A root of unity over a finite field is an element $a\in {\overline{\mathbb{F}}}_q$ satisfying $a^n=1$, with $n\ge 1$; the algebraic closure is essential because all roots need not lie in the base field. Write $p=\mathrm{char}{\mathbb{F}}_q$ and $n=p^{a}m$, where $p\nmid m$. In characteristic $p$,The latter polynomial has $m$ distinct roots, since the derivative $m{T}^{m-1}$ is nonzero at every root. Those roots are closed under multiplication and inversion and contain $1$, so $E(n,q)$ is a finite multiplicative subgroup of a field. The proof in the preceding solution makes it a cyclic group. In particularThis also explains why the coprimality qualification is needed.

The multiplicative group of a finite field ${\mathbb{F}}_{q^s}^{\times }$ is cyclic of order $q^s-1$. It contains the entire group of $m$th roots precisely when $m\mid q^s-1$. For necessity, a generator of $E(n,q)$ has order $m$ and Lagrange's theorem applies. For sufficiency, a cyclic group whose order is divisible by $m$ contains $m$ roots of $T^m-1$; since there are exactly $m$ roots in the algebraic closure, these are all of them. Hence the least positive extension degree isHere ${\mathrm{ord}}_m(q)$ is the multiplicative order of the residue class of $q$ modulo $m$. If $n$ is coprime to $q$, replace $m$ by $n$. Existence of this order follows because $q$ is a unit modulo $m$.

A primitive $n$th root of unity over a finite field has multiplicative order exactly $n$. If $p\mid n$, no such element exists, since all roots have order dividing $m<n$. If $\gcd (n,q)=1$, choose a generator $\zeta$ of $E(n,q)$. The element ${\zeta }^j$ has order $n/\gcd (n,j)$, so it is primitive precisely when $\gcd (n,j)=1$. Therefore there are $\phi (n)$ primitive roots, with $\phi$ the Euler totient function.

If $\omega$ is primitive of order $n$, then $\omega \in {\mathbb{F}}_{q^{\ell }}$ exactly when ${\omega }^{q^{\ell }}=\omega$, equivalently $n\mid q^{\ell }-1$. The implication from that equality to membership follows because the roots of $T^{q^{\ell }}-T$ form ${\mathbb{F}}_{q^{\ell }}$, as proved in the preceding solution. Consequentlyfor $n>1$, and the degree is one for $n=1$. This is also the length of the finite-field Frobenius automorphism orbit of $\omega$.

Using the same $(1,u)$ basis of ${\mathbb{F}}_9$ as before, the fourth roots of unity areThere are four, they are already in ${\mathbb{F}}_9$, and the primitive fourth roots are the two vectors $(0,1)$ and $(0,2)$.

A finite field is a field with finitely many elements. Its characteristic is the least positive integer $p$ for which $p\cdot 1=0$; such an integer exists because the additive group is finite. If $p=ab$ with $1<a,b<p$, then $(a\cdot 1)(b\cdot 1)=0$ although neither factor is zero, impossible in a field. Hence $p$ is prime, and the multiples of $1$ form its prime field ${\mathbb{F}}_p$. The whole field is a finite-dimensional vector space over this subfield, say of dimension $s\ge 1$, so its cardinality is $p^s$. Thus $q=p^s$ and its characteristic is $p$.

For existence, let $K$ be a splitting field over ${\mathbb{F}}_p$ of $P(T)=T^{p^s}-T$. Since $P^{\prime }(T)=-1$, it has exactly $p^s$ distinct roots. Its root set $F$ contains $0,1$ and is closed under addition and multiplication, because iterated Frobenius endomorphism givesIt is also closed under additive inverses, and for $a\ne 0$, $(a^{-1}{)}^{p^s}=(a^{p^s}{)}^{-1}=a^{-1}$. Thus $F$ is a field with exactly $p^s$ elements. Since it contains every root, it is the entire splitting field. This proves existence.

