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Over the finite field ${\mathbb{F}}_{13}$,The discriminant of the quadratic is $9+24\equiv 7(\mathrm{mod}13)$. The nonzero squares are $1,3,4,9,10,12$, so the quadratic is irreducible. Its two roots lie in ${\mathbb{F}}_{13^2}$, with the Frobenius automorphism interchanging them. Hence

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.

The characteristic of a field is the least positive integer $r$ with $r{1}_K=0$, or zero if none exists. In a finite field a positive such integer exists because the sequence $0,1_K,2\cdot 1_K,\dots$ must repeat.

The prime subfield is the intersection of all subfields of $K$, equivalently the subfield generated by $1_K$. In characteristic $p$ it is the copy of ${\mathbb{F}}_p$ formed by the multiples of $1_K$. Every field automorphism preserves $1_K$ and addition, so it fixes every element of this subfield.

Finally, once the order-$n$ Frobenius automorphism below is established, all its $n$ powers are distinct. Any automorphism fixes $F$ and is determined by its value at the generator $\lambda$. That value must be a root of the degree-$n$ minimal polynomial of $\lambda$, so there are at most $n$ automorphisms. Hence

By the allowed finite multiplicative-subgroup theorem, the group $K^{\times }$ is cyclic. Let $z$ generate it. Every nonzero element of $K$ is a power of $z$, hence belongs to $F(z)$; zero belongs as well. Since the reverse inclusion is automatic,Thus a finite extension of a finite field is a simple field extension.

The finite field $K$ is a finite-dimensional vector space over its prime subfield $F$ of size $p$. A basis has some positive length $n$, and every element is uniquely specified by $n$ coefficients in $F$. There are $p$ choices per coefficient, so $\boxed{|K|=p^n}$ and $[K:F]=n$.

By the allowed cyclicity theorem, $K^{\times }$ has a generator $z$. Every nonzero element of $K$ is a power of $z$, and zero also belongs to $F(z)$. Thus $K\subseteq F(z)\subseteq K$, giving $\boxed{K=F(z)}$. This is a primitive element argument for finite fields using the multiplicative group, not a claim that the additive group is cyclic.

No field has six elements. A finite field has positive characteristic: the sequence $0,1,2\cdot 1,\dots$ must repeat. By part (a), this characteristic is a prime number $p$. Translation by $1$ partitions the field into cycles of length $p$, since returning to a starting point after $k$ steps is equivalent to $k1=0$. Thus $p$ divides six and $p=2$ or $p=3$.

The multiples of $1$ form a copy of the prime field ${\mathbb{F}}_p$. For completeness, the inverse of a nonzero residue $a$ lies among these multiples: the integer Bezout identity $ab+cp=1$ gives $(a1)(b1)=1$. The entire field is a vector space over this subfield. A maximal linearly independent set is finite and spans, so it is a basis of some positive size $d$. Unique basis coordinates give exactly $p^d$ elements. This proves finite field cardinality is a prime power here. But six is neither a power of two nor a power of three, a contradiction.

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

The given unitriangular group of degree three over F3 has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare element orders.

Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the finite field ${\mathbb{F}}_3$, the binomial theorem for commuting matrices givesEvery nonidentity element of $H$ consequently has element order $3$. In contrast, $x\in G$ has element order $9$, by part (b). An isomorphism preserves element order, soEquivalently, the exponent of a finite group is $3$ for $H$ and $9$ for $G$.

There are six functions: the four power maps $x\mapsto x^k$ for $k=1,2,3,4$, and the two constant maps $0,1$. All calculations in this part are in the finite field ${\mathbb{Z}}_5$.

Put $e=f(1)$. Multiplicativity gives $e^2=e$, so the absence of zero divisors forces $e=0$ or $e=1$. If $e=0$, then $f(x)=f(x)f(1)=0$ for every $x$, giving only the constant-zero map.

Suppose $e=1$. For nonzero $x$, the equality $f(x)f(x^{-1})=1$ shows $f(x)\ne 0$. The nonzero elements form a cyclic group generated by $2$, in the order $1,2,4,3$. Choosing $a=f(2)$ determines every value on this group by $f(2^j)=a^j$. Each $a\in \{1,2,3,4\}$ satisfies $a^4=1$, so all four choices give consistent group homomorphisms.

Finally, $z=f(0)$ satisfies $z^2=z$ and $z=zf(x)$ for every $x$. The choice $z=0$ works for all four group maps. The choice $z=1$ forces every nonzero value to be $1$, giving one extra map, the constant-one map. Thus the count is $1+4+1=6$. The four zero-preserving possibilities are exactly the four displayed power maps. This is the special case with five elements of multiplicative self-maps of a finite field.

