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Use the PDF's covector-first convention: its space ${\mathcal{T}}_l^k$ consists of tensor fields inThis reverses the order in which some texts list tensor type. For a finite-dimensional real vector space $V$, define the tensor contraction first on decomposable tensors:The hats mean omission, with the other factors left in their original order. The formula is a multilinear map of its individual factors, so the universal property of a tensor product gives a unique linear map with this formula.

Apply it with $V=T_{p}M$ at every $p$. Evaluation of a covector on a vector is basis independent: under a change of frame, one factor transforms by a matrix and the other by its inverse transpose, and the matrices cancel in the pairing. Thus the fiber maps agree on overlapping vector bundle trivializations. In a local dual basis, the coefficients of the contracted tensor are finite sums of coefficients with the selected covariant and contravariant indices set equal. They remain smooth. Henceis a globally defined smooth contraction for $k,l\ge 1$. It is also linear over $C^{\infty }(M)$.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

Put $L={\mathcal{O}}_X(D)$ and let $s_D$ be the canonical global section cutting out the effective Cartier divisor $D$. The divisor restriction exact sequence givesThe hypothesis says $L{|}_D$ is an ample line bundle. By Serre vanishing, $H^1(D,L^m{|}_D)=0$ for all sufficiently large $m$, so the long exact sequence in sheaf cohomology makessurjective for all such $m$. These are finite-dimensional vector spaces. Their dimensions form a nonincreasing sequence of nonnegative integers, so the maps are isomorphisms from some point on. Exactness then implies that the restrictionis surjective for all sufficiently large $m$.

Choose such an $m$ for which $L^m{|}_D$ is also a globally generated line bundle. Lift a generating collection of its global sections to $X$. At each point of $D$, one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of $L^m$ by Nakayama lemma. Outside $D$, $s_D^m$ is nowhere zero and generates $L^m$. Together these sections generate it everywhere. ThereforeThis proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If $D$ is empty, $L\cong {\mathcal{O}}_X$ and the conclusion is immediate. Semiampleness is the conclusion: the pullback of a line avoiding the centre of a blowup of a smooth algebraic surface of ${\mathbb{P}}^2$ satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.

Write $|F|=p^n$, where $n=[F:{\mathbb{F}}_p]$ since $F$ is a finite-dimensional vector space over its prime field. Every nonzero element satisfies $x^{p^n-1}=1$ by Lagrange theorem, so $F$ is exactly the splitting field of $X^{p^n}-X$ over ${\mathbb{F}}_p$. The derivative is $-1$, making it a separable polynomial. Thus the extension is a normal field extension and a separable field extension, hence a Galois extension.

The finite-field Frobenius automorphism $\sigma (x)=x^p$ fixes ${\mathbb{F}}_p$ and has ${\sigma }^n=1$. If ${\sigma }^d=1$ with $0<d<n$, all $p^n$ elements would be roots of the degree-$p^d$ polynomial $X^{p^d}-X$, impossible. It has order $n$, equal to the degree of the Galois extension. Therefore

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

A quadratic form on a finite-dimensional real vector space $V$ is a function $q(v)=B(v,v)$ for some symmetric bilinear form $B$. A positive-definite quadratic form satisfies $q(v)>0$ for every $v\ne 0$.

Completing squares givesSet $u=x+y$, $v=y+z$, $w=z$. The corresponding ordered basis in the original coordinates isand the diagonal form is $u^2+v^2+2w^2$. All diagonal coefficients are positive, so the form is positive definite.

A basis of a vector space $V$ is a family of vectors that is both linearly independent and a spanning set of $V$.

Write the given finite basis as $\mathcal{B}=\{v_1,\dots ,v_n\}$. First, any other basis ${\mathcal{B}}^{\prime }$ must also be finite. Each $v_i$ is a finite linear combination of elements of ${\mathcal{B}}^{\prime }$, so the union $S$ of the finitely many elements of ${\mathcal{B}}^{\prime }$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in {\mathcal{B}}^{\prime }\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the linear independence of ${\mathcal{B}}^{\prime }$. Hence ${\mathcal{B}}^{\prime }=S$; write $|{\mathcal{B}}^{\prime }|=m$.

It remains to prove the elementary Steinitz exchange lemma directly. If independent vectors $w_1,\dots ,w_r$ lie in the span of $v_1,\dots ,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\dots ,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\dots ,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\le n$.

Apply this argument first to the independent family ${\mathcal{B}}^{\prime }$ and the spanning family $\mathcal{B}$ to obtain $m\le n$, and then with the two bases interchanged to obtain $n\le m$. Therefore

Two norms $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$ on one vector space are Lipschitz equivalent when constants $c,C>0$ exist such that

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

Let $\mathfrak{m}$ be the maximal ideal and $k=A/\mathfrak{m}$ the finite residue field. The maximal ideal of an Artinian local ring is nilpotent, so ${\mathfrak{m}}^r=0$ for some $r$. Every quotientis both an Artinian $A$-module and a vector space over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a finite set. The finite filtrationtherefore proves that the underlying set of $A$ is finite. This is the Finiteness criterion for an Artinian local ring.

If $y_1\le y_2$, then the utility function is increasing, so for every $X\in \mathcal{X}$,Taking suprema gives $F(y_1)\le F(y_2)$.

For $0\le t\le 1$, let $X_1$ and $X_2$ attain the suprema at $y_1$ and $y_2$. Because $\mathcal{X}$ is a vector space,The concavity of $U$ now givesThus $F$ is increasing and concave, as stated by optimized affine shift of concave utility.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

An $R$-module is free if it has a basis: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.

Choose a maximal ideal $\mathfrak{m}$ of the nonzero ring $R$. If $R^n\cong R^m$, quotienting by $\mathfrak{m}$ givesThese are vector spaces over the field $R/\mathfrak{m}$, so equality of dimension gives $n=m$. This proves invariant basis number for a commutative ring.

A direct summand of a free module need not be free. Take $R=\mathbb{Z}/6\mathbb{Z}$. Its ideals $P=(2)$ and $Q=(3)$ satisfyso $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.