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Hölder's inequality states that for conjugate exponents $1/p+1/q=1$, $f\in L^p$ and $g\in L^q$ imply $fg\in L^1$ andIt includes $p=1,q=\infty$ (and the reversed case), with the essential-supremum norm for $L^{\infty }$.

For $1\le p<\infty$, $L^p(E,\mathcal{E},\mu )$ consists of measurable functions with ${\int }_E|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^p{)}^{1/p}$. Thus membership is a condition on an almost-everywhere equivalence class, not on individual point values.

The equality is false. In $A=k[x,y]$, take $I=(y)$ and $J=(y-x^2)$. Both are prime ideals, since their quotient rings are isomorphic to $k[x]$, so both are radical. ButThe element $x$ belongs to the latter ideal and not the former. This also shows why taking the radical of an ideal can enlarge a sum of already radical ideals.

If a power of $a$ lies in $I\cap J$, then $a\in \sqrt{I}\cap \sqrt{J}$. Conversely, if $a^m\in I$ and $a^n\in J$, the power $a^{\max (m,n)}$ lies in both ideals. Therefore $\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$ for any commutative ring.

If $f\in \sqrt{I}$, a power $f^m$ vanishes at every point of $Z(I)$, so $f$ itself vanishes there. This gives $\sqrt{I}\subseteq I(Z(I))$.

For the reverse inclusion, take $f$ vanishing on $Z(I)$ and use the Rabinowitsch trick. In $k[x_1,\dots ,x_n,t]$ form $J=(I,1-tf)$. A common zero would have $x\in Z(I)$, whence $f(x)=0$ and $1-tf(x)=1$, a contradiction. The assumed Weak Hilbert Nullstellensatz gives $J=(1)$. Thus there is a finite expressionIf $f=0$ the conclusion is immediate. Otherwise substitute $t=f^{-1}$ in the localized polynomial ring. The last term vanishes, and clearing a sufficiently high power of $f$ gives $f^N\in I$. Hence $f\in \sqrt{I}$, proving the Strong Hilbert Nullstellensatz:

The radical of an ideal $I$ in a commutative ring $A$ is $\sqrt{I}=\{a\in A:a^m\in I\text{ for some }m\ge 1\}$. It is an ideal: if $a^m,b^n\in I$, every term of $(a+b{)}^{m+n}$ has either at least $m$ factors $a$ or at least $n$ factors $b$, so the sum belongs to $I$. Also $(ca{)}^m\in I$ for every $c\in A$, and additive inverses preserve the defining condition.

A surjective inverse-polynomial germ projection without even covering comes from the inverse of $P(w)=w^3-3w$ over $G=\mathbb{C}$. Consider $R=\mathbb{C}\setminus \{-1,1\}$, with $\pi (w)=P(w)$. Since $P^{\prime }(w)=3(w^2-1)\ne 0$ on $R$, each $w$ defines a holomorphic inverse germ of $P$ at $P(w)$. This identifies $R$ with a component of the space of germs of holomorphic functions: inverse germs continue along paths in $R$, and the holomorphic identity $P(f(z))=z$ persists under continuation. A germ with value $\pm 1$ cannot exist, because differentiating that identity would give $P^{\prime }(\pm 1)f^{\prime }(z)=1$. Since $R$ is connected, all its inverse germs belong to one complete analytic function.

The projection is surjective. A generic value has three distinct regular preimages; the critical values $2,-2$ each retain a simple regular preimage, respectively $2,-2$, even after their double critical preimages $-1,1$ are removed. Thus fibre size drops from $3$ to $1$ at these values, making an evenly covered neighbourhood impossible. This is surjective but not a regular topological covering.

If “regular” is specifically used for deck transitivity among genuine coverings, there is also a genuine nonnormal example. Remove the critical values from the base and all their preimages from the source:This is a connected three-sheeted analytic covering. A small loop about each critical value swaps the two roots meeting there. The two swaps are distinct and generate $S_3$ (connectedness also forces transitive monodromy). The stabilizer of one sheet is an order-$2$ subgroup, which is not normal in $S_3$. The classification of connected covering spaces therefore shows this covering is not regular in the deck-transitive sense. Either convention has a fully explicit example.

Take $G={\mathbb{C}}^{\ast }$ and the complete analytic function generated by a logarithm germ. Its continuation surface identifies with $\mathbb{C}$ via $w\mapsto [\log {]}_z$ with $z=e^w$ and logarithm value $w$. Thus $\pi (w)=e^w$. It is a universal cover; the deck transformations $w\mapsto w+2\pi in$ act transitively on each fibre, so it is also a regular covering map in the deck-transitive sense. It is not bijective, since every nonzero $z$ has infinitely many logarithm values. It also meets the alternative convention in which “regular” merely requires an evenly covered neighbourhood at each base point.