For uniqueness, let $L$ be any field with $p^s$ elements. Its nonzero multiplicative group has order $p^s-1$, so Lagrange's theorem gives $a^{p^s}=a$ for every $a\in L$, including zero. Hence $L$ is a splitting field of the same polynomial. Here is the embedding argument giving uniqueness, rather than merely invoking it by name. Extend the identity on ${\mathbb{F}}_p$ from one root at a time. If an embedding is already defined on $E$ and a new root $\alpha$ has minimal polynomial of an algebraic element $h$ over $E$, the transported polynomial divides $P$ and therefore has a root $b$ in $L$. Evaluation at $b$ induces an embedding $E[T]/(h)\to L$, extending the existing embedding to $E(\alpha )$. Repeating for the finitely many roots embeds the constructed field $F$ in $L$. They have the same finite cardinality, so this embedding is an isomorphism. Thus the field of order $p^s$ is unique up to isomorphism.

We prove cyclicity of the multiplicative group, also establishing the more general finite multiplicative subgroup of a field is cyclic result. Let $H$ be a finite subgroup of a field's multiplicative group, and let $e$ be the least common multiple of all element orders. For every prime $\ell$ dividing $e$, choose an element whose order has the maximal $\ell$-power ${\ell }^{a_{\ell }}$; raising it to the factor of its order coprime to $\ell$ gives an element of order ${\ell }^{a_{\ell }}$. Since the group is abelian, the product of these elements has order ${\prod }_{\ell }{\ell }^{a_{\ell }}=e$: for commuting elements of coprime orders, an equation $(ab{)}^t=1$ forces $a^t=b^{-t}$ into the intersection of their cyclic subgroups, which is trivial, so both orders divide $t$. Every member of $H$ is a root of $T^e-1$. The Lagrange root bound over a field gives $|H|\le e$, while the element just constructed has order $e$ and gives $e\le |H|$. Thus equality holds and that element generates $H$. In particular

For the field with nine elements, use ${\mathbb{F}}_3[u]/(u^2+1)$. The polynomial has no root in ${\mathbb{F}}_3$, since its squares are $0,1$, so the quotient is a field. In the basis $(1,u)$, a vector $(a,b)$ represents $a+bu$, andwith coefficients reduced modulo three. Put $\beta =1+u$. Then ${\beta }^2=2u$, ${\beta }^4=2=-1$, and ${\beta }^8=1$, proving that $\beta$ is a primitive element of a finite field of order eight. Its complete power/vector table isFor completeness, the full addition and multiplication tables in these labels are displayed below. The subscripts are element labels; they are not the scalar residues of ${\mathbb{F}}_3$.

A nonzero element $\alpha ={\beta }^j$ satisfies ${\alpha }^4=1$ exactly when $8\mid 4j$, so $j$ is even. Thus, with $e=1$,

For a linear code $X\le {\mathbb{F}}^n$, its dual code isThe standard bilinear form is nondegenerate. The map ${\mathbb{F}}^n\to X^{\ast }$ obtained by restricting the dot-product functional to $X$ is surjective: extend a functional from a basis of $X$ to a basis of ${\mathbb{F}}^n$, then use the nondegenerate dot product to represent it. Its kernel is $X^{\perp }$. The rank-nullity theorem therefore gives

The first proposed description of the dual code is false without self-orthogonality. At length three take $X=⟨(1,0,0)⟩$. It has the specified dimension, butwhereas $⟨X,(1,1,1)⟩$ contains $(1,0,0)$ and $(1,1,1)$, neither of which is in $X^{\perp }$. The natural corrected statement is the odd-length binary self-orthogonal dual extension: if also $X\subseteq X^{\perp }$, every word of $X$ has even Hamming weight, the all-ones word lies in $X^{\perp }$ but not in $X$, and dimensions then give $X^{\perp }=X\oplus ⟨1⟩$.

For a binary self-dual code, $x\cdot x=0$ for every $x\in X$. In ${\mathbb{F}}_2$, this equals ${\sum }_i{x}_i$, so every codeword has even Hamming weight. Thus $1\cdot x=0$ for all $x$, giving $1\in X^{\perp }=X$. The dimension identity yields $2\dim X=n$, so $n$ is even. Alternatively $1\in X$ and $1\cdot 1=0$ itself gives evenness. Hence the second assertion is true.

For every $n=2m$, useThis linear code has dimension $m$, and $(u,u)\cdot (v,v)=2(u\cdot v)=0$. Therefore $X\subseteq X^{\perp }$; both have dimension $m$, so equality holds. This constructs a binary self-dual code at every even length. Conversely the dimension identity already proves every self-dual length-$n$ linear code has dimension $n/2$.

