Consider the composition of the embeddings \( K \hookrightarrow {\mathbb{R}}^{{r}_{1}} \times {\mathbb{C}}^{{r}_{2}}\overset{ \sim }{ \rightarrow }{\mathbb{R}}^{n} \) . Then as before, \( {\mathcal{O}}_{K} \) becomes a lattice \( L \) in \( {\mathbb{R}}^{n} \), with mesh equal to\n\n\[ \frac{1}{{2}^{{r}_{2}}}\sqrt{\lef... | Proof. Fun linear algebra problem (you just need to manipulate determinants). Left as Problem 49D. | No |
Theorem 49.6.1 (Minkowski bound)\n\nLet \( \mathfrak{a} \subseteq {\mathcal{O}}_{K} \) be any nonzero ideal. Then there exists \( 0 \neq \alpha \in \mathfrak{a} \) such that\n\n\[{\mathrm{N}}_{K/\mathbb{Q}}\left( \alpha \right) \leq {\left( \frac{4}{\pi }\right) }^{{r}_{2}}\frac{n!}{{n}^{n}}\sqrt{\left| {\Delta }_{K}\r... | Proof. This is a matter of putting all our ingredients together. Let's see what things we've defined already: \n\nPick a value of \( M \) such that the mesh of \( {L}_{\mathfrak{a}} \) equals \( {2}^{-n} \) of the vo... | Yes |
Let \( K \) be a number field, and pick a fractional ideal \( J \) . Then we can find \( \alpha \) such that \( \mathfrak{b} = \left( \alpha \right) \cdot J \) is integral and \( \mathrm{N}\left( \mathfrak{b}\right) \leq {M}_{K} \) . | Proof. For fractional ideals \( I \) and \( J \) write \( I \sim J \) to mean that \( I = \left( \alpha \right) J \) for some \( \alpha \) ; then \( {\mathrm{{Cl}}}_{K} \) is just modding out by \( \sim \) . Let \( J \) be a fractional ideal. Then \( {J}^{-1} \) is some other fractional ideal. By definition, for some \... | Yes |
Theorem 49.8.4 (Gaussian integers \( \mathbb{Z}\left\lbrack i\right\rbrack \) form a UFD) | The field \( \mathbb{Q}\left( i\right) \) has class number 1 . Proof. This is \( {\mathcal{O}}_{K} \) where \( K = \mathbb{Q}\left( i\right) \), so we just want \( {\mathrm{{Cl}}}_{K} \) to be trivial. We have \( {M}_{K} = \frac{2}{\pi }\sqrt{4} < 2 \) . So every class has an integral ideal of norm \( \mathfrak{b} \) s... | Yes |
Proposition 49.8.5 \( \left( {\mathbb{Z}\left\lbrack \sqrt{7}\right\rbrack \text{is a UFD}}\right) \) | The field \( \mathbb{Q}\left( \sqrt{7}\right) \) has class number 1 . Proof. First we compute the Minkowski bound. | No |
Lemma 49.8.8 (Ideals divide their norms)\n\nLet \( \mathfrak{b} \) be an integral ideal with \( \mathrm{N}\left( \mathfrak{b}\right) = n \) . Then \( \mathfrak{b} \) divides the ideal \( \left( n\right) \) . | Proof. By definition, \( n = \left| {{\mathcal{O}}_{K}/\mathfrak{b}}\right| \) . Treating \( {\mathcal{O}}_{K}/\mathfrak{b} \) as an (additive) abelian group and using Lagrange's theorem, we find\n\n\[ 0 \equiv \underset{n\text{ times }}{\underbrace{\alpha + \cdots + \alpha }} = {n\alpha }\;\left( {\;\operatorname{mod}... | Yes |
Proposition 49.8.9 (Class group of \( \\mathbb{Q}\\left( \\sqrt{-{17}}\\right) \) ) | The number field \( K = \\mathbb{Q}\\left( \\sqrt{-{17}}\\right) \) has class group \( \\mathbb{Z}/4\\mathbb{Z} \). Proof. Since \( {\\Delta }_{K} = - {68} \), we compute the Minkowski bound \[ {M}_{K} = \\frac{4}{\\pi }\\sqrt{17} < 6 \] Now, it suffices to factor with (2),(3),(5). The minimal polynomial of \( \\sqrt{-... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.