Fields Construction

Adjoining Roots

Proposition

Any ring homomorphism between two fields is injective.

Proof Suppose that \newcommand{\Z}{\mathbb{Z}}\phi: F \to K is a ring homomorphism between two fields $F$ and $K$ such that $\varphi (a)=\varphi (b)$ for some $a,b\in F$. Then, we have $\varphi (a-b)=0$, which implies that $a-b=0$ since $F$ is a field. Thus, $a=b$, and hence $\varphi$ is injective. $\square$

Lemma

Let $F$ be a field, and let $f$ be an irreducible polynomial in $F[x]$. Then the ring $K=F[x]/(f)$ is an extension field of $F$, and the residue $\bar{x}$ of $x$ is a root of $f(x)$ in $K$.

Proof Since, the principle ideal $(f)$ in $F[x]$ is maximal if and only if $f$ is irreducible, we have that $K$ is a field. The residue $\bar{x}$ of $x$ in $K$ is the equivalence class of $x$ modulo $(f)$. By definition, $\bar{x}$ is a root of $f(x)$ in $K$. $\square$

Complete Splitting

A polynomial $f(x)\in F[x]$ is said to be completely split over a field $F$ if it can be factored into linear factors in $F[x]$.

Proposition

Let $F$ be a field, and let $f(x)$ be a monic polynomial in $F[x]$ of positive degree. There exists a field extension $K$ of $F$ such that $f(x)$ splits completely in $K$.

Proof We use induction on the degree of $f(x)$. If $\mathrm{deg}(f)=1$, then $f(x)$ is linear and has a root in $F$. Thus, we can take $K=F$. Now, suppose that the statement holds for all polynomials of degree less than $n$, and let $f(x)$ be a monic polynomial of degree $n$. By the lemma, there exists a field extension $F_1$ of $F$ such that $f(x)$ has a root $\alpha$ in $F_1$. Suppose that $f(x)$ can be factored as $f(x)=g(x)f_1(x)$, where $g(x)$ is a irreducible polynomial with $g(\alpha )=0$. By the induction hypothesis, there exists a field extension $F_2$ such that $g(x)$ splits completely in $F_2$, and a field extension $F_3$ of $F_1$ such that $f_1(x)$ splits completely in $F_3$. Then, the polynomial $f(x)$ splits completely in the field extension $F_2\cup F_3$. $\square$

Theorem

Let $f$ and $g$ be polynomials with coefficients in a field $F$, with $f\ne 0$, and let $K$ be an extension field of $F$. Then the following statements hold:

1. Division with remainder of $g$ by $f$ gives the same answer, whether carried out in $F[x]$ or in $K[x]$.
2. $f$ divides $g$ in $K[x]$ if and only if $f$ divides $g$ in $F[x]$.
3. The (monic) greatest common divisor $d$ of $f$ and $g$ is the same, whether computed in $F[x]$ or in $K[x]$.
4. If $f$ and $g$ have a common root in $K$, they are not relatively prime in $F[x]$. If $f$ and $g$ are not relatively prime in $F[x]$, there exists an extension field in which they have a common root.
5. If $f$ is an irreducible element of $F[x]$ and if $f$ and $g$ have a common root in $K$, then $f$ divides $g$ in $F[x]$.

Proof We prove each statement one by one:

1. In $F[x]$, suppose we have $g=fq+r$, this is also true in $K[x]$ since $F\subset K$. Meanwhile, division with remainder is unique, so we have $g=fq+r$ in $K[x]$ as well. Thus, the statement holds. $\square$
2. Directly follows from (1). $\square$
3. Let $d$ and $d^{\prime }$ be the (monic) greatest common divisor of $f$ and $g$ in $F[x]$ and $K[x]$ respectively. Then, by definition, $d\mid d^{\prime }$. Note that there exist polynomials $a$ and $b$ in $F[x]$ such that $d=af+bg$, and $d^{\prime }$ divides both $f$ and $g$ in $K[x]$. Thus, $d^{\prime }\mid d$. So $d$ and $d^{\prime }$ are associates in $K[x]$. Since $d$ is monic, we have $d=d^{\prime }$. $\square$
4. Suppose that $f$ and $g$ are not relatively prime in $F[x]$. Then, there exists a non-zero greatest common divisor $d$ such that $d\mid f$ and $d\mid g$. By the lemma, there exists a field extension in which $d$ has a root, this will also be a root of $f$ and $g$. Conversely, if $f$ and $g$ have a common root $\alpha$ in $K$, then $x-\alpha$ is a common divisor of $f$ and $g$ in $K[x]$, so the greatest common divisor of $f$ and $g$ is not a unit in $F[x]$. Thus, $f$ and $g$ are not relatively prime in $F[x]$. $\square$
5. If $f$ is irreducible, its only monic divisors in $F[x]$ are $1$ and $f$. (4) tells us that the greatest common divisor of $f$ and g in $F[x]$ isn’t $1$. Therefore it is $f$. $\square$

Multiple Roots of a Polynomial

Theorem

Let $f$ be a polynomial with coefficients in a field $F$. An element $\alpha$ in an extension field $K$ of $F$ is a multiple (double) root, meaning that $(x-\alpha {)}^2$ divides $f$, if and only if it is a root of $f$ and also a root of $f^{\prime }$.

