# Questions tagged [extension-field]

Use this tag for questions about extension fields in abstract algebra. An extension field of a field K is just a field containing K as subfield, but interesting questions arise with them. Use this topic for dimension of extension fields, algebraic closure, algebraic/transcendental extensions, normal/separable extensions... This tag often goes along with the abstract-algebra and/or field-theory tags.

3,077 questions
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### Splitting field over K of an infinite set of polynomial

Suppose $F$ is a finite splitting field over $K$ of $X=\lbrace f_i(x)\rbrace_{i\in I}$, some infinite set. Is there necessarily a finite set $Y\subseteq X$ such that $F$ is a finite splitting field of ...
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### $K$ is intermideate field between $F$ and $F(x)$. then $\operatorname{dim}F(x)$ over $K$ is finite.

Let $F$ be a field and let $F(X)$ be the field of rational functions with coefficients in $F$. Let $K$ be any field such that $F\subseteq K\subseteq F(X)$ and $K\neq F$. Prove that $[F(X):K]\lt\infty$....
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### If $u\in F$ is Transcendental over $K$, $F$ an extension field of $K$, Show every element in $K(u)$ not in $K$ is transcendental over $K$.

Doing some problems out of Beachy’s Algebra text, I came across that problem, and I’m at a loss how to show it without a bit of hand waving. Do I make some statement about spaces, and prove by ...
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### Extension of Splitting Fields over An Arbitrary Field

Let $F$ be a field in which $0 \neq2$ in $F$, and consider $f=x^4+1$. If $E$ is the splitting field for $f$ over $F$, it turns out that $E$ is a simple extension of $F$. How does one realize this fact?...
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### Kernel is a group of order $q+1$

Hello I hope you can help me with this doubt I have. Let $F_{q}$ a field and consider its quadratic extension $F_{q^{2}}=F_{q}(\sqrt{d})$ for $d$ square free in $F_{q}$ now we consider the next ...
### Determine all algebraic extensions of $\Bbb Q$ contained in $\Bbb Q(\sqrt{2},\pi)$
Determine all algebraic extensions of $\Bbb Q$ contained in $\Bbb Q(\sqrt{2},\pi)$ Assuming $\pi$ to be transcendental over $\Bbb Q$ , it seems to me that the answer must be only $\Bbb Q(\sqrt{2})$ . ...