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A field extension $L/K$ is algebraic if every element of $L$ is an algebraic element over $K$. Every finite field extension is algebraic, but algebraic extensions can have infinite degree.

The algebraic conjugates of an algebraic element over a field $K$ are the polynomial roots of its algebraic minimal polynomial over $K$ in an algebraic closure.

The algebraic conjugates of an algebraic number are the polynomial roots of its algebraic minimal polynomial, equivalently its images under field embeddings of its number field into an algebraic closure.

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

An element $x$ of an extension $L/K$ is an algebraic element over $K$ when some nonzero polynomial in $K[t]$ vanishes at $x$. Let its monic minimal polynomial have degree $d$. Polynomial division shows $K[x]$ is spanned by $1,x,\dots ,x^{d-1}$, and minimality makes these vectors independent. The minimal polynomial is irreducible; for any nonzero $q(x)$, Bezout's identity between $q$ and that polynomial gives a polynomial representative for $q(x{)}^{-1}$. Thus $K[x]$ is a field, equals $K(x)$, and has degree $d$.

Conversely, if $K(x)$ has finite vector-space dimension $d$, then $1,x,\dots ,x^d$ are linearly dependent over $K$, giving a nonzero annihilating polynomial. HenceAn algebraic extension is one in which every element is algebraic. Suppose $M/L$ and $L/K$ are algebraic, and take $\alpha \in M$. A polynomial relation for $\alpha$ over $L$ uses only finitely many coefficients $a_1,\dots ,a_r$. Each is algebraic over $K$, so adjoining them successively gives a finite extension $E=K(a_1,\dots ,a_r)$, by the proved criterion and the tower law. The same relation makes $\alpha$ algebraic over $E$, so $E(\alpha )/E$ is finite. Therefore $E(\alpha )/K$ is finite, and its subspace $K(\alpha )$ is finite-dimensional. The criterion makes $\alpha$ algebraic over $K$. Since $\alpha$ was arbitrary, $M/K$ is algebraic.

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.

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.

An element $\alpha$ is algebraic over $\mathbb{Q}$ when it is a root of a nonzero polynomial in $\mathbb{Q}[X]$. Clearing denominators givesFor $\alpha \ne 0$, take $\beta =-(a_1+a_2\alpha +\cdots +a_d{\alpha }^{d-1})\in \mathbb{Z}[\alpha ]$; then $\alpha \beta =a_0\ne 0$.

Since $K=\mathrm{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne 0$. Applying the first result to $b$ gives $b\beta =m\in \mathbb{Z}\setminus \{0\}$, whence $x=a\beta /m$. Changing signs makes $m>0$.

Choose a $\mathbb{Z}$-basis $e_1,\dots ,e_d$ of ${\mathcal{O}}_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. ThenThus $R$ is a free abelian group of rank $d=[K:\mathbb{Q}]$ and has finite index in ${\mathcal{O}}_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m{\mathcal{O}}_K\subseteq R$ is an ideal of $R$.

Put $h=({\mathcal{O}}_K:R)$ and $I=m{\mathcal{O}}_K$. ThenThe assumed multiplicativity of indices givesso $h=1$. Therefore $\boxed{R={\mathcal{O}}_K}$.

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

For an algebraic element $\alpha$ over a field $F$ of characteristic $p>0$, suppose first that $\alpha$ is a separable algebraic element. Its minimal polynomial over $E=F({\alpha }^p)$ divides its original separable polynomial and also $T^p-{\alpha }^p=(T-\alpha {)}^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha \in E$. Conversely, if the irreducible minimal polynomial $m$ is inseparable, its formal derivative vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\mathrm{deg}(m)/p$. Thus $[F({\alpha }^p):F]\le \mathrm{deg}(m)/p<[F(\alpha ):F]$, proving the converse.