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To prove that a field $F$ with characteristic $0$ contains $\mathbb Q$, the following lemma is used.

Lemma: Let $R$ be a ring with unity. If the characteristic of $R$ is $0$, then $R$ contains a subring isomorphic to $\mathbb Z$.

Solution: Let $S$ be a subring of $F$ that is $\approx \Bbb Z$. Let $T = \{ab^{-1}~~|~~a,b \in S, b \neq 0\}$.

Then it is stated that $T \approx \mathbb Q$. If we take a map $: ab^{-1} \rightarrow a/b$ .

Then, clearly, $T \approx \mathbb Q$ under multiplication. Does isomorphism under addition need to be shown as well?

My book further states that the intersection of the subfields of a field is a subfield (called the prime subfield), so there exists a subfield (EDIT: and a prime subfield) isomorphic to the rationals.

I don't clearly understand this paragraph, did they just try to show that $T$ is a subfield of $F$?

The subfield test says that a subset $A$ of a field $F$ forms a subfield iff

$(i) ~~a-b \in A ~~\forall~~a,b \in A$

$(ii)~~ab^{-1} \in A ~~\forall~~a,b \in A$

But, we don't know if $ef^{-1}-gh^{-1} = e (f^{-1}-e^{-1}gh^{-1})\in T?~~|~~ e,f,g,h \in S?$. I don't think $T$ should be a sub field. What is the book trying to say?

Thank you for your help ..

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4 Answers 4

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If you think the subfield test is wrong then you should go back and review why it's sound. Indeed, the test tells us $0=a-a\in A$, hence $0-a\in A$ for all $a\in A$, so it's closed under additive inverses, and $1=aa^{-1}\in A$ for any nonzero $a\in A$, hence $1\in A$ and $1a^{-1}\in A$ for all nonzero $a\in A$, so it's also closed under multiplicative inverses. Addition is $a-(0-b)$ and multiplication $a(1b^{-1})^{-1}$, so the subset is closed under addition, subtraction, multiplication, division, it has $0$ and $1$, so it is a subfield. Hence we do know $ab^{-1}+cd^{-1}\in A$ if $a,b,c,d\in A$.

You say you don't think $T$ should be a subfield, but your intuition should suggest the opposite conclusion: $T$ is the set of all ratios between sums of $1$ and their additive inverses, which is exactly what the rationals $\Bbb Q$ are! This doesn't require the subfield test per se: one may argue that $T$ is a field in the exact same way that one argues $\Bbb Q$ is a field! Should be a trip down memory lane.

As for the intersection discussion: the prime subfield of a field is the minimal field, which has no smaller subfields, and one may prove that it is equal to the intersection of all subfields. Since $1$ is contained in all subfields, so are its sums and additive inverses, and their ratios, all by the closure property of the operations, so it follows that $T$ is contained in all subfields hence their intersection, hence the intersection equals $T$, hence $T$ is a prime subfield. Since $T\cong\Bbb Q$, this means we have shown that the field contains a prime subfield which is just a copy of $\Bbb Q$.

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  • $\begingroup$ Thank you . I realize where I went wrong. $\endgroup$
    – MathMan
    Commented Jul 16, 2014 at 1:48
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Well, that subring $S$ is not only isomorphic to $\Bbb Z$, but it is generated by the unit element $1$ of the field $F$. More concretely $S$ consists of $\pm(1+1+\dots+1)$ which are operations in $F$.

Yes, we have to verify that the given map also respects addition. But, using commutativity of multiplication we can imitate the common denominator: $$ab^{-1}\,+\,cd^{-1}=adb^{-1}d^{-1}+bcb^{-1}d^{-1}=(ad+bc)b^{-1}d^{-1}\,.$$

I think the cited paragraph wants to say that $F$ not only contains $\Bbb Q$, but it is its smallest subfield, i.e. the intersection of all of its subfields.

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  • $\begingroup$ That means that $T$ is an isomorphism under both addition and multiplication. Also, $T$ is a sub field of $F$ as well? $\endgroup$
    – MathMan
    Commented Jul 15, 2014 at 23:20
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    $\begingroup$ $T$ is the $F$-internal instance of the rational numbers, it contains exactly the numbers of the form $\pm(1+\dots+1)(1+1+..+1)^{-1}$ where $1$ is the unit of $F$. So, $T$ is isomorphic to $\Bbb Q$, and indeed $T$ is the subfield of $F$, which might be technically (as set) different from $\Bbb Q$, but they are usually identified any way. $\endgroup$
    – Berci
    Commented Jul 15, 2014 at 23:26
  • $\begingroup$ Okay .. So, $T \approx \mathbb Q$ . Now, as per the paragraph in my book, the prime sub field $P$ is the smallest sub field and also must be a sub field of $T$ as well then. But, how, can $P \approx \mathbb Q$ when it is actually $T \approx \mathbb Q$? $\endgroup$
    – MathMan
    Commented Jul 15, 2014 at 23:31
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Hint The unique morphism $f:\Bbb Z\to F$ has $f(n)$ invertible for each $n\in\Bbb Z$ nonzero. Localize.

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Hint $ $ Prove the universal property of fraction fields: $ $ if $\,D\,$ is a domain with fraction field $\,F\,$ then a ring injection $h:D\to K\,$ into field $\,K\,$ (uniquely) extends to $\,\hat h : F\to K,\,$ $\,\hat h(a/b) := h(a)h(b)^{-1}\!$

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