From fd816fb8c10182c7a55152d3587385df1419f0ed Mon Sep 17 00:00:00 2001 From: semapheur Date: Thu, 9 May 2024 15:47:15 +0200 Subject: [PATCH] Deploy --- content/notes/math/linear_algebra.mdx | 164 +++++++++++++------------- 1 file changed, 82 insertions(+), 82 deletions(-) diff --git a/content/notes/math/linear_algebra.mdx b/content/notes/math/linear_algebra.mdx index c1aa407..20c8b66 100644 --- a/content/notes/math/linear_algebra.mdx +++ b/content/notes/math/linear_algebra.mdx @@ -2880,17 +2880,17 @@ $$ Let $R$ be a commutative ring with identity $1$, whose elements are called scalars. A *module* is a set $M$ over $R$ equipped with the two closed operations -- $+: M\times M \ni (\mathbf{u},\mathbf{v}) \mapsto \mathbf{u} + \mathbf{v} \in M$ **(addition)** -- $\cdot: R\times M \ni (\rho, \mathbf{u}) \mapsto \rho\mathbf{u} \in M$ **(scalar multiplication)** +- $+: M\times M \ni (\boldsymbol{u},\boldsymbol{v}) \mapsto \boldsymbol{u} + \boldsymbol{v} \in M$ **(addition)** +- $\cdot: R\times M \ni (\rho, \boldsymbol{u}) \mapsto \rho\boldsymbol{u} \in M$ **(scalar multiplication)** and has the following properties - $M$ is an abelian group under addition -- Scalar multiplication is compatible, satisfying for all $\alpha,\beta\in R$ and $\mathbf{x}\in M$ - - $(\alpha\beta)\mathbf{x} = \alpha(\beta\mathbf{x})$ - - $1\mathbf{x} = \mathbf{x}$ +- Scalar multiplication is compatible, satisfying for all $\alpha,\beta\in R$ and $\boldsymbol{x}\in M$ + - $(\alpha\beta)\boldsymbol{x} = \alpha(\beta\boldsymbol{x})$ + - $1\boldsymbol{x} = \boldsymbol{x}$ - Addition and scalar multiplication are related by distributivity for all $r,s\in R$ and $x, y\in M$ - - $\alpha(\mathbf{x} + \mathbf{y}) = \alpha\mathbf{x} + \alpha\mathbf{y}$ - - $(\alpha + \beta)\mathbf{x} = \alpha\mathbf{x} + \beta\mathbf{x}$ + - $\alpha(\boldsymbol{x} + \boldsymbol{y}) = \alpha\boldsymbol{x} + \alpha\boldsymbol{y}$ + - $(\alpha + \beta)\boldsymbol{x} = \alpha\boldsymbol{x} + \beta\boldsymbol{x}$ The ring $R$ is called the *base ring* of $M$. @@ -2903,7 +2903,7 @@ A submodule of an $R$-module $M$ is a nonempty subset $N\subseteq M$ that is an Let $M$ be an $R$-module. A nonempty subset $N\subseteq M$ is a submodule if and only if it is closed under the taking of linear combinations, i.e. $$ - \alpha,\beta \in R, \mathbf{u},\mathbf{v}\in N \implies \alpha\mathbf{u} + \beta\mathbf{v} \in N + \alpha,\beta \in R, \boldsymbol{u},\boldsymbol{v}\in N \implies \alpha\boldsymbol{u} + \beta\boldsymbol{v} \in N $$ @@ -2919,22 +2919,22 @@ When we consider $R$ as an $R$-module rather than as a ring, multiplication is t The submodule spanned or generated by a subset $N$ of an $R$-module $M$ is the set of all linear combinations of elements of $N$ $$ - \langle N \rangle = \Set{\sum_{i=1}^n \alpha_i \mathbf{v}_i | \alpha_i \in R, \mathbf{v}_i \in N, n\geq 1} + \langle N \rangle = \Set{\sum_{i=1}^n \alpha_i \boldsymbol{v}_i | \alpha_i \in R, \boldsymbol{v}_i \in N, n\geq 1} $$ A subset $N\subseteq M$ spans or generates $M$ if $M = \langle N \rangle$. -Note that if a nontrivial linear combination of the elements $\mathbf{v}_1,\dots,\mathbf{v}_n$ in an $R$-module $M$ is $0$, i.e. $\sum_{i=1}^n = \rho_i \mathbf{v}_i = 0$, where not all of the coefficients are $0$, then we cannot conclude that one of the elements $\mathbf{v}_i$ is a linear combination of the others. After all, this involves dividing by one of the coefficients, which may not be possible in a ring. +Note that if a nontrivial linear combination of the elements $\boldsymbol{v}_1,\dots,\boldsymbol{v}_n$ in an $R$-module $M$ is $0$, i.e. $\sum_{i=1}^n = \rho_i \boldsymbol{v}_i = 0$, where not all of the coefficients are $0$, then we cannot conclude that one of the elements $\boldsymbol{v}_i$ is a linear combination of the others. After all, this involves dividing by one of the coefficients, which may not be possible in a ring. Let $M$ be an $R$-submodule. A submodule of the form $$ - \langle \mathbf{v} \rangle = R\mathbf{v} = \Set{\rho\mathbf{v} | \rho\in R} + \langle \boldsymbol{v} \rangle = R\boldsymbol{v} = \Set{\rho\boldsymbol{v} | \rho\in R} $$ -for $\mathbf{v}\in M$ is the *cyclic submodule* generated