I have solved that $T$={$a+b\sqrt2$|$a,b\in Q$} is a subfield of $R$. No I am having to show that the only subfields of $T$ are $T$ and $Q$ itself.
I'm lost on where to start on this. Any assistance would be appreciated.
Show this field has no subfields but itself and the rationals
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abstract-algebra
field-theory
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0By the way: Showing this includes showing (or at least recalling) that the rationals have no proper subfield – 2017-02-23
2 Answers
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If $L$ is a subfield of $T$ distinct of $Q$, $x=a+b\sqrt2, b\neq 0$, the inverse of $x$ is ${1\over{a^2+b^2}}(a-b\sqrt2)$, this implies that $a-b\sqrt2\in L$ since $L$ is stable by multiplication by elements of $Q$. We deduce that $2b\sqrt2=a+b\sqrt2-(a-b\sqrt2)\in L$ and $\sqrt2\in L $ since $b\neq 0$, this implies that $L=T$.
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If you are familiar with the concept of the degree of an extension, there's a very easy way to do this. The degree of $T$ over the rationals, $[T:{\bf Q}]$, is $2$, so for any field $F$ "in between", either $T$ has degree $1$ over $F$, or $F$ has degree $1$ over the rationals. But one field having degree $1$ over another immediately implies that the two fields are equal.