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The ring form of the Artin–Wedderburn theorem states that the following conditions on a unital ring $R$ are equivalent: $R$ is a semisimple ring; $R$ satisfies the descending chain condition on left ideals and $J(R)=0$; andwhere $s$ is finite, each $n_i\ge 1$, and each $D_i$ is a division ring. The factors are unique up to permutation and isomorphism. The zero ring may be included as the empty product. This theorem also implies the corresponding right-handed statements.

First prove the implication from the chain condition and zero Jacobson radical. Among finite intersections of maximal left ideals, choose a minimal one, say $L=L_1\cap \cdots \cap L_t$. Such a minimal intersection exists by the descending chain condition. Intersecting with any other maximal left ideal cannot make $L$ smaller, so $L$ is contained in every maximal left ideal. Thus $L\subseteq J(R)=0$. It follows that the map of left modulesis injective. Each quotient is a simple module.

For completeness, a submodule $N$ of a finite direct sum $V={⨁}_j{T}_j$ of simple modules is itself a finite direct sum of simple modules and is a direct summand of $V$. Choose a sum $W$ of coordinate summands, maximal among those with $N\cap W=0$. For any omitted coordinate summand $T_j$, simplicity gives $T_j\cap (N+W)=0$ or $T_j\subseteq N+W$. In the first case $N\cap (W\oplus T_j)=0$, contradicting maximality; hence every coordinate summand lies in $N+W$. Therefore $V=N\oplus W$. Projection onto $N$ shows $N$ is a finite sum of images of the $T_j$, each image zero or simple. Add these images successively, discarding any image already in the sum: simplicity makes each new intersection zero. This yields a direct-sum decomposition. Applying this argument to the displayed injection proves that the left regular module is a finite sum of simple modules.

Conversely, if the regular module is a direct sum of simple modules, that sum is finite because $1$ lies in a finite subsum and generates the whole regular module. Every left ideal, being a submodule, has the preceding decomposition and splitting property. Its chain lengths are bounded by the number of simple summands, so the descending chain condition holds. The Jacobson radical annihilates every simple summand of the regular module, hence annihilates $1$ and is zero. This identifies the first two conditions.

Write the finite decomposition using pairwise nonisomorphic simple left modules:Every simple module is a quotient of $R$, by sending $1$ to any nonzero vector, so this list includes every simple isomorphism class. The Schur lemma gives ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$, and $E_i={\mathrm{End}}_R(S_i)$ is a division ring: any nonzero endomorphism has zero kernel and full image by simplicity. ConsequentlyEvery endomorphism of the left regular module is right multiplication by its value on $1$. Composition reverses multiplication of those values, so the left side is $R^{\mathrm{op}}$. Taking the opposite ring, and using transpose to identify $M_n(E{)}^{\mathrm{op}}$ with $M_n(E^{\mathrm{op}})$, gives the theorem's decomposition with $D_i=E_i^{\mathrm{op}}$. This keeps the left/right conventions explicit.

Conversely, the standard column module $D^n$ over $M_n(D)$ is simple: a nonzero coordinate of a vector can be inverted to construct a matrix sending it to any prescribed vector. The left regular module decomposes as the sum of its $n$ column ideals, each isomorphic to $D^n$. A finite product of these matrix rings is therefore a semisimple ring. This proves all implications. For uniqueness, the isomorphism classes of simple modules and their multiplicities in the regular module determine the numbers $n_i$ and the division rings $E_i^{\mathrm{op}}$. The multiplicity can also be recovered as the dimension of ${\mathrm{Hom}}_R(S_i,R)$ as a right $E_i$-module, so it does not depend on the chosen decomposition.

Now let $A=\mathbb{C}[G]$ for the finite group $G$. It has dimension $|G|$, with the elements of $G$ as a basis, so its left ideals satisfy the descending chain condition by their dimensions. The permitted assumption $J(A)=0$ makes the Artin–Wedderburn theorem applicable. Its decomposition is a complex-algebra decompositionwith each $D_i$ finite-dimensional over the central copy of $\mathbb{C}$. Each such division algebra is just $\mathbb{C}$. Indeed, for $d\in D_i$, finite dimensionality gives a nonzero polynomial relation $p(d)=0$ with complex coefficients. Factor $p$ into linear factors over $\mathbb{C}$. Their product at $d$ is zero, so the absence of zero divisors forces $d-\lambda =0$ for some scalar root $\lambda$. Thus every element is scalar.

