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There is a sequence of real numbers for which every finite initial segment consists of algebraically independent elements over the rationals. Given an independent initial segment, specialize each nonzero integer polynomial in one extra variable at those values. At least one coefficient remains nonzero, so each resulting polynomial has finitely many roots. There are countably many such polynomials by countability of integer polynomial rings. Choosing the next real outside the countable union of their roots continues the construction. Clearing denominators makes independence over the integers equivalent to independence over the rationals.

A Cartier divisor on an integral projective variety is big if and only if some multiple has a complete linear system of a divisor giving a rational map birational onto its image. Kodaira's lemma embeds a very ample subsystem in a suitable multiple. Conversely, $n$ algebraically independent elements among the section ratios give $\left(\genfrac{}{}{0ex}{}{q+n}{n}\right)$ independent degree-$q$ section monomials, where $n=\dim X$. Smoothness is not necessary.

We use these countability facts: the integers are countable; a finite Cartesian product of countable sets is countable; a countable union of countable sets is countable; and every subset of a countable set is countable. Finite products can be enumerated by listing index tuples in increasing maximum coordinate, and countable unions by diagonal enumeration of pairs of indices.

Fix $d\ge 1$. For each degree bound $D$, there are finitely many monomials $X_1^{{\alpha }_1}\cdots X_d^{{\alpha }_d}$ with ${\alpha }_1+\cdots +{\alpha }_d\le D$. An integer polynomial of total degree at most $D$ is specified by an integer coefficient for each of them, so these polynomials form a finite Cartesian product of copies of $\mathbb{Z}$ and are countable. Every polynomial has some finite degree bound. Taking the union over $D=0,1,2,\dots$ proves the countability of integer polynomial rings:It is infinite because it contains all integer constant polynomials.

For $d=1$, each nonzero polynomial has only finitely many real roots: repeated application of the factor theorem bounds their number by its degree. The real algebraic numbers are the union of these finite root sets over the countably many nonzero integer polynomials. Thus they are countable. If the real transcendental numbers were countable too, their union with the algebraic numbers would make $\mathbb{R}$ countable, contrary to the allowed uncountability assumption. Hence there are uncountably many real transcendental numbers.

To construct an algebraically independent real sequence, choose $x_1$ outside the algebraic numbers. Suppose $x_1,\dots ,x_d$ already satisfy the required nonvanishing condition. For each nonzero $P\in \mathbb{Z}[X_1,\dots ,X_d,T]$, writeAt least one coefficient polynomial $P_j$ is nonzero, and the inductive hypothesis makes its value at $(x_1,\dots ,x_d)$ nonzero. Consequently $P(x_1,\dots ,x_d,T)$ is a nonzero one-variable polynomial and has only finitely many real roots. There are countably many choices of $P$, so the union of all these forbidden roots is countable. Choose $x_{d+1}$ outside it, which is possible because $\mathbb{R}$ is uncountable. This extends the nonvanishing condition to $d+1$. Recursion therefore givesIn particular the chosen numbers are distinct, since a relation $X_i-X_j$ is forbidden. Clearing rational denominators shows that every finite initial segment consists of algebraically independent elements over $\mathbb{Q}$.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

The actions on the ordered pair $(x,y)$ satisfyThus every element of $G$ has the form ${\sigma }^i$ or $\tau {\sigma }^i$ for $0\le i<4$, so $|G|\le 8$. Their actions areBecause $x,y$ are algebraically independent elements over $\mathbb{Q}$, these eight substitutions are distinct. Henceand the presentation identifies $G$ with the dihedral group $D_8$.

PutBoth generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb{Q}(u,v)$,The elements $x^2$ and $y^2$ are the two roots ofso $[F(x^2,y^2):F]\le 2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the tower law givesPart (b) gives $8=|G|\le [K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law givesAll inequalities are equalities and $[K^G:F]=1$. Thereforethe dihedral fixed field of a two-variable rational function field.