2020年2月27日 · Then $\mathbb{F}$ is a subfield of $\mathbb{K}$ if $\mathbb{F}$ is a unital subring of $\mathbb{K}$. Your definition, on the other hand, is also missing a bit of context: it is not merely "a subset of a field which is itself a field", but rather "a subset of the field which is itself a field under the induced operations ."

$\begingroup$ Consider the following. When you do calculations involving $\sqrt2$ you, at least more often than not, don't use its "numerical value".

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If $\sigma : \mathbb{C} \to \mathbb{C}$ is any automorphism whatsoever, then $\sigma(\mathbb{R})$ is another subfield of index $2$ in $\mathbb{C}$, and there are uncountably many such automorphisms (assuming the axiom of choice). If such an automorphism preserves $\mathbb{R}$ then it is either the identity or complex conjugation. $\endgroup$

2014年5月22日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2020年12月1日 · Let K ⊆ L is a field extension and S ⊆ L be a non empty set. We define the subfield of L generated by K ∪ S, denoted by K(S), to be the smallest subfield of L containing K ∪ S. Moreover, K(S) is the intersection of all the subfields of L containing K ∪ S. In this case, we say that K(S) is the field obtained from K by adjoining the set S.

There is an important link with geometry as follows. The group $\text{GL}_2^+(\Bbb R)$ of matrices with positive determinant acts on the complex upper halfplane $$ \cal H=\{z=x+iy\in\Bbb C\text{ such that }y>0\} $$ by linear fractional transformations $$ \left(\begin{array}{cc}a&b \\ c&d \end{array}\right)\cdot z=\frac{az+b}{cz+d} $$ Parenthetically, these transformations are …

2020年10月28日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

A subfield that is not properly contained in a larger subfield of ‎$‎A ‎$‎ is called a maximal subfield of ‎$‎A‎$‎. 1 Could there exist some subfield of $\mathbb{R}$ such that $\pi$ is algebraic over that subfield?