The pairwise nonisomorphic simple modules of ${\prod }_i{M}_{n_i}(\mathbb{C})$ are the column modules ${\mathbb{C}}^{n_i}$, with all other factors acting as zero. Hence ${\dim }_{\mathbb{C}}{S}_i=n_i$. Taking complex dimensions gives

A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,

The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate ring is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A global section is therefore an element in the intersectioninside $R[t,t^{-1}]$: a Laurent polynomial with both only nonnegative and only nonpositive exponents is constant.

Let $k=R/(\pi )$. Reducing the presentation of $B$ givesAs a vector space and as a ring extension, this iswhere $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne 0$: the free basis of $B$ in part ii remains a basis after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.

On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart ring is $k[t^2,t^3]$, andin $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. HenceThe map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,is not surjective. This is global-section base-change failure in a flat projective family, despite the properness and flatness proved above.

On $D_{+}(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. ThereforeThe other chart isThere is a homomorphism $B\to R[t]$ defined by $a\mapsto \pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination ofTheir images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi \ne 0$, these images are $R$-linearly independent. Hence they are a basis, the map is injective, andIn particular $B$ is torsion-free, indeed free, as an $R$-module.

The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate rings are torsion-free over the discrete valuation ring $R$, hence flat: over a principal ideal domain a torsion-free module is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to \mathrm{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper projective space ${\mathbb{P}}_R^3$. Consequently $X$ is both proper and flat over $R$.

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

Write $s=\sqrt{1-{\rho }^2}>0$. The inverse change of variables is $y=\rho x+sz$, whose Jacobian determinant has absolute value $s$. Completing the square givesThe change of variables formula therefore transforms the joint probability density function toIt factors into two normalized standard normal distribution densities. Thus $X$ and $Z$ are independent $N(0,1)$ variables; factorization proves the requested independence, rather than simply zero covariance.

The standard Gaussian random vector $(X,Z)$ has a rotation-invariant density. In polar coordinates, $X=r\cos \theta$, $Z=r\sin \theta$, its density with the area factor is $(2\pi {)}^{-1}r{e}^{-r^2/2}$, so the angular coordinate is uniform on an interval of length $2\pi$.

Put $\alpha =\arcsin \rho \in (-\pi /2,\pi /2)$, so $s=\cos \alpha$ andIn the right half-plane, choose $-\pi /2<\theta <\pi /2$. The additional condition $Y>0$ then requires $-\alpha <\theta <\pi /2$. This wedge has angle $\pi /2+\alpha$; its boundary has probability zero. ConsequentlyAt $\rho =0$ this is $1/4$, as required by independence. The limits as $\rho \to 1$ and $\rho \to -1$ are $1/2$ and zero, respectively. The angular calculation is also the geometric basis of the Gaussian sign-correlation identity.

With the indicated order of the two solutions, define the Wronskian byWhere $y_1\ne 0$, this gives the first-order linear differential equationIntegrating produces the reduction of order formulaThis formula is local on an interval where the known solution does not vanish; an arbitrary nonzero scaling of the linearly independent solution changes the normalization of $W$.

Differentiate the Wronskian and use the differential equation for each solution:Thus $\boxed{W^{\prime }+pW=0}$, proving the Abel identity, and $W=C\exp (-\int pdx)$. For continuous coefficients on a nonsingular interval, linearly independent solutions have $C\ne 0$.

For the particular equation, write $s=x-1$. The proposed solution is $y_1=s^2$, with $y_1^{\prime }=2s$ and $y_1^{\prime \prime }=2$, so substitution gives $2s^2+2s^2-4s^2=0$. Away from the singular point $x=1$, division by $s^2$ gives $p=1/s$. The Abel identity therefore giveswith a separate constant on each interval lying on one side of the singular point. The reduction of order integral is $\int W/y_1^{2}dx=C\int s^{-5}ds=-C/(4s^4)$, soChoosing $C_2=0$, $C=-4$ givesThe nonzero Wronskian proves linear independence. Direct substitution also verifies the second solution, since $s^2(6s^{-4})+s(-2s^{-3})-4s^{-2}=0$. These solutions form a basis on either nonsingular interval; the second solution is not defined at $x=1$.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a cyclic vector for an operator algebra for both $A$ and $A^{\prime }$. They also make it a separating vector for each: if $a\Omega =0$, then $a$ vanishes on $A^{\prime }\Omega =V$, and similarly for $A^{\prime }$. Thus the Tomita operatoris well defined, antilinear, invertible and satisfies $S^2=I$. Define the modular operator and modular conjugation byWe calculate them explicitly to prove the commutant assertion.