A field is perfect when every algebraic extension is separable. In positive characteristic $p$, this is equivalent to surjectivity of the Frobenius endomorphism $x\mapsto x^p$; every finite field is perfect.

A polynomial function is the function obtained by evaluating a polynomial on a specified ring or field. Distinct polynomials can induce the same polynomial function over a finite field; for example, $x^p$ and $x$ agree on ${\mathbb{F}}_p$.

Over ${\mathbb{F}}_2$, the nonzero formal polynomial $X^2-X$ evaluates to zero at both possible inputs, zero and one. It therefore induces the same polynomial function as the zero polynomial, despite having different coefficients and polynomial degree. Over an infinite field, equality at every input forces equality of polynomials by the root bound for a polynomial. The distinction matters whenever a representation is supposed to identify the function uniquely.

Work in the finite field ${\mathbb{F}}_5=\{0,1,2,3,4\}$, using modular arithmetic modulo five. Consider the two original polynomial examplesThe distributive property gives $(x+1{)}^2=x^2+2x+1$, so all coefficients of $F$ cancel: $F$ is identically zero. In contrast, $G=x^2-3x+2=x^2+2x+2$ in this finite field, with nonzero leading coefficient. Its roots of a polynomial are exactly one and two: a product in a field is zero only if a factor is zero.

The complete polynomial evaluations, displayed as representatives modulo five, areIn particular, $G(3)=(3-1)(3-2)=2$ is a nonzero witness, which disproves identity without examining other points. At one or two, a zero value fails to reveal that $G$ is nonzero. A random variable $a$ with a uniform distribution on a finite set ${\mathbb{F}}_5$ hits those two polynomial roots with probability $2/5$.

A crucial limit appears with $H(x)=x^5-x$. Its coefficients make it a nonzero polynomial of polynomial degree five in ${\mathbb{F}}_5[x]$, yet direct polynomial evaluation gives zero at all five field elements: the values before reduction are $0,0,30,240,1020$, all divisible by five. Thus $H$ induces the same polynomial function as the zero polynomial. Any number of trials restricted to ${\mathbb{F}}_5$ will falsely accept this identity. This is compatible with the Schwartz-Zippel lemma: the ratio of polynomial degree to sampling-set size is $5/5=1$, which gives no useful guarantee.

When a polynomial's representation permits polynomial evaluation over a larger field extension, a larger sampling set can restore the useful degree ratio. Access only to the induced polynomial function on the original finite field cannot distinguish $H$ from zero.^[ref]

For $a\ne 0$ in the finite field ${\mathbb{F}}_p$, put $d=\gcd (k,p-1)$. The multiplicative group of a finite field is cyclic, and character orthogonality expresses the number of $k$th roots of $y\ne 0$ as ${\sum }_{{\chi }^d=1}\chi (y)$. The trivial character contributes $-1$, cancelling the $x=0$ term. Each of the remaining $d-1$ Gauss sums of Dirichlet characters has modulus $\sqrt{p}$, by the prime-field character Gauss sum identity. Consequently $|G_{a,p}|\le (d-1)/\sqrt{p}\le k/\sqrt{p}$.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

Quadratic-equation satisfiability over ${\mathbb{F}}_2$ asks whether a finite system of polynomial equations of degree at most two over the finite field ${\mathbb{F}}_2$ has a common solution.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

A quadratic polynomial in one variable over a field has polynomial degree two and the displayed form. Its roots of a polynomial satisfy a quadratic equation. The polynomial and its associated polynomial function are distinct objects: over a finite field, different polynomials can define the same function.

In the language naming addition, zero and every scalar multiplication over a fixed field, the theory of infinite underlying vector spaces eliminates quantifiers. In a conjunction of linear equations and inequations, an equation with a nonzero coefficient solves for the quantified variable. Otherwise the variable is excluded from only finitely many values, so infinitude supplies a witness. Disjunction and induction finish elimination. This works over finite fields with infinite dimension and over infinite fields with every nonzero dimension.

A nonzero element $a$ of a finite field is a quartic residue when $a=b^4$ for some nonzero $b$. If $q\equiv 1(\mathrm{mod}4)$, the multiplicative group of a finite field is cyclic of order divisible by four, so the quartic residues form a subgroup of index four; equivalently $a^{(q-1)/4}=1$.