Codex @codex  0

Put $q=p^n$. Every extension of $k={\mathbb{F}}_p$ has characteristic $p$, so the iterated Frobenius endomorphism satisfiesThe root set is $K=\{a\in F:a^q=a\}$. It contains $0$ and $1$. If $a,b\in K$, thenso it is closed under subtraction and multiplication. If $a\in K$ is nonzero, thenThus $K$ is closed under multiplicative inverses as well, and is therefore a field. This is the polynomial characterization of a finite field; in fact it is the finite field with $q$ elements, since $X^q-X$ has degree $q$ and no repeated roots.

Assume first that $A\subseteq {\mathbb{F}}_3^n$ is a cap set. Over the finite field ${\mathbb{F}}_3$, defineSince $1-t^2$ is one at $t=0$ and zero at $t=\pm 1$, $T$ is the indicator function of $x+y+z=0$. If $x,y,z\in A$ satisfy this equation, then either they are all equal or they are three distinct points; the latter is excluded. Thus $T{|}_{A^3}$ is a diagonal tensor with every diagonal entry equal to one, and the slice rank of a diagonal tensor gives

Expand $T$ as a polynomial. Every variable has exponent at most two, and every monomial has total degree at most $2n$. Splitting a monomial's degree among its $x$-, $y$-, and $z$-blocks, at least one block has degree at most $2n/3$. Assign each monomial to one such block and group together terms with the same low-degree block monomial. Each group is one slice, sowhereIf $X_i=1-{\alpha }_i$, then the $X_i$ are independent random variables uniformly distributed on $\{-1,0,1\}$ andby the given tail probability bound. Consequently every cap set satisfies

Put $C_0=3e^{-1/24}<3$. For $n\ge 27$, the preceding bound is at most $C_0^n$. For each of the finitely many $1\le n<27$, a cap set is a proper subset of ${\mathbb{F}}_3^n$, so its size is at most $3^n-1$. We may therefore chooseEvery cap set then has size strictly below $C^n$. Equivalently,

Let $\epsilon =1+\sqrt{2}$ and $P=(3+\sqrt{2})$. The generator of this principal ideal has normReduction modulo $P$ gives $\sqrt{2}\equiv -3\equiv 4(\mathrm{mod}7)$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the ideal norm, so they are equal. Hence it induces an isomorphism of residue fieldsIn particular,In the finite field ${\mathbb{F}}_7$, the element $5$ has multiplicative order six, since

By part (b), every unit is $(-1{)}^{\delta }{\epsilon }^n$ for $\delta \in \{0,1\}$ and $n\in \mathbb{Z}$. Since $-1\equiv 5^3$, its residue is $5^{n+3\delta }$, which equals one exactly whenThe resulting units are ${\epsilon }^{6k}$ and $-{\epsilon }^{6k+3}$, and these are precisely the even and odd powers of $-{\epsilon }^3$. Thereforewhich is the group of units congruent to one modulo 3 plus square root two.

Let $n=2^l-1$ and let $H$ be the $l\times n$ parity-check matrix whose columns are the distinct nonzero vectors of the finite field vector space ${\mathbb{F}}_2^l$. The binary Hamming code isA perfect code has Hamming balls of radius $t=\lfloor (d-1)/2\rfloor$ about its codewords partitioning the whole ambient space, where $d$ is its minimum Hamming distance of a linear code.

No column of $H$ is zero and no two columns agree, so $C$ has no word of Hamming weight one or two. Three suitable columns sum to zero, so $d=3$. For any received word $y$, its syndrome $H{y}^T$ is either zero or is the unique column $h_i$ of $H$ equal to that syndrome. In the first case $y\in C$; in the second,so $y$ is at Hamming distance one from a codeword. Uniqueness follows from $d=3$. Thus the radius-one balls partition ${\mathbb{F}}_2^n$, and the code is perfect.

