2 votes

Quadratic field extension of finite field $\mathbb{F}_{q}$.

In general, this question falls under the umbrella of Kummer theory. The basic result says: Let $n\geq 1$ be any integer. If $K$ is a field of characteristic coprime to $n$ (or $0$) that contains all $...
Tim Seifert's user avatar
1 vote

Quadratic field extension of finite field $\mathbb{F}_{q}$.

The case of finite fields is easier because an extension of degree $n$ in this case is unique. So if you can find any element $\alpha\in\mathbb{F}_q$ such that $x^3-\alpha$ is irreducible over $\...
Mark's user avatar
  • 38.6k
1 vote

$f = X^{3} + 2$ irreducible in $\mathbb{F}_{49}[X]$.

For the first part, suppose $u$ is a root of $f$ in $\mathbb{F}_{49}$. Then $-2=u^3$ and so $(-2)^{16}=u^{48}=1$. But $(-2)^{16} \equiv 2 \bmod 7$, contradiction.
lhf's user avatar
  • 216k
1 vote
Accepted

$f = X^{3} + 2$ irreducible in $\mathbb{F}_{49}[X]$.

It is correct, since the minimal polynomial $m_\alpha \in \mathbb{F}_7[X]$ $\alpha$ divides $X^3+2$, because $\alpha^3+2 = 0$. It has to be $\deg m_\alpha = 3$, since otherwise $X^3 +2 = m_\alpha g$ ...
psl2Z's user avatar
  • 2,306

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