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Consider the universal property for the fraction field of an integer domain:

Let $R$ be a integral domain, $F(R)$, its fraction field, $K$ some field and $f:R\rightarrow K$ a injective homomorphism, i.e. $R$ is embedded via $f$ in $K$. Then there exists a unique homomorphism $g$ such that $g\circ \varepsilon =f$, where $ \varepsilon $ is the map that embeds $R$ in $F(R)$.

Now my question is: Can we drop the requirement that $f$ is injective ? If we do that, the proof I have in mind has to be modified, since in it $g$ is defined as $g(\frac{a}{b})=f(a)f(b)^{-1}$ and the fact that $f$ is injectiv implies that $f(b)$ is nonzero, which is necessary for $f(b)$ to be invertible.

Although I couldn't come up with a new proof where I define my $g$ in a different way, I also couldn't come p with a counterexample, i.e. some $R,K$ and $f$ such that for every homomorphisms $g$ we have $ g\circ \varepsilon \neq f$. And looking through the internet and books makes me think that the "injectiveness" of $f$ is necessary.

EDIT Since I didn't put enough thought to this, as the first comment shows, I'm modifying my question to consider $g$ only as a ring homomorphism.

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Every map of fields is injective, and $\epsilon$ is injective, so in fact ANY non-injective map $f$ gives a counterexample. –  Julian Rosen Dec 2 '12 at 17:41
    
@PinkElephants Ok. But what would happen, if we only require $g$ to be a ring homomorphism ? (I know, this slightly distorts the original meaning of my question) –  temo Dec 2 '12 at 17:45
    
@temo: "ring homomorphism between fields" is the same thing as "field homomorphism". –  Chris Eagle Dec 2 '12 at 18:00
    
@ChrisEagle Why is that so ? I thought for field homomorphisms we require that $1$ gets mapped to $1$, but for ring homomorphisms we drop that condition (and therefor injectivity gets lost). –  temo Dec 2 '12 at 19:01

3 Answers 3

No, we cannot. For example, take $R=\mathbb{Z}$, $K=\mathbb{Z}/2\mathbb{Z}$, and $f$ to be the canonical map. Since $f(2)=0$, there's no way to define $f(1/2)$, and so $f$ cannot be extended to $\mathbb{Q}$.

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If I don't require that my $g$ is a field hom., but only a ring hom., then $g$ doesn't automatically have to be injective anymore, since (generally) a ring hom. $h$ only has to satisfy $h(a+b)=h(a)+h(b)$ and $h(ab)=h(a)h(b)$ but not necessarily $h(1)=1$. So we could define in the above example $g$ (I think you meant $g$ not $f$ the last two times you used that symbol) to be the zero map, which is a ring homomorphism. Probably uniqueness gets lost, but I wonder the theorem would otherwise still hold. [Only if we think of $g$ to be $g(a/b)=f(a)f(b)^{-1}$ there no way to define $g(1/2)$.] –  temo Dec 2 '12 at 19:12
    
$0 \circ \epsilon \neq f$. –  Chris Eagle Dec 2 '12 at 19:53
    
Not that it's relevant here, but the correct definition of ring homomorphism does require $1$ to map to $1$. –  Chris Eagle Dec 2 '12 at 19:54
    
Sorry, but your use of the symbol $f$ for $g$ confused me, although I can see that from the viewpoint of extending $g$ that makes sense. You're right that the zero map isn't working here, but that doesn't prove that there aren't other maps, such that $g\circ \varepsilon = f$, where $g$ is merely a ring homomorphism! –  temo Dec 2 '12 at 20:09
    
And, according to mathworld.wolfram.com/RingHomomorphism.html, the definition of ring homomorphism doesn't require that $1$ maps to $1$ unless we're talking about unit rings (and I'm viewing $F(R)$ and $K$ - although I can see that it may not be very clear from the edit of my question - not as unit rings). –  temo Dec 2 '12 at 20:11

Several people have already answered your question to show that injectivity is necessary in the universal property of the fraction field. However, there is a more general construction which works when the map is not injective, namely localization.

If $f: R \to K$ is a ring homomorphism, with $R$ a domain and $K$ a field, then let $P = f^{-1}(0)$ be the kernel of the map. Note that $P$ is necessarily prime, as $R/P \hookrightarrow K$. Then the localization of $R$ at $P$ is the ring $R_P$ with all elements not contained in $P$ inverted. In fact, it's the subset of $\mathrm{Frac}(R)$ where the denominators can be any element not in $P$. Hence $R \subseteq R_P \subseteq \mathrm{Frac}(R)$. Moreover, $R_P$ is exactly characterized by this fact! Just like the fraction field it has a universal property, namely that if every element not contained in a prime ideal $P$ becomes invertible in the image, then the map factors uniquely through $R_P$.

For a concrete example, let $R = \mathbf{Q}[x,y],$ and $K = \mathbf{Q}(x)$. Since any map from $R$ is uniquely determined by where it maps $x$ and $y$, we define $f: R \to K$ to be the map given by $x \mapsto x$, $y \mapsto 0$. It is not hard to see that the kernel of this map is exactly the prime ideal $(y) \subset R$, so $f$ factors uniquely through $R_{(y)}$. The elements of $R_{(y)}$ are exactly fractions of $R$ with denominators not divisible by $y$, so $\frac{x^2y}{x^2+y^2-1} \in R_{(y)}$ but $\frac{3}{xy+y^2}$ is not.

In the special case that the map is injective, then $P = (0)$. Since every nonzero element of $R$ lies outside of $P$, then every element is inverted and we see that $R_{(0)} = \mathrm{Frac}(R)$.

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Dear @BrandonCarter: You may want $R/P$ to be isomorphic to a subring of $K$ and not $K$ itself in the second paragraph where you deduce that $P$ is prime? –  Rankeya Dec 2 '12 at 19:34
    
@Rankeya: Of course. Fixed. –  Brandon Carter Dec 2 '12 at 20:16

First observe that the kernel of any ring homomorphism is always an ideal, regardless of whether you require ring homomorphism to map $1$ to $1$ or not (I should add that it is standard to require this when one deals exclusively with rings with $1$). So for a ring homomomorphism $g: F(R) \to K$ between fields, its kernel will either be $0$ (in which case $g$ is injective) or $F(R)$ (in which case $g$ is the constant zero-map). Consequently, the composition $g\circ\varepsilon$ will either be injective or zero. Therefore, if a ring homomorphism $f: R\to K$ is neither injective nor constant it cannot possibly factor as $f = g\circ \varepsilon$.

Take for instance the map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$.

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