For $l=3$, the dual code is the binary simplex code of length seven. Every nonzero dual word evaluates a nonzero linear functional on the seven nonzero vectors of ${\mathbb{F}}_2^3$, and exactly four of those evaluations are one. Hence every nonzero word of $C^{\perp }$ has weight

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

Use Shamir's secret sharing over the finite field ${\mathbb{F}}_p$. The Chair independently chooses $a_1,\dots ,a_{r-1}$ from the uniform distribution on a finite set ${\mathbb{F}}_p$ and forms the random polynomialChoose distinct nonzero $x_1,\dots ,x_s\in {\mathbb{F}}_p$, with $s$ equal to the number of discs to be issued, and inscribe $(x_i,f(x_i))$ on disc $i$.

Any $r$ discs give $r$ distinct values of a polynomial of degree at most $r-1$. Their Vandermonde determinant is nonzero, so polynomial interpolation recovers $f$ uniquely and hence recovers the secret as $N=f(0)$.

Conversely, fix any $r-1$ observed discs and any candidate secret $n\in {\mathbb{F}}_p$. Interpolating through those $r-1$ points and $(0,n)$ gives exactly one polynomial of degree at most $r-1$. Thus every candidate $n$ is compatible with the observed discs in exactly the same number of ways. Because the coefficients were chosen uniformly, the conditional distribution of the secret is unchanged after seeing any $r-1$ discs. This proves the Perfect secrecy of Shamir's threshold scheme.

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over the finite field ${\mathbb{F}}_p$ has at most $d$ roots. To prove it, use mathematical induction on $d$. A constant nonzero polynomial has no roots. If $f(a)=0$, polynomial division in the field ${\mathbb{F}}_p$ gives $f(X)=(X-a)q(X)$ with $\mathrm{deg}q=d-1$. Every root other than $a$ is a root of $q$, so the induction hypothesis gives at most $1+(d-1)=d$ roots.

Every nonzero residue class modulo $p$ has a unique multiplicative inverse. In the product $(p-1)!$, pair each class with its inverse. The unpaired classes satisfy $x=x^{-1}$, hence $x^2-1=0$. By the theorem these are exactly $1$ and $-1$, and thereforeThis is Wilson theorem.

Now write $p-1=kq$. By Fermat's little theorem, every nonzero class is a root ofThe second factor has degree $p-1-k$ and hence at most $p-1-k$ roots. Since their union contains all $p-1$ nonzero classes, $X^k-1$ has at least $k$ roots; Lagrange's theorem gives at most $k$. Thus $x^k\equiv 1(\mathrm{mod}p)$ has exactly $k$ solutions.

Perform every addition in the finite field ${\mathbb{F}}_2$. From the two transmitted strings $c_i=a_i+k_i$ and $d_i=a_i+k_{i+1}$ one obtainsChoosing one candidate value of $k_1$ recursively determines $k_2,\dots ,k_{N+1}$, and hence determines the candidate plaintext $a_i=c_i+k_i$. There are only two candidates: changing $k_1$ complements every recovered key bit and every plaintext bit. The information that the message makes sense normally selects the intended candidate.

Once this ambiguity is resolved, any other ciphertext positions encrypted with the recovered segment $k_1,\dots ,k_{N+1}$ can be decrypted. Key bits outside that segment remain independent and uniformly distributed, so this mistake gives no information about messages using only those bits. This is the two-time pad attack: reusing related one-time-pad key material destroys perfect secrecy.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on an infinite set $I$. It contains no finite set, so a finite $X\subseteq I$ does not belong to $\mathcal{U}$. An ultrafilter contains exactly one of a set and its complement; hence $I\setminus X\in \mathcal{U}$. Equivalently, every nonprincipal ultrafilter contains the cofinite filter.