For a nonbinary example, over ${\mathbb{F}}_5$ take $X=⟨(1,2)⟩$. Its generator has self-product $1+2^2=0$ in ${\mathbb{F}}_5$, so $X\subseteq X^{\perp }$. Both are one-dimensional, proving

For a memoryless quantum channel $\mathcal{N}$, successive uses are described by ${\mathcal{N}}^{\otimes n}$. Its entanglement-assisted classical capacity is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared quantum entanglement, independent of the message. Only the channel is used to transmit the message.

For input state ${\rho }_A$, choose a purification $|{\psi }_{\rho }{⟩}_{RA}$ and set ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The capacity formula isPurifications differ only by an isometry on the reference, so this expression depends only on the input density matrix. The quantity optimized is quantum mutual information.

The binary quantum erasure channel sends a qubit to a qubit subspace plus an orthogonal erasure flag:On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho )$. The entropy of an orthogonal quantum mixture givesThe reference-output state has two orthogonal branches:The first branch is pure; the second has entropy $S({\rho }_R)=s$. Thus $S({\omega }_{RB})=h_2(p)+ps$ andA qubit has entropy at most one bit, with equality at $\rho =I_2/2$. We concludeThe endpoint capacities are two and zero. As an operational check, superdense coding encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased qubit yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p){\log }_{2}4=2(1-p)$, agreeing with the entropy calculation.

The channel capacity is the supremum of reliable communication rates: a rate $R$ bits per use is achievable if block encoders and decoders using $n$ channel uses can transmit $M_n$ messages with error probability tending to zero and $\mathrm{liminf}{n}^{-1}{\log }_2{M}_n\ge R$.

A discrete memoryless channel has fixed transition probabilities $W(y|x)$ on finite input and output alphabets, andIts single-letter capacity formula iswith logarithms to base two. The Shannon second coding theorem identifies this maximum with the operational reliable rate.

For the binary erasure channel, use output symbols $0,1,e$, where $e$ is a recognizable erasure flag. Given input $x$, the output is $x$ with probability $1-p$ and $e$ with probability $p$. If $\mathrm{Pr}(X=1)=a$, the output probabilities are $(1-p)(1-a),(1-p)a,p$. Thus, writing $h_2$ for binary entropy,and $I(X;Y)=(1-p)h_2(a)$. Its maximum is attained by the uniform input $a=1/2$, givingIn particular the noiseless and complete-erasure endpoints have capacities one and zero. The erasure flag reveals which positions were lost; it does not reveal the erased input values.

For a classical error-correcting code of length $n$, alphabet size $q$, minimum Hamming distance $d$ and $M$ codewords, the Singleton bound isFor a linear $[n,k,d{]}_q$ code this is $n-k\ge d-1$. Deleting any $d-1$ coordinates is injective on the codewords, since two words agreeing on the remaining coordinates would differ in fewer than $d$ positions; this also explains the classical bound.

For an n-qubit quantum code carrying $k>0$ logical qubits and having quantum code distance $d$, the quantum Singleton bound isMore generally replace $k$ by ${\log }_{2}K$ for code dimension $K>1$. The proof applies to degenerate codes too.

Let $r=d-1$. Purify the maximally mixed encoded input by a reference $R$ of dimension $K$:Distance $d$ makes erasure of any subset of at most $r$ qubits correctable. The allowed quantum erasure correction condition says that a correctable erased set $A$ is decoupled from $R$:This is the matrix-element correction condition in another form: all operators on $A$ have constant diagonal matrix elements and zero off-diagonal matrix elements in the logical basis, so tracing the other physical qubits in $|\Omega ⟩⟨\Omega |$ yields the displayed product.

First justify the sizes needed for the partition. If $2r>n$, partition all physical qubits into $A,B$, both of size at most $r$. Both sets are correctable. Purity of $RAB$ and the Schmidt decomposition giveAdding forces $k=0$, a contradiction. Thus $2r\le n$, without assuming the bound we aim to prove.

