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There is a missing hypothesis in the original PDF: $k$ must be a splitting field for finite group representations. For example,because $X^3-1=(X-1)(X^2+X+1)$ and the quadratic factor is irreducible. This algebra has two simple modules, whereas $C_3$ has three conjugacy classes, all $2$-regular. Thus the printed assertion for an arbitrary field is false.

Under the intended splitting hypothesis, fix compatible lifts defining Brauer characters. Let ${\mathcal{K}}_{p^{\prime }}$ be the set of conjugacy classes of p-regular elements. Character additivity and the tensor-product formula define the unital algebra homomorphismWe use the Brauer character basis theorem: over a splitting field the irreducible Brauer characters form a complex basis of the class functions on the p-regular conjugacy classes. Therefore $\Phi$ maps the basis $1\otimes [S]$ bijectively to a basis and is an algebra isomorphism. Comparing dimensions givesIsomorphism classes are understood in the count.

Proceed by induction on $d=\mathrm{deg}f$. A nonconstant polynomial over a field has an irreducible factor $h$. Part (a) gives an extension $k_1/k$ containing a root ${\alpha }_1$ of $h$, hence of $f$. The factor theorem giveswhere $f_1$ is monic of degree $d-1$. Apply the induction hypothesis over $k_1$ and compose the resulting field extensions. The final field $F$ contains roots ${\alpha }_1,\dots ,{\alpha }_d$ andEquivalently, every polynomial has a splitting field.

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.

Set $n=0$ in part (ii). The constants give an injection $k\hookrightarrow H^0(Y,{\mathcal{O}}_Y)$ because the projective hypersurface $Y$ is nonempty, while part (ii) gives surjectivity. ThusA disconnected scheme has a nontrivial idempotent global regular function, equal to zero and one on its two clopen pieces. A field has no such idempotent, so the connectedness from global regular functions criterion gives

Every field is an integral domain, so if $R/(p)$ is a field then it is an integral domain.

The Weierstrass equation of an elliptic curve over a field $K$ has the general formwith nonzero elliptic-curve discriminant. Admissible change of Weierstrass coordinates relates two such equations defining isomorphic pointed curves over $K$:where $u\in K^{\times }$ and $r,s,t\in K$.

If $\mathrm{char}K\ne 2,3$, completing the square and translating $x$ simplify every equation. Short Weierstrass form isTwo short equations are $K$-isomorphic precisely when, for some $u\in K^{\times }$,the isomorphism from the first curve to the second is $(x,y)\mapsto (u^{2}x,u^{3}y)$.

The first isomorphism theorem for rings states that a ring homomorphism $\varphi :R\to S$ induces an isomorphismBecause $J$ is an ideal, sums and products of elements of $R+J$ remain in $R+J$, so it is a subring. The surjective homomorphismhas kernel $R\cap J$. The theorem givesEvaluation at $-2$ givesso this quotient is a field of characteristic zero. The other quotient isIt has characteristic $3$ but is not a field, since the nonzero class of $X-1$ is nilpotent.

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.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

True. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

The Lagrange root bound over a field says that a nonzero polynomial of degree $r$ over a field has at most $r$ roots. In particular, a polynomial modular congruence of degree $r$ modulo a prime number has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive integer divisor $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative subgroup $H$ of ${\mathbb{F}}_p^{\times }$. By the fact that every finite multiplicative subgroup of a field is cyclic, $H=⟨g⟩$ is a cyclic group of order $d$. The solutions in $H$ of $x^e=1$ areso there are at least $e$ of them. The Lagrange root bound over a field gives at most $e$ roots in all of ${\mathbb{F}}_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the Chinese remainder theorem for unit groups, a base $b$ is a Fermat-pseudoprime base precisely when its two components satisfyThe power roots in a finite field andshow that each component has ten choices. Therefore there areFermat-pseudoprime bases.

To impose the strong pseudoprime condition, write $n-1=2^{s}m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

Choose $a\in K\setminus F$. The multiplication homomorphismhasThis element is nonzero. Since $1,a$ are linearly independent over $F$, there is an $F$-linear functional $\lambda :K\to F$ with $\lambda (1)=0$ and $\lambda (a)=1$. Applying $\lambda \otimes {\mathrm{id}}_K$ sends the displayed tensor to $1$.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

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.