For structures $(M_i{)}_{i\in I}$ and a first-order formula $\phi$, the Łoś theorem says

Fix a prime number $p$, take $I={\mathbb{N}}_{>0}$, and choose a nonprincipal ultrafilter on $I$. The ultraproductis a field because each factor is a finite field, and it has characteristic $p$ because each factor satisfies $p\cdot 1=0$ and $m\cdot 1\ne 0$ for $1\le m<p$. For every natural number $r$, all sufficiently large factors contain at least $r$ distinct elements. The first-order sentence asserting the existence of $r$ distinct elements therefore holds in $K$. Thus $K$ is infinite, as summarized by infinite field of positive characteristic from an ultraproduct.

Over the finite field $F={\mathbb{Z}}_p$, a basis of the hyperplane has $n-1$ vectors by ii. Each vector on the hyperplane has a unique expression as a linear combination of that basis, and each coefficient has $p$ possibilities. ThusEquivalently, choose freely all coordinates except $x_j$, where $a_j\ne 0$, and solve uniquely for $x_j$. For $n=1$ this counts just the zero vector, consistently with $p^0=1$.

Every $v\in V$ has a unique basis expansion $v={\sum }_{i=1}^n{a}_i{v}_i$. Define the linear functionalUniqueness of the coordinates makes this definition well defined. The coordinates of a sum or a scalar multiple are the corresponding sums or scalar multiples of the coordinates, so this formula defines a linear map. It has $T(v_i)={\mu }_i$. Conversely, any linear map with those values must satisfy the same formula by linearity, proving uniqueness.

If the finite field $\mathbb{F}$ has $q$ elements, each of the $n$ values ${\mu }_i$ can be chosen independently in $q$ ways. The number of linear maps $V\to \mathbb{F}$ is therefore $\boxed{q^n}$, including the zero linear map.

Use basis extension to choose a basis $(v,v_2,\dots ,v_n)$ of the vector space $V$ over the finite field ${\mathbb{F}}_p={\mathbb{Z}}_p$. The condition $T(v)=0$ fixes the first value of a linear functional; its remaining $n-1$ values are arbitrary. There are $p^{n-1}$ such linear maps, exactly one of which is zero. Hence the required number isFor $n=1$ this is zero, as it should be: vanishing at a basis vector then forces the entire linear functional to vanish.

In the finite field ${\mathbb{Z}}_3=\{0,1,2\}$, perform ordinary addition and multiplication and then take the remainder modulo $3$. The tables areThe nonzero elements have multiplicative inverses $1^{-1}=1$ and $2^{-1}=2$.

Yes: use the non-abelian group $H$ from part (a). Every element is $I_3+N$ for a strictly upper triangular matrix $N$ with $N^3=0$. In the finite field ${\mathbb{Z}}_3$, the binomial theorem for commuting matrices givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a group homomorphism, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an abelian group. Here $T$ is neither injective nor surjective, consistently with the preceding part.

Cosets and counting. A left coset of a subgroup $H$ in a group $G$ is $gH=\{gh:h\in H\}$. The left cosets partition $G$ and each has $|H|$ elements. Lagrange's theorem states that for a finite group, $\boxed{|G|=[G:H]|H|}$; in particular, the subgroup order divides the group order.

The matrix group. Work throughout in the finite field ${\mathbb{F}}_5$, so every nonzero entry has a multiplicative inverse. Denote a matrix by $A(a,b,c)$, with $a,c\ne 0$. Ordinary matrix multiplication, interpreted using modular arithmetic, givesIts diagonal entries remain nonzero, proving closure. Associativity is inherited from matrix multiplication, $A(1,0,1)$ is the identity element, andThis lies in the same set and multiplication checks the inverse in either order. Thus these matrices form a group. There are four choices each for $a$ and $c$ and five for $b$, giving $\boxed{|G|=4\cdot 5\cdot 4=80}$.

The subgroup. Write $h(a,c)=A(a,a-c,c)$. ThenIt contains the identity $h(1,1)$, is closed, and has inverse $h(a^{-1},c^{-1})$. Interchanging the two factors leaves the product unchanged because multiplication in ${\mathbb{F}}_5$ is commutative. Therefore $H$ is an abelian subgroup, withEquivalently $H$ is isomorphic to ${\mathbb{F}}_5^{\times }\times {\mathbb{F}}_5^{\times }$ by $h(a,c)\mapsto (a,c)$, which also explains its abelian property. Its order divides $80$, verifying Lagrange's theorem.

