Constructing finite extensions of fields

In this section, we generalize the construction of §2.1.1.

Theorem 2.6.3   Let $F$ be a field. Then $F[x]/(m(x))$ is a field if $m(x)$ is an irreducible polynomial in $F[x]$.

Remark 2.6.4   The converse to this thorem also holds.

proof: We have already seen that $F[x]/(m(x))$ is a commutative ring with unity. We need only show that each non-zero element in $F[x]/(m(x))$ has an inverse modulo $m(x)$, provided $m(x)$ is irreducible.

Let $a(x)\in F[x]/(m(x))$ be non-zero. Since $m(x)$ is irreducible, we have $gcd(m(x),a(x))=1$ (otherwise, it would be a non-trivial factor of $m(x)$). By Lemma 2.5.5, $a(x)$ has an inverse modulo $m(x)$. $\Box$

Definition 2.6.5   If $F$ is a finite field, if $m(x)$ is an irreducible polynomial in $F[x]$ of degree $d$, and if the powers of $x$ ($1$, $x$, $x^2$, ..., $x^{d-1}$) yield all the non-zero elements of the field $F[x]/(m(x))$, then we call $m(x)$ a primitive polynomial. If $a\in F$ and if the powers of $a$ ($1$, $a$, $a^2$, ..., $a^{d-1}$) yield all the non-zero elements of the field $F$ then $a$ is called a primitive element of $F$.

Example 2.6.6 (a)   $m(x)=x^2+x+1$ is a primitive polynomial over $\mathbb{F}_2$.

(b) $m(x)=x^4+x+1$ is a primitive polynomial over $\mathbb{F}_2$.

Example 2.6.7   $\mathbb{R}[x]/(x^2+1)$ is a field. As a set, we may identify $\mathbb{R}[x]/(x^2+1)$ with a set of residue class representatives

$$\mathbb{R}[x]/(x^2+1)=\{a+bx\ \vert\ a,b\in \mathbb{R}\}.$$

In fact, if $p(x)=a_0+a_1x+a_2x^2+...+a_nx^n\in \mathbb{R}[x]$ then

$$\overline{p(x)}\equiv (a_0-a_2+-...)+ (a_1-a_3+-...)x\, ({\rm mod}\, x^2+1),$$

by Example 2.5.4.

As long as you remember that in $\mathbb{R}[x]/(x^2+1)$, we have $x^2=-1$ (since everything is modulo $x^2+1$), addition and multiplication in $\{a+bx\ \vert\ a,b\in \mathbb{R}\}$ is easy:

$$(a+bx)+(c+dx)=a+c+(b+d)x, \ \ \ \ \ \ (a+bx)\cdot (c+dx)=ad-bd+(ad+bc)x.$$

This defines addition $+$ and multiplication $\cdot$ on $\{a+bx\ \vert\ a,b\in \mathbb{R}\}$. With these binary operations, $\{a+bx\ \vert\ a,b\in \mathbb{R}\}$ is a field.

This looks just like complex multiplication, where $i=\sqrt{-1}$ has been replaced by $x$. In other words, there is a direct correspondence between the addition and multiplication formulas for $\mathbb{C}$ and for $\{a+bx\ \vert\ a,b\in \mathbb{R}\}$. More abstractly said, if we define $\phi:\{a+bx\ \vert\ a,b\in \mathbb{R}\}\rightarrow \mathbb{C}$ by $\phi(a+bx)=a+ib$ then

$$\phi(a)\phi(b)=\phi(ab),\ \ \ \ \ \ \phi(a)+\phi(b)=\phi(a+b),$$

for all $a,b\in \mathbb{R}$. This means that $\{a+bx\ \vert\ a,b\in \mathbb{R}\}$ and $\mathbb{C}$ are ``isomorphic fields.''

Example 2.6.8   $\mathbb{F}_3[x]/(x^2+1)$ is a field. As a set, we may identify $\mathbb{F}_3[x]/(x^2+1)$ with a set of residue class representatives

$$\mathbb{F}_3[x]/(x^2+1)=\{a+bx\ \vert\ a,b\in \mathbb{F}_3\}.$$

In fact, if $p(x)=a_0+a_1x+a_2x^2+...+a_nx^n\in \mathbb{F}_3[x]$ then

$$\overline{p(x)}\equiv (a_0-a_2+-...)+ (a_1-a_3+-...)x\, ({\rm mod}\, x^2+1),$$

as in the above example.

Addition and multiplication in $\{a+bx\ \vert\ a,b\in \mathbb{F}_3\}$ is easy:

$$(a+bx)+(c+dx)=a+c+(b+d)x, \ \ \ \ \ \ (a+bx)\cdot (c+dx)=ac-bd+(ad+bc)x.$$

With these binary operations, $\{a+bx\ \vert\ a,b\in \mathbb{F}_3\}$ is a field.

This looks just like complex multiplication, where $i=\sqrt{-1}$ has been replaced by $x$. In other words, there is a direct correspondence between the addition and multiplication formulas for $\mathbb{F}_3(i)=\{a+bi\ \vert\ a,b\in \mathbb{F}_3\}$ and $\mathbb{F}_3[x]$.

In this example, $x^2+x$ is a primitive element in $\mathbb{F}_3[x]$ but $x$ is not (so $x^2+1$ is not a primitive polynomial over $\mathbb{F}_3$). Also, $i-1\in \mathbb{F}_3(i)$ is a primitive element but $i$ is not.

Motivated by the facts in the above examples, we make the following definition.

Definition 2.6.9   Let $F,F'$ be fields. Denote the binary oprations on both these fields by $+$ and $\cdot$ (even though they are probably different operations, it is assumed that the reader can distinguish which is which by the context). If $\phi:F\rightarrow F'$ is a map such that

$$\phi(a)\phi(b)=\phi(ab),\ \ \ \ \ \ \phi(a)+\phi(b)=\phi(a+b),$$

for all $a,b\in F$ then we call $\phi$ a field isomorphism. We also say that $F$ and $F'$ are isomorphic (as fields) and write $F\cong F'$.

Example 2.6.10   The field $\mathbb{R}[x]/(x^2+1)$ is isomorphic to $\mathbb{C}$. The field isomorphism is induced by sending $x\in \mathbb{R}[x]/(x^2+1)$ to $i\in\mathbb{C}$ and then extending it to all of $\mathbb{R}[x]/(x^2+1)$. In other words, define $\phi :\mathbb{R}[x]/(x^2+1)\rightarrow \mathbb{C}$ by $\phi(1)=1$, $\phi(x)=i$, and then

$$\phi(a)\phi(b)=\phi(ab),\ \ \ \ \ \ \phi(a)+\phi(b)=\phi(a+b),$$

for all $a,b\in \mathbb{R}[x]/(x^2+1)$. This defines a field isomorphism.