by $\mathbf{v}$. +for $\boldsymbol{v}\in M$ is the *cyclic submodule* generated by $\boldsymbol{v}$. Any finite-dimensional vector space is the direct sum of cyclic submodules, i.e. one-dimensional subspaces. @@ -2963,10 +2963,10 @@ The last sum does not involve $x_k$ and so it must equal $0$. Hence, the first s ## Linear independence -A subset $N$ of an $R$-module $M$ is *linearly independent* if for any distinct $\mathbf{v}_1,\dots,\mathbf{v}_n \in N$ and $\rho_1,\dots,\rho_n\in R$, we have +A subset $N$ of an $R$-module $M$ is *linearly independent* if for any distinct $\boldsymbol{v}_1,\dots,\boldsymbol{v}_n \in N$ and $\rho_1,\dots,\rho_n\in R$, we have $$ - \sum_{i=1} \rho_i \mathbf{v}_i = 0 \implies \rho_i = 0,\; \forall i + \sum_{i=1} \rho_i \boldsymbol{v}_i = 0 \implies \rho_i = 0,\; \forall i $$ A set $N$ that is not linearly independent is *linearly dependent*. @@ -2979,16 +2979,16 @@ The abelian group $\Z_n = \Set{0,1,\dots,n-1}$ is a $\Z$-module, with scalar mul -Let $M$ be an $R$-module. A nonzero element $\mathbf{v}\in M$ for which $\rho\mathbf{v} = 0$ for some nonzero $\rho\in R$ is a *torsion element* of $M$. A module that has no nonzero torsion elements is *torsion free*. If all elements of $M$ are torsion elements, then $M$ is a *torsion module*. The set of all torsion elements of $M$, together with the zero element, is denoted $M_{\mathrm{tor}}$. +Let $M$ be an $R$-module. A nonzero element $\boldsymbol{v}\in M$ for which $\rho\boldsymbol{v} = 0$ for some nonzero $\rho\in R$ is a *torsion element* of $M$. A module that has no nonzero torsion elements is *torsion free*. If all elements of $M$ are torsion elements, then $M$ is a *torsion module*. The set of all torsion elements of $M$, together with the zero element, is denoted $M_{\mathrm{tor}}$. ## Annihilators -Let $M$ be an $R$-module. The *annihilator* of an element $\mathbf{v}\in M$ is +Let $M$ be an $R$-module. The *annihilator* of an element $\boldsymbol{v}\in M$ is $$ - \operatorname{ann}(\mathbf{v}) = \Set{\rho\in R | \rho\mathbf{v} = 0} + \operatorname{ann}(\boldsymbol{v}) = \Set{\rho\in R | \rho\boldsymbol{v} = 0} $$ and the annihilator of a submodule $N$ of $M$ is @@ -2997,24 +2997,24 @@ $$ \operatorname{ann}(N) = \Set{\rho\in R | \rho N = \Set{0}} $$ -where $\rho N = \Set{\rho\mathbf{v} | \mathbf{v}\in N}$. Annihilators are also called *order ideals*. +where $\rho N = \Set{\rho\boldsymbol{v} | \boldsymbol{v}\in N}$. Annihilators are also called *order ideals*. -It is easy to see that $\operatorname{ann}(\mathbf{v})$ and $\operatorname{ann}(N)$ are ideals of $R$. Clearly, $\mathbf{v}\in M$ is a torsion element if and only if $\operatorname{ann}(\mathbf{v}) \neq\Set{0}$. Also, if $A$ and $B$ are submodules of $M$, then +It is easy to see that $\operatorname{ann}(\boldsymbol{v})$ and $\operatorname{ann}(N)$ are ideals of $R$. Clearly, $\boldsymbol{v}\in M$ is a torsion element if and only if $\operatorname{ann}(\boldsymbol{v}) \neq\Set{0}$. Also, if $A$ and $B$ are submodules of $M$, then $$ A \leq B \implies\operatorname{ann}(B) \leq\operatorname{ann}(A) $$ -Let $M = \langle \mathbf{u}_1,\dots,\mathbf{u}_n \rangle$ be a finitely generated module over an integral domain $R$ and assume each of the generators $\mathbf{u}_i$ is torsion, i.e. there is a nonzero $\alpha_i \in\operatorname{ann}(\mathbf{u}_i)$ for each $i$. Then the nonzero product $\alpha = \prod_{i=1}^n \alpha_i$ annihilates each generator of $M$ and therefore every element of $M$, i.e. $\alpha\in\operatorname{ann}(M)$. This shows that $\operatorname{ann}(M) \neq\Set{0}$. On the other hand, this may fail if $R$ is not an integral domain. +Let $M = \langle \boldsymbol{u}_1,\dots,\boldsymbol{u}_n \rangle$ be a finitely generated module over an integral domain $R$ and assume each of the generators $\boldsymbol{u}_i$ is torsion, i.e. there is a nonzero $\alpha_i \in\operatorname{ann}(\boldsymbol{u}_i)$ for each $i$. Then the nonzero product $\alpha = \prod_{i=1}^n \alpha_i$ annihilates each generator of $M$ and therefore every element of $M$, i.e. $\alpha\in\operatorname{ann}(M)$. This shows that $\operatorname{ann}(M) \neq\Set{0}$. On the other