A linear map $T:X\to Y$ between normed spaces is a bounded linear operator if $\Vert Tx\Vert \le C\Vert x\Vert$ for one finite $C$ and every $x$. This implies $\Vert Tx-Ty\Vert \le C\Vert x-y\Vert$, hence continuity. Conversely, continuity at zero supplies $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$ for nonzero $x$. Thus boundedness and continuity are equivalent for linear maps.

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$, equivalently ${\sup }_{x\ne 0}\Vert Tx\Vert /\Vert x\Vert$. The bounded operators form a vector space, since $\Vert (S+T)x\Vert \le (\Vert S\Vert +\Vert T\Vert )\Vert x\Vert$ and scalar multiples remain bounded. Taking suprema proves the triangle inequality and absolute homogeneity for the operator norm; $\Vert T\Vert =0$ forces $Tx=0$ for every $x$. Thus $B(X,Y)$ is a normed space, whether or not $X,Y$ are complete.

Write $K=\mathbb{Q}({\mu }_m)$, $L=\mathbb{Q}({\mu }_n)$. For coprime $m,n$, Bézout's identity and powers of a primitive $mn$-th root give $KL=\mathbb{Q}({\mu }_{mn})$. The degrees of these cyclotomic fields are $\phi (m)$, $\phi (n)$ and $\phi (mn)=\phi (m)\phi (n)$. The natural restriction mapis injective and, by these orders, bijective. Under this product description the subgroups fixing $K$ and $L$ are the two complementary factors. Together they generate the whole group, so their common fixed field is $\mathbb{Q}$. By the Fundamental theorem of Galois theory, $K\cap L=\mathbb{Q}$. This also includes $m=1$ or $n=1$.

The cyclotomic polynomial is ${\Phi }_8(t)=t^4+1$. Over ${\mathbb{F}}_7$,since the middle coefficient in the product is $2-9=-7=0$. Both factors have positive degree, proving reducibility. In fact their discriminant $5$ is not a square modulo $7$, so these are the two irreducible quadratic factors.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.

Let $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The splitting field is $L=\mathbb{Q}(\alpha ,\zeta )$. A splitting field in characteristic zero is normal and separable, hence Galois. Eisenstein criterion at $2$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic polynomial. Thus $[L:\mathbb{Q}]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The subgroup fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition subgroups, so it is not normal.

Put $s=\sqrt{-14}$. Then ${\mathcal{O}}_K=\mathbb{Z}[s]$ and the field discriminant is $-56$. Reduction of $X^2+14$ modulo $2$ and $3$ givesTheir ideal norms are $2,3,3$, respectively. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so every ideal class has an integral representative of norm at most $4$. The only possible representatives are the unit ideal, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$, ${\overline{\mathfrak{p}}}_3$, and the norm-$4$ ideal ${\mathfrak{p}}_2^2=(2)$. Thus there are at most four classes.

The class of ${\mathfrak{p}}_2$ is nontrivial, since a principal ideal of norm $2$ would require $x^2+14y^2=2$. Also ${\mathfrak{p}}_3^2$ is not principal: norm $9$ forces $x+ys=\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Now $N(2+s)=18$. It is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$, but not by ${\overline{\mathfrak{p}}}_3$, as substitution of $s=-1$ modulo $3$ gives $1$. Unique ideal factorization and norms therefore give $(2+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Consequently $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$, a nontrivial class of order $2$. The class of ${\mathfrak{p}}_3$ has order $4$, and the four-class bound is attained:

Use the customary squarefree convention for the quadratic radicand $d$. For $d\equiv 2,3(\mathrm{mod}4)$, the ring of integers of a number field is $\mathbb{Z}[\sqrt{d}]$, soThe ideal $(p)$ is prime exactly when this quotient is an integral domain. A polynomial quotient of a field is an integral domain exactly when the defining polynomial is irreducible. This proves the assertion, including the possibility of ramified primes where the reduction has a repeated root.

Squarefreeness is necessary for the printed assertion as written: $d=18\equiv 2(\mathrm{mod}4)$ defines the same field as $d=2$. The prime $3$ is inert in $\mathbb{Q}(\sqrt{2})$ because $X^2-2$ is irreducible over ${\mathbb{F}}_3$, although $X^2-18=X^2$ is reducible. Thus the result requires a squarefree quadratic radicand.

The character of a representation is its trace. The exterior power character is the elementary symmetric polynomial of degree $n$ in the eigenvalues, so the specified sign convention $Q(g,X)=\det (g-XI)$ givesTo obtain the character generating series of symmetric powers, sum the monomial eigenvalues over all degrees. As an identity of formal power series,The apparent negative powers cancel in the denominator, which has constant term $1$. Equivalently, this series is the reciprocal of ${\sum }_{j=0}^d(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g)t^j$. Comparing coefficients gives ${\sum }_{j=0}^{\min (d,n)}(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g){\chi }_{S^{n-j}V}(g)=0$ for $n>0$, an explicit recurrence computing the symmetric-power characters from $Q$.