Choose disjoint sets $A,B$ of size $r$, and let $C$ contain the remaining $n-2r$ qubits. The joint state $RABC$ is pure, so decoupling and Subadditivity of Von Neumann entropy giveAdding and cancelling gives $k\le S(C)$. The Von Neumann entropy of $D$-dimensional states is at most ${\log }_{2}D$: for positive eigenvalues $p_i$, concavity of the logarithm gives ${\sum }_i{p}_i{\log }_2(1/p_i)\le {\log }_2{\sum }_i{p}_i/p_i\le {\log }_{2}D$. Here $D=2^{n-2r}$, so $k\le n-2r$, proving the quantum bound. The code is assumed to encode information ($K>1$); a one-dimensional code requires a separate distance convention, since simply re-preparing a known state makes its erasure-correction condition vacuous.

The Pauli group includes scalar phases $\mathcal{Z}=\{\pm I,\pm iI\}$. For a nonzero stabilizer code, its Abelian subgroup must satisfy $-I\notin S$. Otherwise a common positive eigenvector would satisfy $-|\psi ⟩=|\psi ⟩$ and vanish; for example $S=\{I,-I\}$ is Abelian but yields no nonzero stabilizer code. We impose this standard admissibility condition.

Every element of such a stabilizer group is Hermitian and squares to $I$: an imaginary-phase Pauli with square $-I$ is excluded. The associated stabilizer code and its orthogonal projector areThe group multiplication proves $P^2=P$, Hermiticity gives $P^{†}=P$, and its range is exactly the common positive eigenspace. If $S$ has $r$ independent generators, $|S|=2^r$. Every nonidentity element has trace zero, while $\mathrm{tr}I=2^n$, soThus it encodes $k=n-r$ logical qubits.

For the symplectic representation of a stabilizer code, label $i^t{X}^a{Z}^b$ by $(a,b)\in {\mathbb{F}}_2^{2n}$, ignoring scalar phase. Products add labels, andThis is the nondegenerate alternating form $\omega ((a,b),(a^{\prime },b^{\prime }))=a\cdot b^{\prime }+b\cdot a^{\prime }$ on the binary symplectic space of Pauli labels. The image $L$ of $S$ is an r-dimensional isotropic subspace of a binary symplectic space, since the generators commute and no nontrivial scalar lies in $S$. Consequently $L\subseteq L^{\perp }$ and $r\le n$. A binary generator matrix $(A\mid B)$ satisfies $A{B}^T+B{A}^T=0$. Conversely an isotropic subspace can be lifted by choosing commuting Hermitian generators with independent labels; their products cannot equal $-I$, and the orthogonal projector formula supplies the stabilizer code. The labels of the centralizer of a stabilizer group are $L^{\perp }$, with nontrivial logical actions represented by $L^{\perp }\setminus L$.

For an example, the three-qubit bit-flip repetition code has $S=⟨Z_1{Z}_2,Z_2{Z}_3⟩$ and ${\mathcal{C}}_S=\mathrm{span}\{|000⟩,|111⟩\}$. Its two binary rows have zero X part and Z parts $110,011$. The errors $I,X_1,X_2,X_3$ have respective error syndromes $00,10,11,01$, so all single bit flips can be corrected. This is not a stabilizer code correcting every single-qubit error: $Z_1$ acts as a nontrivial logical phase error.

Let $C(S)$ denote the Pauli centralizer. The exact Pauli error criterion for a stabilizer code isEquivalently, no difference of error labels lies in $L^{\perp }\setminus L$. The scalar phases in $\mathcal{Z}S$ are essential: an error product $-I$ is harmless even though it is not in $S$.

We prove the criterion. For $T=E_a^{†}{E}_b$ outside $C(S)$, some $s\in S$ anticommutes with $T$. Since $sP=Ps=P$, we have $PTP=PsTsP=-PTP$, hence $PTP=0$. If $T=\lambda s\in \mathcal{Z}S$, then $PTP=\lambda P$. If $T\in C(S)\setminus \mathcal{Z}S$, it restricts to a unitary operator on the stabilizer code, butno $sT$ is scalar. A scalar restriction would have scalar coefficient zero by this trace equation, contradicting unitarity. Thus this third case is not scalar.