The finite-dimensional structure of a unital star-algebra gives an orthogonal decompositionThis is the usual matrix-block structure: minimal central projections separate the simple summands, and each simple matrix algebra acts as its defining representation with a multiplicity space. Identify each vector block with an $n_j$-by-$m_j$ matrix $D_j$ with its Hilbert-Schmidt inner product. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_j{r}_j$ and the right orbit dimension $m_j{r}_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.

Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary matrix $X=aD$, the Tomita operator isIts polar factors areIndeed, $J$ is an antiunitary involution and $J{\Delta }^{1/2}(X)=D^{-1}{X}^{\ast }D$. The fact that ${\Delta }^{1/2}$ is a positive operator is transparent in a basis diagonalizing $D$: it multiplies the matrix unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.

Finally, if $L_a$ denotes left multiplication,As $a$ ranges over the full matrix block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications provesThis also establishes the required modular definitions intrinsically, independent of the chosen matrix coordinates.

Write $M=\lambda (\Gamma {)}^{\prime \prime }$, the group von Neumann algebra, and define right translations by $\rho (g){\delta }_h={\delta }_{h{g}^{-1}}$. Left and right translations commute, so every $T\in M$ commutes with every $\rho (g)$. A Von Neumann factor means that the center of $M$ is exactly $\mathbb{C}I$.

Suppose every nonidentity conjugacy class is infinite and $T$ is central. Put $\eta =T{\delta }_e$. Since $T$ commutes with both regular actions and $\lambda (g)\rho (g){\delta }_e={\delta }_e$,The action on a basis vector is ${\delta }_h\mapsto {\delta }_{gh{g}^{-1}}$, so the coefficients of $\eta \in {\ell }^2(\Gamma )$ are constant on conjugacy classes. A nonzero constant on an infinite class would have infinite squared sum. Thus all coefficients away from the identity vanish and $\eta =c{\delta }_e$.

Commutation with the right action now gives $T{\delta }_h=c{\delta }_h$ for every $h$, because those vectors form the right orbit of ${\delta }_e$. Hence $T=cI$ and $M$ is a factor. This argument also shows that ${\delta }_e$ is separating for $M$.

Conversely, if a nonidentity class $C$ is finite, thencommutes with every $\lambda (g)$, since conjugation permutes $C$. Therefore $Z\in M\cap \lambda (\Gamma {)}^{\prime }=Z(M)$. It is not scalar: $Z{\delta }_e={\sum }_{h\in C}{\delta }_h$ is nonzero and orthogonal to ${\delta }_e$. Consequently $M$ is not a factor. We have provedThe condition is vacuous for the trivial group, whose algebra $\mathbb{C}I$ is indeed a factor.

Include all fixed resource states and all measurement records in an environment, so the entire physical protocol has an linear isometry $U$ from the unknown input to a larger pure output. If every recipient receives a perfect copy, its reduced density matrix is the pure projector $|\psi ⟩⟨\psi |$. A pure marginal forces factorization: write a joint vector in a basis beginning with $|\psi ⟩$; its squared components along every orthogonal basis vector vanish because their marginal probabilities are zero. Applying this successively to all recipients giveswhere $|R⟩$ is the fixed initial resource and $|E_{\psi }⟩$ may depend on the input. This includes Alice's discarded systems and every possible outcome record.

Choose two distinct nonorthogonal pure states, for example $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, and write their overlap as $s$. Preservation of the inner product by the linear isometry requiresNormalized environment vectors have overlap of absolute value at most one. Thus $0<|s|<1$ would imply $|s|\le |s{|}^N$, impossible for $N>1$. Hence perfect transmission of a copy to every recipient is impossible for $N>1$. This proves the no-cloning theorem obstruction even though Alice keeps no copy herself.

Define $E|0⟩=|D_k^N⟩$ and $E|1⟩=|D_{k+1}^N⟩$. From the branch calculation in part (b), outcome $R_0=0,R_1=1$ leavesThus the effective input teleported was $X|\psi ⟩=\beta |0⟩+\alpha |1⟩$. A Pauli X gate flips every bit and hence maps a Dicke state with $j$ excitations to one with $N-j$ excitations. Since $N-k=k+1$, applying $X$ at every receiving party interchanges the two logical basis states and restores $E|\psi ⟩$. ThereforeIn particular Bob's uncorrected reduced density matrix is $X\rho X$, and his correction returns it to the matrix in part (c). The remaining branches can likewise be corrected: $Z^{\otimes N}$ implements logical $Z$ up to the common global phase $(-1{)}^k$, while $X^{\otimes N}$ implements logical $X$. Thus the outcome dependence can be removed by local Pauli gates even though the resulting copies remain imperfect.