Skip to main content

New answers tagged

0 votes

Degree of splitting field of $x^3-5$ over $\mathbb{F}_7$

From "Lehrbuch der Algebra (Gerd Fischer)": Let $p$ be prime, $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p, n \in \mathbb{N}\setminus\{0\}$ and $q := p^n$ ...
Raiden's user avatar
  • 9
2 votes

Find Galois group of polynomials when char F=2

(A) Let us first deal with the case when $F=\mathbb{F}_2$. Your $\alpha$ is a root of $x^3+x+1$. This is a cubic polynomial. It is irreducible. (If it were reducible it would have a linear factor, and ...
ancient mathematician's user avatar
1 vote

Proving the irreducibility of a polynomial in a field extension.

For each $q$, there is at most one field of size $q$ in any field, since they are all the roots of the polynomial $x^q-x$. Therefore you don't have to do any calculation to conclude $x^3+x^2+2$ is ...
Just a user's user avatar
  • 15.5k
0 votes

Every field extension generated by elements of degree two is normal.

Yes. The general statement is that $F/K$ is a normal extension if and only if $F$ is the splitting field of a collection of polynomials over $K$. In the finite case this is equivalent to $F$ being a ...
Mark's user avatar
  • 40.4k
1 vote

Splitting field of $X^{p^n}-1$ over $\mathbb{F}_p$

Theorem (Kummer) Let $p$ be a prime, and $m,n$ be positive integers. The highest power of $p$ that divides $\binom{n+m}{n}$ is the number of carries when adding $n$ and $m$ in base $p$. Corollary. Let ...
Arturo Magidin's user avatar
2 votes

Proving that $r = \displaystyle\sum_{i=1}^k \frac{1}{\alpha_i} \in \mathbb{Q}$

Let us first assume that $0$ is not a root of $f(x)$. You can easily see that $c=\sum_{i=0}^k \frac{1}{\alpha_i}$ is invariant under any element of Gal$(K/\mathbb{Q})$. Thus $c$ belongs to the fixed ...
Mike's user avatar
  • 20.5k
0 votes

Extending homomorphism from from splitting field of a polynomial to another field where that polynomial splits into linear factor

As $K$ contians all roots of $f$, the roots generate a splitting field $E\subset K$ of $f$ by definition. As splitting fields are unique up to isomorphism, the composition $F\xrightarrow{\sim}E\subset ...
Covariant's user avatar
  • 482

Top 50 recent answers are included