To justify why scalar compressions are exactly what correction requires, suppose one recovery corrects every $E_a$. Dilate that recovery to an isometry $W$ including its auxiliary system. Correct recovery of every pure stabilizer code state gives $W{E}_a|\psi ⟩=|\psi ⟩\otimes |{\eta }_a⟩$. The auxiliary state is independent of $\psi$: linearity on basis states and their superpositions forces their auxiliary vectors to coincide. Preservation of inner products then giveswhich is the Knill--Laflamme condition $P{E}_a^{†}{E}_{b}P=c_{ab}P$. The three cases above prove necessity of the boxed criterion.

For sufficiency, put errors in the same class when $E_a^{†}{E}_b\in \mathcal{Z}S$. They act identically on the stabilizer code up to scalar phase, so give the same error-image subspace. Under the criterion, different classes give mutually orthogonal subspaces $E_a{\mathcal{C}}_S$, because their cross-compressions vanish. Measure which of these subspaces is occupied and apply the inverse of a representative error. This returns every stabilizer code state unchanged; extend the recovery on the orthogonal complement by preparing any fixed stabilizer code state. Linearity also corrects any valid noise channel whose Kraus operators are linear combinations of the specified errors: within each syndrome all amplitudes multiply the same logical state, and trace preservation normalizes the recovered state. This proves sufficiency, including degenerate error classes.

For active-route per-flow rates $x_r>0$, feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$ for every resource. The defining condition for a proportionally fair allocation is that every feasible alternative $\tilde{x}$ satisfiesThis is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible set. Its gradient is $(n_r/x_r)$, so the displayed inequality is precisely the first-order condition for this concave function. Strict concavity gives unique active-flow rates. Routes with $n_r=0$ have no utility term or actual per-flow rate to determine; set their aggregate service to zero. This is the inactive routes in proportional fairness convention.

Abbreviate the route populations as $n_1,n_2,n_3,n_4$ in cyclic order and let $z_r=n_r{x}_r$ denote aggregate route service. The resource constraints areEquivalently, $\max (z_1,z_3)+\max (z_2,z_4)\le 1$. Fix these two maxima as $u,v$. Because utility is increasing in each active allocation, every active odd route attains $u$ and every active even route attains $v$. Dropping terms constant in the populations, the objective becomesIf both groups are nonempty, the maximum has $u+v=1$ and derivative $P/u-Q/(1-u)=0$, giving $u=P/M,v=Q/M$. If one group is empty, every route in the other group gets aggregate service one. For $M>0$, all cases are summarized, on active routes, byAt $M=0$ there are no flows and all aggregate services are zero. In particular $n_1{x}_1=(n_1+n_3)/M$ whenever $n_1>0$, as printed in the PDF. The per-flow rate is not generally $1/M$: opposite routes can use the same resource capacities in parallel.

An exponential document size with parameter ${\mu }_r$ has completion hazard ${\mu }_r{x}_r$ when a flow is served at rate $x_r$. Summing over its $n_r$ flows gives the transition ratesThus an active odd route departs at rate ${\mu }_{r}P/M$, and an active even route at rate ${\mu }_{r}Q/M$. An empty route has departure rate zero even if its opposite route is populated.

Put ${\alpha }_r={\nu }_r/{\mu }_r$ andFor an active odd route and an active even route respectively,Hence aggregate service is the neighboring-state ratio of this balance function of a flow-level network. Consider weights $w(n)=\Phi (n){\prod }_r{\alpha }_r^{n_r}$. For every birth edge,These are detailed balance equations. The total departure rate is bounded by ${\sum }_r{\mu }_r$ and total arrival rate by ${\sum }_r{\nu }_r$, so the process is nonexplosive. Provided the normalizing sum is finite, its stationary distribution is thereforeThe large parentheses in the original PDF represent a binomial coefficient, not a ratio. The verified adjacent-state ratios explain why that distinction matters.