hand, this may fail if $R$ is not an integral domain. ## Homomorphisms -Let $M$ and $N$ be $R$-modules. A function $f:M\to N$ is an $R$-homomorphism or $R$-map if it preserves the module operations, i.e. for all $\rho,\sigma\in R$ and $\mathbf{u},\mathbf{v}\in M$ +Let $M$ and $N$ be $R$-modules. A function $f:M\to N$ is an $R$-homomorphism or $R$-map if it preserves the module operations, i.e. for all $\rho,\sigma\in R$ and $\boldsymbol{u},\boldsymbol{v}\in M$ $$ - f(\rho\mathbf{u} + \sigma\mathbf{v}) = \rho f(\mathbf{u}) + \sigma f(\mathbf{v}) + f(\rho\boldsymbol{u} + \sigma\boldsymbol{v}) = \rho f(\boldsymbol{u}) + \sigma f(\boldsymbol{v}) $$ The set all $R$-homomorphisms from $M$ to $N$ is denoted $\hom_R (M, N)$. The following terms are used to classify $R$-maps @@ -3027,7 +3027,7 @@ The set all $R$-homomorphisms from $M$ to $N$ is denoted $\hom_R (M, N)$. The fo It is east to see that $\hom_R (M, N)$ is itself an $R$-module under addition of functions and scalar multiplication defined by $$ - (\rho f)(\mathbf{v}) = \rho(f(\mathbf{v})) = f(\rho\mathbf{v}) + (\rho f)(\boldsymbol{v}) = \rho(f(\boldsymbol{v})) = f(\rho\boldsymbol{v}) $$ @@ -3035,8 +3035,8 @@ Let $f:M\to N$ be an $R$-map. The *kernel* and *range* or *image* of $f$ are def $$ \begin{align*} - \ker(f) =& \Set{\mathbf{v} \in M | f(\mathbf{v}) = 0} \\ - \operatorname{ran}(f) =& \Set{f(\mathbf{v}) | \mathbf{v} \in M} + \ker(f) =& \Set{\boldsymbol{v} \in M | f(\boldsymbol{v}) = 0} \\ + \operatorname{ran}(f) =& \Set{f(\boldsymbol{v}) | \boldsymbol{v} \in M} \end{align*} $$ @@ -3045,7 +3045,7 @@ $$ If $f\in\hom_R (M, N)$ is an $R$-map, the kernel and range of $f$ are submodules of $M$ and $N$ respectively. Moreover, $f$ is a monomorphism if and only if $\ker(f) = \Set{0}$. -If $N$ is a submodule of the $R$-module $M$, the map $j:N\to M$ defined by $j(\mathbf{v}) = \mathbf{v}$ is evidently an $R$-monomorphism, called the *injection* of $N$ into $M$. +If $N$ is a submodule of the $R$-module $M$, the map $j:N\to M$ defined by $j(\boldsymbol{v}) = \boldsymbol{v}$ is evidently an $R$-monomorphism, called the *injection* of $N$ into $M$. ## Direct sums and direct summands @@ -3059,15 +3059,15 @@ $$ is the $R$-module whose elements are ordered $n$-tuples $$ - M = \Set{(\mathbf{v}_i)_{i=1}^n | \mathbf{v}_i \in M_i, i=1,\dots,n} + M = \Set{(\boldsymbol{v}_i)_{i=1}^n | \boldsymbol{v}_i \in M_i, i=1,\dots,n} $$ with componentwise operations $$ \begin{align*} - (\mathbf{u}_1,\dots,\mathbf{u}_n) + (\mathbf{v}_1,\dots,\mathbf{v}_n) =& (\mathbf{u}_1 + \mathbf{v}_1, \dots, \mathbf{u}_n + \mathbf{v}_n) \\ - \rho(\mathbf{v}_1,\dots,\mathbf{v}_n) =& (\rho\mathbf{v}_1,\dots,\rho\mathbf{v}_n) + (\boldsymbol{u}_1,\dots,\boldsymbol{u}_n) + (\boldsymbol{v}_1,\dots,\boldsymbol{v}_n) =& (\boldsymbol{u}_1 + \boldsymbol{v}_1, \dots, \boldsymbol{u}_n + \boldsymbol{v}_n) \\ + \rho(\boldsymbol{v}_1,\dots,\boldsymbol{v}_n) =& (\rho\boldsymbol{v}_1,\dots,\rho\boldsymbol{v}_n) \end{align*} $$ @@ -3092,7 +3092,7 @@ In this case, each $S_i$ is called a *direct summand* of $M$. If $M = S \oplus T If $\mathcal{F} = \Set{S_i}_{i\in I}$ is a family of distinct submodules of an $R$-module $M$, the following are equivalent 1. **Independence of the family:** For each $i\in I$, then $S_i \cap \left(\sum_{j\neq i} S_j \right) = \Set{0}$ 2. **Uniqueness of zero element:** The zero element $0$ cannot be written as a sum of nonzero elements from distinct submodules in $\mathcal{F}$ -3. **Uniqueness of expression:** Every nonzero $\mathbf{v}\in M$ has a unique, except for order of terms, expression as a sum $\mathbf{v} = \sum_{i=1}^n \mathbf{s}_n$ of nonzero elements from distinct submodules in $\mathcal{F}$. +3. **Uniqueness of expression:** Every nonzero $\boldsymbol{v}\in M$ has a unique, except for order of terms, expression as a sum $\boldsymbol{v} = \sum_{i=1}^n \boldsymbol{s}_n$ of nonzero elements from distinct submodules in $\mathcal{F}$. @@ -3118,7 +3118,7 @@ $$
