Subfield of a field F

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Let $x \in K.\,$ Then $x-x\in K$, hence $0 \in K$.

Let $x \in K.\,$ Then since $0 \in K$, we get $0 - x \in K$, hence $-x \in K$.

Let $x,y \in K.\,$ Then $-y \in K$, hence $x-(-y) \in K$, so $x+y \in K$.

Since $K$ has at least two elements, $K$ contains at least one nonzero element, $x$ say. Then $(x)(x^{-1}) \in K$, hence $1 \in K$.

Let $x \in K,\;x \ne 0$. Then since $1 \in K$, we get $(1)(x^{-1}) \in K$, hence $x^{-1} \in K$.

Let $x,y \in K$.

$\qquad$If $y \ne 0$, then $y^{-1} \in K$, so $(x)((y^{-1})^{-1}) \in K$, hence $xy \in K$.

$\qquad$If $y=0$, then $xy=0$, hence $xy \in K$.

Either way, $xy \in K$.

It follows that $K$ is a field.

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Updated on August 01, 2022

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  • Mathematicing
    Mathematicing over 1 year

    Question: Let $F$ be a field and let $K$ be a subset of $F$ with at least two elements. Prove that $K$ is a subfield of $F$, if for any $a,b \left ( b\neq 0 \right )$ in $K$, $a-b$ and $ab^{-1}$ belong to $K$.

    This is an odd question but let's see.

    It is necessary to investigate the case where $a=0$ and $a\neq 0$.

    Let's proceed with the case where $a=0$ first.

    Suppose $a=0, b\neq 0 \in K$. Let $K=\left \{ a,b \right \}$ Then, $a-b=0-b=-b$. It doesn't seem as though I can proceed on with what I have here. The additive inverse of $b$ is contained in $F$ by definition of a field but it is not necessary true that this additive inverse must be contained in $K$.

    Any help is appreciated. Thanks in advance.

    • fleablood
      fleablood over 6 years
      $a,b \in K$ so $a-b \in K$ so $(a-b) - a=-b \in K$. And $a-(-b) = a+b \in K$. So $K$ is closed under addition. This is assuming that statement $a,b \in K \implies a-b \in K$ does not specify that $a-b$ must be distinct. I think you are onto something if $a,b$ must be distinct $0, b$ mean we have $0b^{-1} = 0$ and $0, b , -b \in K$ but that's it.
    • quasi
      quasi over 6 years
      The wording is tricky, but I think the problem is OK. You can get $1$ by using a = b. You can get 0 by b-b. You can get -1 by 0 - 1. You can get 2 by 1 - (-1). You can get 1/b if you have b. etc.
    • quasi
      quasi over 6 years
      My comment above is just exploratory. Using some of those ideas, you should be able to show K has 0,1, is closed under addition, negation, mutliplication, and all nonzero elements have multiplicative inverses.
    • fleablood
      fleablood over 6 years
      The wording is ambiguous. One can argue that if we only have $a-b$ and $a/b$ if $a\ne b$ then you don't nescessarily have a field. We do have 0 and inverses and, i think, closed under addition. but if $b \ne b^{-1}$ and $a=0$ we do not have multiplication closed. I think we have a field even with distinction if $a \ne 0$ but I'm not sure. I think we can assume distinction is not required.