8 votes

Is there a maximal ordered field? What about $\mathbb R$?

Let me give you two seemingly contradictory answers: There is no Maximal Ordered Field We can prove that there is no maximal ordered field. Based on your post, I think some of this will be new to you, ...
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  • 2,006
7 votes
Accepted

Understanding the field extension diagram of $x^3-2$

Note that the minimal polynomial of $\zeta_3$ is $x^2+x+1$, not $x^3-1$ (indeed $x^3-1$ is reducible as $(x-1)(x^2+x+1)$). This is where the $2$ comes from. As for the equation $6\mid[\mathbb E:\...
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3 votes
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Subfield fixed by subgroup of Galois group if and only if subfield is fixed by generators

Your proof is correct. Here is also a shorter proof. Let: $H=\{\sigma\in G: \sigma(x)=x \ \ \text{for all} \ \ x\in E\}$ It's very easy to show that $H$ is a subgroup of $G$. Since by assumption it ...
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  • 33.3k
2 votes
Accepted

About the degree of a field extension of the field of rational functions

If the function field setting is adding confusion, let $L=\mathbb{Q}(x)$ and $K=\mathbb{Q}(y)$. Then $L=K(x)$, and here, as the author shows, $x$ satisfies a quadratic equation over $K$, so $[L:K]\leq ...
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1 vote

Example of a complete, non-archimedean ordered field

The set of functions $$\mathbb{R}((\mathbb{Q})):=\{f:\mathbb{Q}\to \mathbb{R}\ |\ supp(f)\mbox{ is well-ordered}\},$$ where $supp(f):=\{x\in \mathbb{Q}\ | \ f(x)\neq 0\}$, is a field under the ...
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  • 3,889
1 vote

Minimal polynomial over $\mathbb{Q}$ for $1 + \sqrt[3]{2} + \sqrt[3]{4}$

Upon further review, your $27$ linear polynomials reduce to three. First off, the $1$ is not under any root signs, so you can just call it $1$ without multiplying in any roots of unity. That leaves $1+...
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  • 30.5k
1 vote

Trace on Finite fields

We have $$Tr_{q^r|q}(\alpha)=\alpha+\alpha^q...+\alpha^{q^{r-1}}\in \mathbb{F}_q.$$ If $\alpha=\beta^q-\beta$ for some $\beta\in \mathbb{F}_{q^r}$, then $$Tr_{q^r|q}(\alpha)=Tr_{q^r|q}(\beta)^q-Tr_{q^...
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  • 11

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