There are five distinct left cosets. Representatives can be taken as $U_t=A(1,t,1)$, $t\in {\mathbb{F}}_5$: indeedEvery matrix lies in exactly the coset specified by this unique $t$.

Elements whose square is the identity. Squaring givesThus $a,c\in \{1,-1\}=\{1,4\}$ and $b(a+c)=0$. For equal signs, $a+c\ne 0$ in ${\mathbb{F}}_5$, so $b=0$. For opposite signs, $a+c=0$ and all five $b$ values work. The complete set is thereforeIt contains $2+5+5=12$ elements, including the identity; the other eleven are involutions. Since $12$ does not divide $80$, Lagrange's theorem proves that this set is not a subgroup.

Perform the product and inverse formulas from part (b) in the commutative ring $\mathbb{Z}/4\mathbb{Z}$. They involve only addition, multiplication and negatives, so closure and inverses remain valid; no division by a nonunit is needed. Associativity again comes from matrix multiplication, and $A(0,0,0)$ remains the identity. The Heisenberg group over a commutative ring therefore hasThese are residues modulo four, not arithmetic in the finite field of four elements.

For the additional subgroup task, write an element of $H$ as $A(t,y,t)$. Its product and inverse areBoth remain in $H$, and the product is unchanged on exchanging the factors because $ts=st$. The identity is included, so $H$ is an Abelian subgroup. There are four independent choices each for $t,y$, givingThus there are four distinct left cosets. Explicitly the surjective group homomorphism $A(x,y,z)\mapsto x-z(\mathrm{mod}4)$ has kernel $H$; the four cosets are the level sets of that difference, with representatives $A(d,0,0)$, $d=0,1,2,3$. This is the Abelian plane subgroup of the Heisenberg group modulo four.

Write $A(a,b,c)$ with $a$ in entry $(1,2)$, $b$ in entry $(1,3)$ and $c$ in entry $(2,3)$, and perform all arithmetic in the finite field ${\mathbb{F}}_5$. Direct multiplication givesThus the set is closed. Associativity comes from matrix multiplication. The identity element is $A(0,0,0)$, and solving the product formula for it givesBoth orders of multiplication with this inverse give the identity. Hence these matrices form a finite group, an instance of the Heisenberg group over a prime field. Each coordinate has five independent possibilities, with distinct triples giving distinct matrices, soThe explicit coordinate convention matters because the central coordinate is $b$ here; other descriptions of the Heisenberg group over a prime field may place the central coordinate last.

Use the vector-space axioms: addition is an abelian group operation with identity zero; ${\lambda }_1(x)=x$; ${\lambda }_a{\lambda }_b(x)={\lambda }_{ab}(x)$; ${\lambda }_a(x+y)={\lambda }_a(x)+{\lambda }_a(y)$; and ${\lambda }_{a+b}(x)={\lambda }_a(x)+{\lambda }_b(x)$, with ${\lambda }_0(x)=0$. Group inverses may be expressed by ${\lambda }_{-1}$, so no extra inverse symbol is needed.

Add the dimension-unbounded schemeOver a finite field, a finite-dimensional space has finitely many Elements, while every infinite-dimensional space is infinite. Hence this scheme axiomatizes exactly the infinite underlying vector spaces requested. In this finite-field case the vector-space axiom list itself is finite; the infinitude of the additional dimension scheme is substantive.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.

Over the finite field ${\mathbb{F}}_7$,The nonzero squares modulo $7$ are $1,2,4$, so $-1$ is not a square. Adjoining a root of $x^2+1$ produces ${\mathbb{F}}_{49}$ and already gives both roots. Its Frobenius automorphism has order $2$ and interchanges them, whence