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I have two questions about the fractional field of an integral domain.

Given an integral domain $D$:

  1. Is there a difference between saying "the fractional field of $D$ is the smallest field containing $D$" or "the fractional field of $D$ is the smallest field containing an embedding of $D$"?

  2. How do you prove that the fractional field is the smallest field containing $D$ (or an embedding of $D$, if there is a difference...)? Specifically, I want to show that if $F$ is any field containing $D$, then $F$ must contain the fractional field of $D$.

Thanks for your help.

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  • $\begingroup$ What is your definition of "fractional field of integer domain"? For me, it is precisely the minimal field containing the domain. $\endgroup$ – Timbuc Oct 11 '14 at 18:52
  • $\begingroup$ @timbuc *integral domain. My definition is based on the construction. First, given an integral domain $D$, we define the set of ordered pairs where the second coordinate is non-zero. Then we define an equivalence relation where $(a,b) \sim (c,d)$ if $ad = bc$. Then we call the equivalence class containing $(a,b)$ as $\frac{a}{b}$. Then we define addition and multiplication on this set of equivalence classes in the same way as it is defined in $\mathbb{Q}$. This forms a field, what we call the fractional field. $\endgroup$ – layman Oct 11 '14 at 18:55
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    $\begingroup$ Ok, but then it is trivial, isn't it? I mean, any field containing $\;D\;$ must contain all the multiplicative inverses of non-zero elements $\;d\in D\;$ and thus their product by any element in $\;D\;$ , and this means (by the definition!) that $\;d_1\cdot\frac1{d_2}:=\frac{d_1}{d_2}\;$ is in the field, for any $d_1,d_2\in D\;,\;\;d_2\neq 0\;$ , which means any such field contains the fractions field of $\;D\;$ . $\endgroup$ – Timbuc Oct 11 '14 at 18:57
  • $\begingroup$ @Timbuc Well now I need to know the answer to my first question. Is there a difference between saying a field contains $D$ and saying it contains an embedding of $D$? $\endgroup$ – layman Oct 11 '14 at 19:00
  • $\begingroup$ Well, formally there is, yet for most usual cases one doesn't usually pay attention to that slight difference. $\endgroup$ – Timbuc Oct 11 '14 at 19:01
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Let $F'$ be a smallest field containing an embedding of $D$ ($f:D\to F'$), and $F$ a field of fraction of $D$.

We can extend $f$ to morphism of field $\tilde f:F\to F'$ by $\tilde f(a/b)=f(a)/f(b)$.

Now we have that $\tilde f(F)\subseteq F'$ and $\tilde f(F)$ containing an embedding of $D$ , by smallest property we have $\tilde f(F)=F'$.

So the tow fields $F$ and $F'$ are isomorphic.

Edit: If $F$ is any field containing $D$. And denote $K$ the field of fraction of $D$.

Let $a/b\in K$, $a\in D$ and $0\neq b\in D$, hence $a,b\in F$, it follow that $a$ and $1/b $ are in $F$ so $a. (1/b)=a/b\in F$. Thus $K\subseteq F$.

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  • $\begingroup$ Is your extension of $f$ well defined? $\endgroup$ – layman Oct 11 '14 at 19:26
  • $\begingroup$ I guess it is since $f$ is a homomorphism, and if $a/b = c/d$, then $ad = bc$, so $f(a)f(d) = f(b)f(c)$ which implies $f(a)/f(b) = f(c)/f(d)$. $\endgroup$ – layman Oct 11 '14 at 19:28
  • $\begingroup$ Yes $b\neq 0$ implies $f(b)\neq 0$ because $f(0)=0\neq f(b)$ ($f$ is injective). $\endgroup$ – Hamou Oct 11 '14 at 19:28
  • $\begingroup$ I guess I would need to do a little bit more work because I would need to show the image of the field $F$ under the extension of $f$ is a field. $\endgroup$ – layman Oct 11 '14 at 19:29
  • $\begingroup$ When $\frac ab\neq 0$, $\tilde f(a/b)\tilde f(b/a)=1$, hence $\tilde f(a/b)$ is invertible is in $\tilde f(F)$. $\endgroup$ – Hamou Oct 11 '14 at 19:32
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The right (i.e. categorical) way to say this (without the ambiguities of words like "smallest", "containing", etc.) ought to be that the inclusion $\iota: D\to Q(D)$ has the following universal property:

If $K$ is a field, and $f: D\to K$ is any morphism of rings, then there is a unique morphism of fields $g : Q(D) \to K$ such that $f = g \circ \iota$.

(In particular, $Q(D)$ embeds into any field that $D$ embeds into.)

This property uniquely determines (up to isomorphism) not only $Q(D)$, but $\iota$ as well.

And it's easily proved, since $g(1/b)g(b)=g(1)$ forces $g(a/b) = f(a)/f(b)$, so this amounts to checking that $a/b \mapsto f(a)/f(b)$ is actually a homomorphism.

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