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Is it by definition that $\varphi(1)=1$ if $\varphi$ is a field homomorphism ?

My field theory lecture said that yes, but now $\varphi\equiv0$ is not a field homomorphism...

What is the 'standard' in mathematics ? (i.e. if someone says that $\varphi$ is a field homomorphism does it imply $\varphi(1)=1$ ?)

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It is standard to assume $\varphi(1) = 1$. It isn't hard to show that if $\varphi(1)\neq 1$, then $\varphi\equiv 0$. Thus assuming $\varphi(1) = 1$ is just to rule out this case. –  froggie May 6 '12 at 14:25
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Why would you want $0$ to be a homomorphism? It's not like you can add field (or ring) homomorphisms. –  Zhen Lin May 6 '12 at 14:39
    
Hi, it seems that my answer is somewhat misleading and possibly even wrong. Please unaccept it so I can delete the answer. Pete Clark has agreed to write an answer of his own, which I am sure will be much better than mine. –  Asaf Karagila May 6 '12 at 17:40
    
@AsafKaragila - done. –  Belgi May 6 '12 at 18:17
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1 Answer

up vote 18 down vote accepted

Upon request, I am leaving an answer.

First, let's acknowledge that this is a question about conventions first and only secondarily about mathematics. Your lecturer's definition of a field homomorphism is whatever she says it is: that's what a definition means. ("When I use a word, it means just what I choose it to mean -- neither more nor less." -- Humpty Dumpty) On the other hand, definitions, while in some formal sense arbitrary, can certainly be good or bad, for both mathematical and sociological reasons.

Your lecturer's definition is a good one both because of its mathematical consequences and because, as she claimed, this is very much the "standard" definition, i.e., the one you will find in the vast majority of contemporary texts and courses, the one which is being used by practitioners both in field theory itself and in areas of mathematics where field theory is applied, and so forth.

The notion of "homomorphism" is so important in mathematics that there has been a lot of effort to make it as systematic and "un-(ad hoc)" as possible. Two very general frameworks for discussing homomorphisms are universal algebra and category theory. The basic objects of study in universal algebra are relational structures, namely sets equipped with various relations (and, what is a special case of that, functions), of various "arities".

From this perspective, the relational structure of a ring is a set $R$, two binary operations called $+$ and $\cdot$ and two "nullary functions" -- i.e., constants $0$ and $1$. There is more to the definition of a ring, namely the axioms that these relations must satisfy. In this sense, model theory is being overlayed on top of universal algebra. However, the definition of a homomorphism of relational structures doesn't depend on the axioms: a homomorphism of relational structures is just a map of the underlying sets which preserves, in a rather straightforward sense which I won't completely write out here (one can easily look it up online) all the relations. In the case of rings, this means that a homomorphism $f: R \rightarrow S$ is a map of the underlying sets such that

$\bullet$ For all $x,y \in R$, $f(x+y) = f(x) + f(y)$,
$\bullet$ For all $x,y \in R$, $f(x \cdot y) = f(x) \cdot f(y)$,
$\bullet$ $f(0) =0$, and
$\bullet$ $f(1) = 1$.

Again, I stress that this does not depend on the axioms (or, in more model-theoretic terminology, the theory, of the structure): a homomorphism of non-associative rings (with distinguished elements $0$ and $1$) is the same definition as a homomorphism of not necessarily commutative rings, which is the same definition as a homomorphism of fields. This uniformity is useful and powerful.

Now let $f: R \rightarrow S$ be a homomorphism of rings. An element $x \in R$ is a unit if there exists $y \in R$ such that $xy = yx = 1$. Two easy facts:

(i) For all $x \in R$, there is at most one $y \in R$ such that $xy = yx = 1$.
(Proof: If also $xz = zx = 1$, then $y = y \cdot 1 = y(xz) = (yx)z = 1 \cdot z = z$.)

Thus if $x$ is a unit, it has a well-defined inverse which we may denote $x^{-1}$. However, $x \mapsto x^{-1}$ is not a completely kosher function from the perspective of universal algebra, since it is not a function on $R$ but only on $R^{\times}$, the set of units of $R$. There are ways to get around this (a keyword is sorts) but the easiest solution is simply not to regard $x \mapsto x^{-1}$ is being part of the relational structure defining a commutative ring.

(ii) If $f: R \rightarrow S$ is a homomorphism of rings and $x \in R^{\times}$, then $f(x) \in S^{\times}$ and $f(x^{-1}) = f(x)^{-1}$.
(Proof: $f(x^{-1}) f(x) = f(x^{-1} x) = f(1) = 1 = f(1) = f(x x^{-1}) = f(x) f(x^{-1})$.)

Thus the preservation of inverses comes for free from this definition of ring homomorphisms.

There is also the categorical perspective, which is more general and thus more permissive. It has the bright idea that in the same breath as we define a mathematical object of a certain kind ("category"!), we should define the notion of homomorphism between objects of that category. In order to be a homomorphism, certain very mild axioms of composition and identities must be satisfied. So in particular, there is a category $\operatorname{Ring}$ in which the objects are rings and the morphisms are defined exactly as above. But there is also a category $\operatorname{Rng}$ in which the objects are "rngs", i.e., what you get by taking the constant $1$ out of the relational structure and taking out all the parts of the axioms for rings which refer to it. And there is even the category -- call it $\operatorname{Ring}'$ -- in which the objects are rings -- i.e., have $1$ -- but in which homomorphisms are not required to carry $1$ to $1$. In particular, for any rings $R$ and $S$, the identically zero map from $R$ to $S$ is a homomorphism in $\operatorname{Ring}'$ but not in $\operatorname{Ring}$ (except in the trivial case where $0 = 1$ in $S$).

It is worth spending a little time getting experience with the category $\operatorname{Ring}'$, both because (i) it is not simply a pathology: these sorts of maps between rings do come up sometimes and (ii) it gives you insight about the much more standard category $\operatorname{Ring}$. In particular, for a homomorphism $f: R \rightarrow S$ in $\operatorname{Ring}'$ $f(1)$ cannot be just any old element of $S$ but must be an idempotent element:

$f(1) = f(1 \cdot 1) = f(1) \cdot f(1)$.

Many rings have only $0$ and $1$ as idempotents -- in particular, this holds for fields -- so either $f(1) = 1$, in which case we're back to the standard case, or $f(1) = 0$, and in this case for all $x \in R$, $f(x) = f(x \cdot 1) = f(x) \cdot f(1) = f(x) \cdot 0 = 0$, so $f$ is identically zero.

So now we know exactly what we're excluding by defining a homomorphism of fields to satisfy $f(1) = 1$: we're excluding the identically zero map between fields. So we can be rather confident that we're not excluding anything important: certainly the zero map is not interesting in its own right, just as the identity element of a group or zero element of a ring is not really interesting in its own right. However, the identity element of a group and zero element of a ring are there for another very important reason: they enable the appropriate algebraic structure of group or ring to exist!

In contrast, if $R$ and $S$ are rings, then I see no intrinsic algebraic structure on the set of homomorphisms from $R$ to $S$: we cannot add them, multiply them (so as to get another homomorphism from $R$ to $S$) and so forth. So adding the zero element is not helping us in any evident way. In fact, if we don't believe this at first, no matter -- keep it in and see what happens. We'll find that in our further study of field theory, we'll keep having to say "except for the zero homomorphism" in results and arguments. For instance, most of field theory is based upon the fact that absolutely every field homomorphism $\iota: E \rightarrow F$ is injective and makes $F$ into an $E$-vector space...except for the zero homomorphism.

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