# Tag Info

Accepted

### Galois group of the irreducible cubic equation $x^3-3x+1$

You don't actually need to find the last root. Consider the field $\mathbb{Q}(\alpha)$, where $\alpha$ is any root of $f$. This field not only contains $\alpha$, but also $2- \alpha^2$ per the field ...
• 33.1k
Accepted

### Galois group of an irreducible , separable polynomial be abelian , then each of the roots of the polynomial generates the splitting field?

The proof goes as follows: Since $\operatorname{Gal}(E/k)$ is abelian, any subgroup is normal, hence any intermediate field of $E/k$ is normal. In particular $k(a)/k$ is normal and by the definition ...
• 31.6k
Accepted

• 355k

### What fields are isomorphic to $\mathbb{Q}[\sqrt{2}]$ other than $\mathbb{Q}[\sqrt{2}]$ itself?

If you define $\Bbb Q[\sqrt 2]$ as the smallest subfield of $\Bbb C$ that contains $\sqrt 2$, then here's a totally different field that is isomorphic to it: $\Bbb Q[X]/(X^2-2)$, one (of two possible) ...
• 375k
Accepted

### Galois number fields that have the imaginary unit.

Items 2 and 4 are incompatible. Assume that $$\Bbb{Q}\subset \Bbb{Q}(i)\subset K_f,$$ where $K_f$ is Galois over $\Bbb{Q}$, $G=Gal(K_f/\Bbb{Q})\simeq A_4$. By Galois correspondence $\Bbb{Q}(i)$ is ...
• 134k

• 2,932

### Treating splitting fields as the same dangerous?

Let's consider the polynomial $f(x) = x^2 + 1$ over $\mathbb{Q}$. Then one way to construct a splitting field for $f$ is to take the subfield of $\mathbb{C}$ generated by its roots, which of course is ...
• 421k

### $\mathbb{C}$ is not the splitting field of any polynomial over $\mathbb{Q}$ (without cardinality)

The existence of a transcendental number provides such a proof.
• 247k

### Separable Extension and Splitting Field

$Q(2^{1/3})$ is separable since it is an extension of a field of characteristic zero, but is not a splitting field, since it does not contain complex roots of $X^3-2$. If a field is not perfect, its ...
• 87.8k
As $\sqrt[6]{7}$ is a zero of the polynomial and $\sqrt[6]{7} \not \in \mathbb{Q}$ we adjoin it to $\mathbb{Q}$ to obtain $\mathbb{Q}(\sqrt[6]{7})$. Now in $\mathbb{Q}(\sqrt[6]{7})$ the polynomial ...