2 votes

Field extension $L/K$ such that $L$ and $K$ are isomorphic, then the degree of the extension need not be 1?

The field $\mathbb{Q}(\pi^2)$ is isomorphic to $\mathbb{Q}(x)$, the field of rational functions in one variable (the field of fractions of the polynomial ring $\mathbb{Q}[x]$), since $\pi^2$ is ...
Arturo Magidin's user avatar
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

Primitive element of Galois extension whose conjugates are linear dependent.

You can consider the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}$, which primitive element (one of them) is $\alpha = \sqrt{2} +\sqrt{3}$. This is a Galois extension with $\left| Gal \left( \...
Adr3W's user avatar
  • 26
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
  • 15.4k
1 vote

Compositum normal when only one field assumed normal

No. Take $K = \mathbb{Q}$, $L_1 = \mathbb{Q}(\sqrt{2})$, and $L_2 = \mathbb{Q}(\sqrt[3]{2})$. Then $L_1 / K$ is normal but $L_2 / K$ is not, and neither is $L_1 . L_2 = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2}...
Ben Steffan's user avatar
  • 2,947

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