Proof -**(1):** If $M = N\oplus H$, then a module epimorphism $f:N\to M_1$ can be extended to an epimorphism $\bar{f}:M\to M_1$ simply be sending the elements of $H$ to zero, i.e. $\bar{f}(\mathbf{n}+\mathbf{h}) = f(\mathbf{n})$. This is easily seen to be an $R$-map with $\ker(\bar{f}) = \ker(f) \oplus H$. Moreover, if $g$ is another extension of $f$ with the same kernel as $\bar{f}$, then $\tau$ and $\bar{f}$ agree on $H$ as well as on $N$, thus $\tau = \bar{f}$. Thus, there is a unique extension of $f$ with kernel $\ker(f)\oplus H$. +**(1):** If $M = N\oplus H$, then a module epimorphism $f:N\to M_1$ can be extended to an epimorphism $\bar{f}:M\to M_1$ simply be sending the elements of $H$ to zero, i.e. $\bar{f}(\boldsymbol{n}+\boldsymbol{h}) = f(\boldsymbol{n})$. This is easily seen to be an $R$-map with $\ker(\bar{f}) = \ker(f) \oplus H$. Moreover, if $g$ is another extension of $f$ with the same kernel as $\bar{f}$, then $\tau$ and $\bar{f}$ agree on $H$ as well as on $N$, thus $\tau = \bar{f}$. Thus, there is a unique extension of $f$ with kernel $\ker(f)\oplus H$. **(2):** Suppose that $f: N\cong M_1$ is an isomorphism. If $N$ is complemented, i.e. if $G = N \oplus H$, then there is a unique extension $\bar{f}$ of $f$ for which $\ker(\bar{f}) = H$. Thus, the correspondence $H\mapsto\bar{f}$, where $\ker(f) = H$, from complements of $N$ to extensions of $f$ is an injection. To see that this correspondence is a bijection, if $\bar{f}:M\to M_1$ is an extension of $f$, then $M = N\oplus\ker(\bar{f})$. To see this, we have @@ -3126,16 +3126,16 @@ $$ N \cap\ker(\bar{f}) = \ker(f) = \Set{0} $$ -and if $\mathbf{a}\in M$, then there is a $\mathbf{b}\in N$, for which $f \mathbf{b} = \bar{f}\mathbf{a}$ and so +and if $\boldsymbol{a}\in M$, then there is a $\boldsymbol{b}\in N$, for which $f \boldsymbol{b} = \bar{f}\boldsymbol{a}$ and so $$ - \bar{f}(\mathbf{a} - \mathbf{b}) = \bar{f}\mathbf{a} - f \mathbf{b} = 0 + \bar{f}(\boldsymbol{a} - \boldsymbol{b}) = \bar{f}\boldsymbol{a} - f \boldsymbol{b} = 0 $$ Thus, $$ - \mathbf{a} = \mathbf{b} + (\mathbf{a} - \mathbf{b}) \in N + \ker(\bar{f}) + \boldsymbol{a} = \boldsymbol{b} + (\boldsymbol{a} - \boldsymbol{b}) \in N + \ker(\bar{f}) $$ which shows that $\ker(\bar{f})$ is a complement of $N$. @@ -3242,7 +3242,7 @@ with basis $\Set{(1,1)}$. However, the submodule $\Z\times\Set{0}$ is not free s Let $M$ and $N$ be $R$-modules where $M$ is free with basis $B = \Set{\mathbb{b}_i}_{i\in I}$. Then we can define a unique $R$-map $f:M\to N$ by specifying the values of $f(\mathbb{b}_i)$, arbitrarily for all $\mathbb{b}_i \in B$ and then extending $f$ to $M$ by linearity, i.e. $$ - f\left(\sum_{i=1} \alpha_i \mathbf{v}_i \right) = \sum_{i=1} \alpha_i f(\mathbf{v}_i) + f\left(\sum_{i=1} \alpha_i \boldsymbol{v}_i \right) = \sum_{i=1} \alpha_i f(\boldsymbol{v}_i) $$ @@ -3260,19 +3260,19 @@ The idea of the proof is to find a vector space $V$ with the property that, for Let $\mathcal{I}$ be a maximal ideal of $R$. Then $R\setminus\mathcal{I}$ is a field. Note that $M$ is not a vector space over $R\setminus\mathcal{I}$. Scalar multiplication using the field $R\setminus\mathcal{I}$, $$ - (\rho + \mathcal{I})\mathbf{v} = \rho\mathbf{v} + (\rho + \mathcal{I})\boldsymbol{v} = \rho\boldsymbol{v} $$ is not well-defined, since this would require that $\mathcal{I}M = \Set{0}$. On the other hand, we can fix this issue by factoring out the submodule $$ - \mathcal{I}M = \Set{\sum_{i=1} \alpha_i \mathbf{v}_i | \alpha_i \in\mathcal{I}, \mathbf{v}_i \in M} + \mathcal{I}M = \Set{\sum_{i=1} \alpha_i \boldsymbol{v}_i | \alpha_i \in\mathcal{I}, \boldsymbol{v}_i \in M} $$ Indeed, $M\setminus \mathcal{I}M$ is a vector space over $R\setminus\mathcal{I}$, with scalar multiplication defined by $$ - (\rho + \mathcal{I})(\mathbf{u} + \mathcal{I}M) = \rho\mathbf{u} + \mathcal{I}M + (\rho + \mathcal{I})(\boldsymbol{u} + \mathcal{I}M) = \rho\boldsymbol{u} + \mathcal{I}M $$ To see that this is well-defined, we must show that the conditions @@ -3280,63 +3280,63 @@ To see that this is well-defined, we must show that the conditions $$ \begin{align*} \rho + \mathcal{I} =& \rho' + \mathcal{I} \\ - \mathbf{u} + \mathcal{I}M =& \mathbf{u}' + \mathcal{I}M + \boldsymbol{u} + \mathcal{I}M =& \boldsymbol{u}' + \mathcal{I}M \end{align*} $$ imply $$ - \rho\mathbf{u} + \mathcal{I}M = \rho'\mathbf{u}' + \mathcal{I}M + \rho\boldsymbol{u} + \mathcal{I}M = \rho'\boldsymbol{u}' + \mathcal{I}M $$ But this follows from the fact that $$ - \rho\mathbf{u} - \rho'\mathbf{u}' = \rho(\mathbf{u} - \mathbf{u}') + (\rho - \rho')\mathbf{u}' \in\mathcal{I}M + \rho\boldsymbol{u} - \rho'\boldsymbol{u}' = \rho(\boldsymbol{u} - \boldsymbol{u}') + (\rho - \rho')\boldsymbol{u}' \in\mathcal{I}M $$ Hence, scalar multiplication is well-defined. It is easy to show that $M\setminus\mathcal{I}M$ is a vector space over $R\setminus\mathcal{I}$. -Consider a set $B = \Set{\mathbf{b}_i}_{i\in I}\subseteq M$ and the corresponding set +Consider a set $B = \Set{\boldsymbol{b}_i}_{i\in I}\subseteq M$ and the corresponding set $$ - B + \mathcal{I}M = \Set{\mathbf{b}_i + \mathcal{I}M | i\in I} \subseteq M\setminus\mathcal{I}M + B + \mathcal{I}M = \Set{\boldsymbol{b}_i + \mathcal{I}M | i\in I} \subseteq M\setminus\mathcal{I}M $$ -If $B$ spans $M$ over $R$, then $B + \mathcal{I}M$ spans $M\setminus\mathcal{I}M$ over $R\setminus\mathcal{I}$. To see this, note that any $\mathbf{v}\in M$ has the form $\mathbf{v} = \sum \rho_i \mathbf{b}_i$ for $\rho_i \in R$ and so +If $B$ spans $M$ over $R$, then $B + \mathcal{I}M$ spans $M\setminus\mathcal{I}M$ over $R\setminus\mathcal{I}$. To see this, note that any $\boldsymbol{v}\in M$ has the form $\boldsymbol{v} = \sum \rho_i \boldsymbol{b}_i$ for $\rho_i \in R$ and so $$ \begin{align*} - \mathbf{v} + \mathcal{I}M =& \left(\sum_j \rho_{i_j} \mathbf{b}_{i_j} \right) + \mathcal{I}M \\ - =& \sum_j \rho_{i_j} (\mathbf{b}_{i_j} + \mathcal{I}M) \\ - =& \sum_j (\rho_{i_j} + \mathcal{I})(\mathbf{b}_{i_j} + \mathcal{I}M) + \boldsymbol{v} + \mathcal{I}M =& \left(\sum_j \rho_{i_j} \boldsymbol{b}_{i_j} \right) + \mathcal{I}M \\ + =& \sum_j \rho_{i_j} (\boldsymbol{b}_{i_j} + \mathcal{I}M) \\ + =& \sum_j (\rho_{i_j} + \mathcal{I})(\boldsymbol{b}_{i_j} + \mathcal{I}M) \end{align*} $$ showing that $B + \mathcal{I}M$ spans $M\setminus\mathcal{I}M$. -Suppose that $B = \Set{\mathbf{b}_i}_{i\in I}$ is a basis for $M$ over $R$. We will show that $B + \mathcal{I}M$ is a basis for $M\setminus\mathcal{I}M$ over $R\setminus\mathcal{I}$. We have seen that $B + \mathcal{I}M$ spans $M\setminus\mathcal{I}M$. Also, if +Suppose that $B = \Set{\boldsymbol{b}_i}_{i\in I}$ is a basis for $M$ over $R$. We will show that $B + \mathcal{I}M$ is a basis for $M\setminus\mathcal{I}M$ over $R\setminus\mathcal{I}$. We have seen that $B + \mathcal{I}M$ spans $M\setminus\mathcal{I}M$. Also, if $$ - \sum_j (\rho_{i_j} + \mathcal{I})(\mathbf{b}_{i_j} + \mathcal{I}M) = \mathcal{I}M + \sum_j (\rho_{i_j} + \mathcal{I})(\boldsymbol{b}_{i_j} + \mathcal{I}M) = \mathcal{I}M $$ -then $\sum_j \rho_{i_j}\mathbf{b}_{i_j} \in\mathcal{I}M$ and so +then $\sum_j \rho_{i_j}\boldsymbol{b}_{i_j} \in\mathcal{I}M$ and so $$ - \sum_j \rho_{i_j}\mathbf{b}_{i_j} = \sum_k \alpha_{i_k} \mathbf{b}_{i_k} + \sum_j \rho_{i_j}\boldsymbol{b}_{i_j} = \sum_k \alpha_{i_k} \boldsymbol{b}_{i_k} $$ where $\alpha_{i_k}\in\mathcal{I}$. From the linear independence of $B$ we deduce that $\rho_{i_j} \in\mathcal{I}$ for all $j$ and so $\rho_{i_j} + \mathcal{I} = \mathcal{I}$. Hence $B + \mathcal{I}M$ is linearly independent and therefore a basis. -To see that $|B| = |B + \mathcal{I}M|$, note that if $\mathbf{b}_i + \mathcal{I}M = \mathbf{b}_k + \mathcal{I}M$, then +To see that $|B| = |B + \mathcal{I}M|$, note that if $\boldsymbol{b}_i + \mathcal{I}M = \boldsymbol{b}_k + \mathcal{I}M$, then $$ - \mathbf{b}_i - \mathbf{b}_k = \sum_j \alpha_{i_j} \mathbf{b}_{i_j} + \boldsymbol{b}_i - \boldsymbol{b}_k = \sum_j \alpha_{i_j} \boldsymbol{b}_{i_j} $$ -where $\alpha_{i_j}\in\mathcal{I}$. If $\mathbf{b}_i \neq \mathbf{b}_k$, then the coefficients of $\mathbf{b}_i$ on the right must be equal to $1$ and so $1\in\mathcal{I}$, which is not possible since $\mathcal{I}$ is a maximal ideal. Thus, $\mathbf{b}_i = \mathbf{b}_k$. +where $\alpha_{i_j}\in\mathcal{I}$. If $\boldsymbol{b}_i \neq \boldsymbol{b}_k$, then the coefficients of $\boldsymbol{b}_i$ on the right must be equal to $1$ and so $1\in\mathcal{I}$, which is not possible since $\mathcal{I}$ is a maximal ideal. Thus, $\boldsymbol{b}_i = \boldsymbol{b}_k$. Hence, if $B$ is a basis for $M$ over $R$, then @@ -3368,16 +3368,16 @@ The ring $\mathcal{L}(V)$ is an $\mathcal{L}(V)$-module and as such, the identit We begin by partitioning $B$ into $n$ blocks. For each $s = 0,\dots,n-1$, let $$ - B_s = \Set{\mathbf{b}_i | i \equiv s\mod n} + B_s = \Set{\boldsymbol{b}_i | i \equiv s\mod n} $$ -Define elements $f_s \in\mathbb{L}(V)$ by $f_s (\mathbf{b}_{kn+t}) = \delta_{ts} \mathbf{b}_k$ where $0\leq t < n$ and $\delta_{ts}$ is the Kronecker delta function. These functions are surjective and have disjoint support. It follows that $C_n = \Set{f_i}_{i=0}^{n-1}$ is linearly independent: if $0 = \sum{i=0}^{n-1} g_i f_i$ for $g_i \in\mathcal{L}(V)$, then applying this to $\mathbf{b}_{kn+t}$ gives +Define elements $f_s \in\mathbb{L}(V)$ by $f_s (\boldsymbol{b}_{kn+t}) = \delta_{ts} \boldsymbol{b}_k$ where $0\leq t < n$ and $\delta_{ts}$ is the Kronecker delta function. These functions are surjective and have disjoint support. It follows that $C_n = \Set{f_i}_{i=0}^{n-1}$ is linearly independent: if $0 = \sum{i=0}^{n-1} g_i f_i$ for $g_i \in\mathcal{L}(V)$, then applying this to $\boldsymbol{b}_{kn+t}$ gives $$ - 0 = g_t f_t (\mathbf{b}_{kn+1}) = g_t (\mathbf{b}_k),\; \forall k + 0 = g_t f_t (\boldsymbol{b}_{kn+1}) = g_t (\boldsymbol{b}_k),\; \forall k $$ -Thus, $g_t = 0$. Also $C_n$ spans $\mathcal{L}(V)$, for if $h \in\mathcal{L}(V)$, we define $g_s \in\mathcal{L}(V)$ by $g_s (\mathbf{b}_k) = h(\mathbf{b}_{kn+1})$ to get +Thus, $g_t = 0$. Also $C_n$ spans $\mathcal{L}(V)$, for if $h \in\mathcal{L}(V)$, we define $g_s \in\mathcal{L}(V)$ by $g_s (\boldsymbol{b}_k) = h(\boldsymbol{b}_{kn+1})$ to get $$ \left(\sum_{i=0}^{n-1} g_i f_i \right)(b_{kn+1}) = g_t f_t (b_{kn+t}) = g_t (b_k) = h(b_{kn+1}) @@ -3437,19 +3437,19 @@ Let $R$ be an integral domain and let $M$ be a free $R$-module. Then all linearl Since $M\cong (R^\kappa)_0$ we need only prove the result for $(R^\kappa)_0$. Let $Q$ be the field of quotients of $R$. Then $(Q^\kappa)_0$ is a vector space. If $$ - B = \Set{\mathbf{v}_i}_{i\in I}\subseteq (R^\kappa)_0 \subseteq (Q^\kappa)_0 + B = \Set{\boldsymbol{v}_i}_{i\in I}\subseteq (R^\kappa)_0 \subseteq (Q^\kappa)_0 $$ is linearly independent over $Q$ as a subset of $(Q^\kappa)_0, then $B$ is clearly linearly independent over $R$ as a subset of $(R^\kappa)_0$. Conversely, suppose that $B$ is linearly independent over $R$ and $$ - \sum_{j=1}^k \frac{\rho_j}{\sigma_j} \mathbf{v}_{i_j} = 0 + \sum_{j=1}^k \frac{\rho_j}{\sigma_j} \boldsymbol{v}_{i_j} = 0 $$ where $\sigma_j \neq 0$ for all $j$ and $\rho_j \neq 0$ for some $j$. Multiplying by $\sigma = \prod_{i=1} \sigma_i \neq 0$ produces a nontrivial linear dependency over $R$ $$ - \sum_{j=1}^k \frac{\sigma}{\sigma_j} \rho_j \mathbf{v}_{i_j} = 0 + \sum_{j=1}^k \frac{\sigma}{\sigma_j} \rho_j \boldsymbol{v}_{i_j} = 0 $$ which implies that $\rho_i = 0$ for all $i$. Thus $B$ is linearly dependent over $R$ if and only if it is linearly dependent over $Q$. However, in the vector space $(Q^\kappa)_0$, all sets of cardinality greater than $\kappa$ are linearly dependent over $Q$ and hence all subsets of $(R^\kappa)_0$ of cardinality greater than $\kappa$ are linearly dependent over $R$. @@ -3481,21 +3481,21 @@ $$ Let $S$ be a submodule of an $R$-module $M$. The binary relation $$ - \mathbf{u} \equiv \mathbf{v} \iff \mathbf{u} - \mathbf{v} \in S + \boldsymbol{u} \equiv \boldsymbol{v} \iff \boldsymbol{u} - \boldsymbol{v} \in S $$ is an equivalence relation on $M$, whose equivalence classes are the *cosets* $$ - \mathbf{v} + S = \Set{\mathbf{v} + \mathbf{s} | \mathbf{s}\in S} + \boldsymbol{v} + S = \Set{\boldsymbol{v} + \boldsymbol{s} | \boldsymbol{s}\in S} $$ of $S$ in $M$. The set $M\setminus S$ of all cosets of $S$ in $M$, called the *quotient module* of $M$ *modulo* $S$, is an $R$-module under the well-defined operations $$ \begin{align*} - (\mathbf{u} + S) + (\mathbf{v} + S) =& (\mathbf{u} + \mathbf{v}) + S \\ - \rho(\mathbf{u} + S) =& \rho\mathbf{u} + S + (\boldsymbol{u} + S) + (\boldsymbol{v} + S) =& (\boldsymbol{u} + \boldsymbol{v}) + S \\ + \rho(\boldsymbol{u} + S) =& \rho\boldsymbol{u} + S \end{align*} $$ @@ -3571,21 +3571,21 @@ $$ S_1 \subseteq S_2 \subseteq \cdots $$ -of submodules. Then the union $S = \bigcup_j S_j$ is easily seen to be a submodule of $M$. Hence, $S$ is finitely generated, say $S = \langle \mathbf{u}_1,\dots, u_n \rangle$. Since $\mathbf{u}_i \in S$, there exists an index $k_i$, such that $\mathbf{u}_i \in S_{k_i}$. Thus, if $k = \max\Set{k_i}_{i=1}^n$, we have $\Set{\mathbf{u}_i}_{i=1}^n \subseteq S_k$ and so +of submodules. Then the union $S = \bigcup_j S_j$ is easily seen to be a submodule of $M$. Hence, $S$ is finitely generated, say $S = \langle \boldsymbol{u}_1,\dots, u_n \rangle$. Since $\boldsymbol{u}_i \in S$, there exists an index $k_i$, such that $\boldsymbol{u}_i \in S_{k_i}$. Thus, if $k = \max\Set{k_i}_{i=1}^n$, we have $\Set{\boldsymbol{u}_i}_{i=1}^n \subseteq S_k$ and so $$ - S = \langle \mathbf{u}_1,\dots,\mathbf{u}_n \rangle \subseteq S_k \subseteq S_{k+1} \subseteq\cdots\subseteq S + S = \langle \boldsymbol{u}_1,\dots,\boldsymbol{u}_n \rangle \subseteq S_k \subseteq S_{k+1} \subseteq\cdots\subseteq S $$ which shows that the chain is eventually stable. -Conversely, suppose that $M$ satisfies the ascending chain condition on submodules and let $S$ be a submodule of $M$. Pick $\mathbf{u}_1 \in S$ and consider the submodule $S_1 = \langle \mathbf{u}_1 \rangle \subseteq S$ generated by $\mathbf{u}_1$. If $S_1 = S$, then $S$ is finitely generated. If $S_1 \neq S$, then there is a $\mathbf{u}_2 \in S - S_1$. Let $S_2 = \langle \mathbf{u}_1, \mathbf{u}_2 \rangle$. If $S_2 = S$, then $S$ is finitely generated. If $S_2 \neq S$, then pick $\mathbf{u}_3 \in S - S_2$ and consider the submodule $S_3 = \langle \mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\rangle$. Continuing this way, we get an ascending chain of submodules +Conversely, suppose that $M$ satisfies the ascending chain condition on submodules and let $S$ be a submodule of $M$. Pick $\boldsymbol{u}_1 \in S$ and consider the submodule $S_1 = \langle \boldsymbol{u}_1 \rangle \subseteq S$ generated by $\boldsymbol{u}_1$. If $S_1 = S$, then $S$ is finitely generated. If $S_1 \neq S$, then there is a $\boldsymbol{u}_2 \in S - S_1$. Let $S_2 = \langle \boldsymbol{u}_1, \boldsymbol{u}_2 \rangle$. If $S_2 = S$, then $S$ is finitely generated. If $S_2 \neq S$, then pick $\boldsymbol{u}_3 \in S - S_2$ and consider the submodule $S_3 = \langle \boldsymbol{u}_1,\boldsymbol{u}_2,\boldsymbol{u}_3\rangle$. Continuing this way, we get an ascending chain of submodules $$ - \langle \mathbf{u}_1 \rangle \subseteq \langle \mathbf{u}_1,\mathbf{u}_2 \rangle \subseteq\cdots\subseteq S + \langle \boldsymbol{u}_1 \rangle \subseteq \langle \boldsymbol{u}_1,\boldsymbol{u}_2 \rangle \subseteq\cdots\subseteq S $$ -If none of these submodules were equal to $S$, we would have an infinite ascending chain of submodules, each properly contained in the next, which contradicts the fact that $M$ satisfies the ascending chain condition on submodules. Hence, $S = \langle \mathbf{u}_1,\dots,\mathbf{u}_n \rangle$ for some $n$ and so $S$ is finitely generated. +If none of these submodules were equal to $S$, we would have an infinite ascending chain of submodules, each properly contained in the next, which contradicts the fact that $M$ satisfies the ascending chain condition on submodules. Hence, $S = \langle \boldsymbol{u}_1,\dots,\boldsymbol{u}_n \rangle$ for some $n$ and so $S$ is finitely generated.
@@ -3597,13 +3597,13 @@ Let $R$ be a commutative ring with identity.
Proof -**(1):** If $M$ is a Noetherian $R$-module, then evidently $R$ is Noetherian. Conversely, assume that $R$ is Noetherian and let $M = \langle \mathbf{u}_1,\dots,\mathbf{u}_n \rangle$ be a finitely generated $R$-module. Consider the epimorphism $f:R^n \to M$ defined by +**(1):** If $M$ is a Noetherian $R$-module, then evidently $R$ is Noetherian. Conversely, assume that $R$ is Noetherian and let $M = \langle \boldsymbol{u}_1,\dots,\boldsymbol{u}_n \rangle$ be a finitely generated $R$-module. Consider the epimorphism $f:R^n \to M$ defined by $$ - f(\rho_1,\dots,\rho_n) = \sum_{i=1}^n \rho_i \mathbf{u}_i + f(\rho_1,\dots,\rho_n) = \sum_{i=1}^n \rho_i \boldsymbol{u}_i $$ -Let $S$ be a