The required stability condition can be established directly from the normalizing sum. Set $a={\alpha }_1,c={\alpha }_3,b={\alpha }_2,d={\alpha }_4$, and $h_p(a,c)={\sum }_{i=0}^p{a}^i{c}^{p-i}$. Grouping opposite-route counts givesIf $A_{\ast }=\max (a,c)$ and $B_{\ast }=\max (b,d)$, then $A_{\ast }^p\le h_p\le (p+1)A_{\ast }^p$, with the same bounds for the other group. Restriction to the routes attaining the maxima bounds $B$ below by ${\sum }_m(A_{\ast }+B_{\ast }{)}^m$. If $A_{\ast }+B_{\ast }<1$, the upper bound is at most ${\sum }_m(m+1{)}^2(A_{\ast }+B_{\ast }{)}^m$, which converges. ThusEquivalently every resource's offered load is strictly below its unit capacity. With positive arrival rates the irreducible chain then has the unique stationary probability distribution above. Without this condition the weights remain an invariant measure but are not normalizable; the question's probability-distribution claim implicitly needs this hypothesis.

For completeness, the normalizing constant has the four-cycle partition function expressionTo derive it when $a\ne c$ and $b\ne d$, substitute $h_p=(a^{p+1}-c^{p+1})/(a-c)$ into the double sum and use ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. The resulting expression iswhich simplifies to the displayed formula. Continuity within the stability region covers equal opposite loads and zero loads.

Take a finite set of source-sink pairs $S$, with fixed demands $b_s\ge 0$, and finitely many permitted routes $R_s$ for each pair. Every positive demand must have at least one available route. Let ${\nu }_r$ be route flow and ${\rho }_j$ link throughput. The source-sink route incidence matrix and link-route incidence matrix arefor simple routes; if a route traverses a link repeatedly, its entry counts those traversals. Thus $H\nu =b$ conserves each demand and $A\nu =\rho$ aggregates link traffic.

At a Wardrop equilibrium, every route carrying positive flow has minimum delay among routes for the same source-sink pair. Ifthere are numbers ${\tau }_s$ such that $c_r(\rho )\ge {\tau }_s$ for $r\in R_s$, with equality whenever ${\nu }_r>0$. This is a nonatomic routing game: one infinitesimal user's route change does not alter the link costs it sees.

For continuous nondecreasing link delays, define the Beckmann potentialEach term is convex because its derivative is nondecreasing. The feasible route-flow set $\mathcal{F}=\{\nu \ge 0:H\nu =b\}$ is nonempty and compact: $0\le {\nu }_r\le b_s$ on each pair's routes. Hence the continuous function $V(A\nu )$ attains a minimum. Its gradient with respect to route flow isIf $r$ carries positive flow and $r^{\prime }$ serves the same pair, moving a small amount of flow from $r$ to $r^{\prime }$ is feasible. At a minimizer its directional derivative $c_{r^{\prime }}-c_r$ is nonnegative. Thus every used route minimizes delay, proving existence of a Wardrop equilibrium.

Conversely, at such an equilibrium, any feasible competitor $\tilde{\nu }$ satisfiesIndeed, the competitor's cost sum for pair $s$ is at least ${\tau }_s{b}_s$, while the current flow's cost sum equals ${\tau }_s{b}_s$. This first-order variational inequality is sufficient for optimality of a differentiable convex function. ThereforeThe argument proves both directions of the claimed optimization equivalence, including unused routes with costs tied at the minimum.

If each $D_j$ is strictly increasing, each integral is strictly convex, hence so is $V$ as a function of link throughputs. The feasible link-throughput set $A\mathcal{F}$ is convex. Two distinct minimizing vectors would have a feasible midpoint with strictly smaller potential, so the equilibrium link-throughput vector is unique. Strictly convex function in $\rho$ need not give strictly convex function in $\nu$, because $A$ can have a nontrivial kernel.

For an explicit route-flow ambiguity, put two parallel links in each of two consecutive stages, with one unit of demand traversing both stages. Order the four routes by the first and second link choices: $(1,1),(1,2),(2,1),(2,2)$. Set all four link delays to $D_j(z)=z$. Theninduces throughput $1/2$ on each link, and every route has delay one. Every member of this family is an equilibrium, though its route-flow vector changes. This is route-flow nonuniqueness at a Wardrop equilibrium with strictly increasing delays.

For Braess paradox, use vertices $s,u,v,t$ and one unit of $s$-to-$t$ demand. The delays are $D_{su}(z)=z$, $D_{ut}(z)=1$, $D_{sv}(z)=1$, and $D_{vt}(z)=z$. Initially the only routes are $sut$ and $svt$. If the upper route carries $p$, their costs are $1+p$ and $2-p$, so equilibrium has $p=1/2$ and common delay $3/2$.