Show that $\mathbb{F}_5[x]$/$x^2+x+1$ is a field.

Question

If someone can help me showing that $\mathbb{F}_5$/$x^2+x+1$ is a field.

Answer 1

Since (as pointed out by @Muralidharan) $f = x^2+x+1$ is irreducible, which can be checked by noting that it has no root in $\mathbb{F}_5$, you know, that the ideal $I=(f)$ is maximal and hence the factor ring $\mathbb{F}_5[x]/I$ is a field.

The original polynomial ring $\mathbb{F}_5[x]$ consisted of polynomials, so the factor consists of their respective equvalence sets. If a polynomial $p$ is of degree higher than $1$, you can divide it by $f$ obtaining $p = q\cdot f + r$, where the part $q\cdot f$ lies in $I$, hence it is (in the factor) equal to zero. As a result, you only need to look at the behaviour of polynomials of degree at most $1$: $(ax + b)\cdot (cx+d) = ac x^2 + (bc + ad) x + bd$ which, when divided by $f$ gives $ac x^2 + (bc + ad) x + bd = ac \cdot f + (bc+ad-ac)x + (bd - ac)$, which is in the factor equal to $(bc+ad-ac)x + (bd - ac)$. Summing is done similarly (and remains the same as in the original polynomial ring.)

(Equivalently, you can take the factorization as imposing the rule "$x^2 = -x-1$".)

Answer 2

if $p$ is a prime, then $\mathbb F_p[x]/(f(x)) \cong \mathbb Z[x]/(p,f(X))$.

In question we have $p=5$,then $\mathbb F_5[x]/(x^2+x+1) \cong \mathbb Z[x]/(5,x^2+x+1)$ and $x^2+x+1$ has no roots in $\mathbb Z_5$ ,then $x^2+x+1$ is an irreducible polynomial in $\mathbb Z_5[x]$, Therefore $(5,x^2+x+1)$ is maximal ideal in $\mathbb Z[x]$ and $\mathbb Z[x]/(5,x^2+x+1)$ is a field ,so $\mathbb F_5[x]/(x^2+x+1)$ is a field.

Answer 3

The ring in your question is isomorphis to the algebra over $\Bbb F_5$ generated bu the matrix $M = \begin{pmatrix} 0 & -1 \\1 & -1 \end{pmatrix}$ (called companionmatrix in the literature). Since $M^2 + M + 1 = 0$ this algebra is two dimensional and since the polynomial $x^2+x+1$ is irreducible over $\Bbb F_5$, it has no zero divisors and is thus a field. Since the algebra has dimension two (a basis is $\{1, M\}$) this field is isomorphic to $\Bbb F_{25}$.