submodule of $M$. Then $f^{-1}(S) = \Set{\mathbf{u} \in R^n | f \mathbf{u} \in S}$ is a submodule of $R^n$ and $f(f^{-1}S) = S$. If every submodule of $R^n$ is finitely generated, then $f^{-1}(S)$ is finitely generated and $f^{-1}(S) = \langle \mathbf{v}_1,\dots,\mathbf{v}_k \rangle$. Then $S$ is finitely generated by $\Set{f \mathbf{v}_i}_{i=1}^k$. Thus, it is sufficient to prove the theorem for $R^n$, which we do by induction on $n$. +Let $S$ be a submodule of $M$. Then $f^{-1}(S) = \Set{\boldsymbol{u} \in R^n | f \boldsymbol{u} \in S}$ is a submodule of $R^n$ and $f(f^{-1}S) = S$. If every submodule of $R^n$ is finitely generated, then $f^{-1}(S)$ is finitely generated and $f^{-1}(S) = \langle \boldsymbol{v}_1,\dots,\boldsymbol{v}_k \rangle$. Then $S$ is finitely generated by $\Set{f \boldsymbol{v}_i}_{i=1}^k$. Thus, it is sufficient to prove the theorem for $R^n$, which we do by induction on $n$. In the base case $n = 1$, we can extract from $S$ something that is isomorphic to an ideal of $R$, and so will be finitely generated. In particular, let $S_1$ be the "last coordinates" in $S$, specifically, let @@ -3611,38 +3611,38 @@ $$ S_1 = \Set{0,\dots,0,\alpha_n} | (\alpha_i)_{i=1}^n \in S,\; \alpha_i \in R $$ -The set $S_1$ is isomorphic to an ideal of $R$ and is therefore finitely generated, say $S_1 = \langle G_1 \rangle$, where $G_1 = \Set{\mathbf{g}_i}_{i=1}^k$ is a finite subset of $S_1$. +The set $S_1$ is isomorphic to an ideal of $R$ and is therefore finitely generated, say $S_1 = \langle G_1 \rangle$, where $G_1 = \Set{\boldsymbol{g}_i}_{i=1}^k$ is a finite subset of $S_1$. Also let $$ - S_2 = \Set{\mathbf{v} \in S | \mathbf{v} = (\alpha_1,\dots,\alpha_{n-1},0), \alpha_1,\dots,\alpha_{n-1}\in R} + S_2 = \Set{\boldsymbol{v} \in S | \boldsymbol{v} = (\alpha_1,\dots,\alpha_{n-1},0), \alpha_1,\dots,\alpha_{n-1}\in R} $$ be the set of all elements of $S$ that have last coordinate equal to $0$. Note that $S_2$ is a submodule of $R^n$ and is isomorphic to a submodule of $R^{n-1}$. Hence, the inductive hypothesis implies that $S_2$ is finitely generated, say $S_2 = \langle G_2 \rangle$ where $G_2$ is a finite subset of $S$. -By definition of $S_1$, each $\mathbf{g}_i \in G_1$ has the form $\mathbf{g}_i = (0,\dots,0,g_{i,n})$ for $g_{i,n}\in R$ where there is a $\bar{\mathbf{b}}_i \in S$ of the form +By definition of $S_1$, each $\boldsymbol{g}_i \in G_1$ has the form $\boldsymbol{g}_i = (0,\dots,0,g_{i,n})$ for $g_{i,n}\in R$ where there is a $\bar{\boldsymbol{b}}_i \in S$ of the form $$ - \bar{\mathbf{g}}_i = (g_{i,1},\dots,g_{i,n-1},g_{i,n}) + \bar{\boldsymbol{g}}_i = (g_{i,1},\dots,g_{i,n-1},g_{i,n}) $$ -Let $\bar{G} = \Set{\bar{\mathbf{g}}_i}_{i=1}^k$. We claim that $S$ is generated by the finite set $\bar{G}_1 \cup G_2$. To see this, let $\mathbf{v} = (\alpha_i)_{i=1}^n \inn S$. Then $(0,\dots,0,\alpha_n)\in S_1$ and so +Let $\bar{G} = \Set{\bar{\boldsymbol{g}}_i}_{i=1}^k$. We claim that $S$ is generated by the finite set $\bar{G}_1 \cup G_2$. To see this, let $\boldsymbol{v} = (\alpha_i)_{i=1}^n \inn S$. Then $(0,\dots,0,\alpha_n)\in S_1$ and so $$ - (0,\dots,0,\alpha_n) = \sum_{i=1}^k \rho_i \mathbf{g}_i,\; \rho_i \in R + (0,\dots,0,\alpha_n) = \sum_{i=1}^k \rho_i \boldsymbol{g}_i,\; \rho_i \in R $$ Consider the sum $$ - w = \sum_{i=1}^k \rho_i \bar{\mathbf{g}}_i \in \langle \bar{G}_1 \rangle + w = \sum_{i=1}^k \rho_i \bar{\boldsymbol{g}}_i \in \langle \bar{G}_1 \rangle $$ -The last coordinate of this sum is $\sum_{i=1}^k \rho_i \mathbf{g}_{i,n} = a_n$ and so the difference $\mathbf{v} - \mathbf{w}$ has last coordinate $0$ and is thus in $S_2 = \langle G_2 \rangle$. Hence +The last coordinate of this sum is $\sum_{i=1}^k \rho_i \boldsymbol{g}_{i,n} = a_n$ and so the difference $\boldsymbol{v} - \boldsymbol{w}$ has last coordinate $0$ and is thus in $S_2 = \langle G_2 \rangle$. Hence $$ - \mathbf{v} = (\mathbf{v} - \mathbf{w}) + \mathbf{w} \in \langle \bar{G}_1 \rangle + \langle G_2 \rangle = \langle \bar{G}_1 \cup G_2 \rangle + \boldsymbol{v} = (\boldsymbol{v} - \boldsymbol{w}) + \boldsymbol{w} \in \langle \bar{G}_1 \rangle + \langle G_2 \rangle = \langle \bar{G}_1 \cup G_2 \rangle $$