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### 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 ...
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### 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 ...
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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 $\left| Gal \left( \... • 26 1 vote Accepted ### Proving the irreducibility of a polynomial in a field extension. For each$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$is ... • 15.4k 1 vote ### Compositum normal when only one field assumed normal No. Take$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}...
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