1.9 Construction of some extensions

Let $f(X)\in F[X]$ be a monic polynomial of degree $m$ , and let $(f)$ be the ideal generated by $f$ . Consider the quotient ring $F[X]/(f(X))$ , and write $x$ for the image of $X$ in $F[X]/(f(X))$ , i.e., $x$ is the coset $X+(f(X))$ .

(a) The map

P(X)\mapsto P(x)\colon F[X]\rightarrow F[x]

is a homomorphism sending $f(X)$ to $0$ . Therefore, $f(x)=0$ .

(b) The division algorithm shows that every element $g$ of $F[X]/(f)$ is represented by a unique polynomial $r$ of degree $<m$ . Hence each element of $F[x]$ can be expressed uniquely as a sum

equation (2) (2)

a_{0}+a_{1}x+\cdots+a_{m-1}x^{m-1},\qquad a_{i}\in F.\hfill

(d) To multiply two elements expressed in the form (2), multiply in the usual way, and use the relation $f(x)=0$ to express the monomials of degree $\geq m$ in $x$ in terms of lower degree monomials.

(e) Now assume that $f(X)$ is irreducible. Then every nonzero $\alpha\in F[x]$ has an inverse, which can be found as follows. Use (b) to write $\alpha=g(x)$ with $g(X)$ a polynomial of degree $\leq m-1$ , and apply Euclid’s algorithm in $F[X]$ to find polynomials $a(X)$ and $b(X)$ such that

a(X)f(X)+b(X)g(X)=d(X)

with $d(X)$ the gcd of $f$ and $g$ . In our case, $d(X)$ is $1$ because $f(X)$ is irreducible and $\deg g(X)<\deg f(X)$ . When we replace $X$ with $x$ , the equality becomes

b(x)g(x)=1.

Hence $b(x)$ is the inverse of $g(x)$ .

We have proved the following statement.

1.25

For a monic irreducible polynomial $f(X)$ of degree $m$ in $F[X]$ ,

F[x]\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}F[X]/(f(X))

is a field of degree $m$ over $F$ . Computations in $F[x]$ come down to computations in $F$ .

Note that, because $F[x]$ is a field, $F(x)=F[x]$ .⁶⁶ 6 Thus, we can denote it by $F(x)$ or by $F[x]$ . The former is more common, but I use $F[x]$ to emphasize the fact that its elements are polynomials in $x$ .

Example 1.26

Let $f(X)=X^{2}+1\in\mathbb{R}[X]$ . Then $\mathbb{R}[x]$ has

elements: $a+bx$ , $a,b\in\mathbb{R};$

addition: $(a+bx)+(a^{\prime}+b^{\prime}x)=(a+a^{\prime})+(b+b^{\prime})x;$

multiplication: $(a+bx)(a^{\prime}+b^{\prime}x)=(aa^{\prime}-bb^{\prime})+(ab^{\prime}+a^{% \prime}b)x;$

inverses: in this case, it is possible write down the inverse of $a+bx$ directly.

We usually write $i$ for $x$ and $\mathbb{C}$ for $\mathbb{R}[x].$

Example 1.27

Let $f(X)=X^{3}-3X-1\in\mathbb{Q}[X]$ . We observed in (1.12) that this is irreducible over $\mathbb{Q}$ , and so $\mathbb{Q}{}[x]$ is a field. It has basis $\{1,x,x^{2}\}$ as a $\mathbb{Q}$ -vector space. Let

\beta=x^{4}+2x^{3}+3\in\mathbb{Q}[x].

Then using that $x^{3}-3x-1=0$ , we find that $\beta=3x^{2}+7x+5$ . Because $X^{3}-3X-1$ is irreducible,

\gcd(X^{3}-3X-1,3X^{2}+7X+5)=1.

In fact, Euclid’s algorithm gives

\textstyle(X^{3}-3X-1)\left(\frac{-7}{37}X+\frac{29}{111}\right)+(3X^{2}+7X+5)% \left(\frac{7}{111}X^{2}-\frac{26}{111}X+\frac{28}{111}\right)=1.

Hence

\textstyle(3x^{2}+7x+5)\left(\frac{7}{111}x^{2}-\frac{26}{111}x+\frac{28}{111}% \right)=1,

and we have found the inverse of $\beta.$

We can also do this in PARI: b=Mod(X^4+2*X^3+3,X^3-3*X-1) reveals that $\beta=3x^{2}+7x+5$ in $\mathbb{Q}[x]$ , and b^(-1) reveals that $\beta^{-1}=\frac{7}{111}x^{2}-\frac{26}{111}x+\frac{28}{111}$ .