Proof Suppose that $\alpha$ is a root of $f$, say $f(x)=(x-\alpha )g(x)$. Then by the product rule of differentiation, we have $f^{\prime }(x)=(x-\alpha )g^{\prime }(x)+g(x),$hence $f^{\prime }(\alpha )=0$ if and only if $g(\alpha )=0$. $\square$

Corollary

Let $f(x)$ be a polynomial with coefficients in $F$. There exists a field extension $K$ of $F$ in which $f$ has a multiple root if and only if $f$ and $f^{\prime }$ are not relatively prime.

Proof If $f$ has a multiple root in $K$, then $f$ and $f^{\prime }$ have a common root in $K$, so they are not relatively prime in $K$ or $F$. Conversely, if $f$ and $f^{\prime }$ are not relatively prime, then they have a common root in some field extension $K$, hence $f$ has a multiple root there. $\square$

Proposition

Let $f$ be an irreducible polynomial in $F[x]$, where $F$ is a field. Then $f$ has no multiple root in any field extension of $F$ unless the derivative $f^{\prime }$ is the zero polynomial.

Proof We shall show that $f$ and $f^{\prime }$ are relatively prime unless $f^{\prime }$ is the zero polynomial. Since $f$ is irreducible, $f$ and $f^{\prime }$ are not relatively prime if and only if $f\mid f^{\prime }$. And unless $f^{\prime }=0$, $f^{\prime }$ should have degree no less than $\mathrm{deg}(f)$, however, $\mathrm{deg}(f^{\prime })\le \mathrm{deg}(f)$. Thus, $f^{\prime }$ must be the zero polynomial. $\square$

Corollary

Let $f$ be an irreducible polynomial in $F[x]$, where $F$ is a field of characteristic zero, then $f$ has no multiple root in any field extension of $F$.

Proof In a field of characteristic zero, the derivative of a polynomial is never the zero polynomial. $\square$

Primitive Elements

Primitive Element

Let $K$ be a field extension of $F$. An element that generates $K$, i.e. $K=F[\alpha ]$, is called a primitive element of $K$ over $F$.

Lemma

Let $F$ be a field of characteristic $0$, and let $K$ be an extension field that is generated over $F$ by two elements $\alpha$ and $\beta$. For all but finitely many $c\in F$, the element $\alpha +c\beta$ is a primitive element of $K$ over $F$.

Proof See Artin’s Lemma 15.8.2. $\square$

Primitive Element Theorem

Let $K$ be a finite extension of a field $F$ of characteristic $0$. Then $K$ contains a primitive element $\alpha$ such that $K=F(\alpha )$. In other words, $K$ is a simple extension of $F$.

Proof Since $K$ is a finite extension, $K$ is generated by a finite set. In particular, a basis $\{{\alpha }_1,\dots ,{\alpha }_k\}$ will generate $K$ over $F$, say $K=F[{\alpha }_1,\dots ,{\alpha }_k]$. We apply induction on $k$. There is nothing to prove when $k=1$. For any $k>1$, assume the theorem holds for $k-1$, that is the extension field $L=F[{\alpha }_1,\dots ,{\alpha }_{k-1}]$. So $L$ is generated by a primitive element $\beta$, $L=F[\beta ]$. Now, we have $K=F[{\alpha }_1,\dots ,{\alpha }_{k-1},{\alpha }_k]=F[\beta ,{\alpha }_k]$. By the lemma, there exists a primitive element $\gamma$ such that $K=F[\gamma ]$. Thus, the theorem holds. $\square$

The Fundamental Theorem of Algebra

Algebraically Closed Field

A field $F$ is said to be algebraically closed if every non-constant polynomial with coefficients in $F$ has a root in $F$.

Fundamental Theorem of Algebra

$\mathbb{C}$ is an algebraically closed field.

Proof

Remark

There are other proofs as well. See proof using Rouché’s theorem, proof using algebraic topology.

Proposition

Let $X$ be a finite set and $R=\mathrm{Func}(X,\mathbb{R})$. Then $I◃R$ is maximal iff $I=I_x=\{f\in R|f(x)=0\}$ for some $x\in X$.

Hilbert’s Nullstellensatz

The maximal ideals in $\mathbb{C}[x_1,\dots ,x_n]$ are in bijection with points in ${\mathbb{C}}^n$.

Theorem

Let $I$ be an ideal of $\mathbb{C}[x_1,\dots ,x_n]$ generated by $f_1,\dots ,f_m\in \mathbb{C}[x_1,\dots ,x_n]$, let $R=\mathbb{C}[x_1,\dots ,x_n]/I$ and let vector space $V=V(f_1,\dots ,f_m)$.Then points in $V$are in bijection with maximal ideals in $R$.