Define homomorphism of fields, and prove that every homomorphism of fields is injective.
I got stuck in the first part actually. I knew homomorphism of groups. How to extend that to a Field? How to use the equation $\varphi(ab)=\varphi(a)\varphi(b)$ (because this is true for only on one group action)?
Well, if $L$ and $K$ are two fields, a homomorphism $\phi:L \to K$ from $L$ to $K$ is first of all a function, but one which preserves the field operations in the sense that, if $a, b \in L$, then
$\phi(a + b) = \phi(a) + \phi(b) \tag 1$
and
$\phi(ab) = \phi(a) \phi(b). \tag 2$
In other words, if we think of a field such as $L$ as having two sub-groups, a multiplicative one and an additive one, then $\phi$ must act as a homomorphism of each of these (sub-)groups.
Because fields have so much structure, it is pretty easy to figure out how field homomorphisms behave. We want to show $\phi$ is injective. First of all, let's look at $\phi(0_L)$, where $0_L$ is the additive identity of $L$. We have, for any $a \in L$,
$\phi(a) + \phi(0_L) = \phi(a + 0_L) =\phi(a), \tag 1$
so
$\phi(0_L) = 0_K; \tag 2$
now if $r \in L$ and $r \ne 0_L$, and
$\phi(r) = 0_K, \tag 3$
then for any $s \in L$ we can write
$s = s1_L = s(r^{-1}r) = (sr^{-1})r, \tag 4$
so
$\phi(s) = \phi((sr^{-1})r) = \phi(sr^{-1}) \phi(r) = \phi(sr^{-1})0_K = 0_K, \tag 5$
which shows that
$\phi(s) = 0_K \tag 6$
for all $s \in L$; such a homomorphism doesn't do very much, so we dub it trivial.
For the remainder of this answer, we will only be concerned with non-trivial homomorphisms. From the above, we see that $r \ne 0_L$ implies
$\phi(r) \ne 0_K \tag 7$
for any non-trivial homomorphism $\phi$.
We also note that
$\phi(-r) = -\phi(r): \tag 8$
$\phi(r) + \phi(-r) = \phi(r + (-r)) = \phi(0_L) = 0_K, \tag 9$
whence
$\phi(-r) = -\phi(r) \tag{10}$
as claimed.
We now show such (non-trivial) $\phi$ are injective. If
$\phi(r_1) = \phi(r_2), \tag{11}$
then
$\phi(r_1 - r_2) = \phi(r_1 + (-r_2))$ $= \phi(r_1) + \phi(-r_2) = \phi(r_1) + (-\phi(r_2)) = \phi(r_1) - \phi(r_2) = 0_K; \tag {12}$
since $\phi$ is assumed non-trivial, we must have
$r_1 - r_2 = 0_L, \tag{13}$
whence
$r_1 = r_2, \tag{14}$
and thus $\phi$ is injective.
Note: some folks avoid the need to show $r \ne 0_L \rightarrow \phi(r) \ne 0_K$ by assuming $\phi(1_L) = 1_K$; observe that since $\phi(1_L) = \phi(1_L^2) = (\phi(1_L))^2$, we have either $\phi(1_L) = 1_K$ or $\phi(1_L) = 0_K$; in this latter case $\phi(s) = \phi(s 1_L) = \phi(s) \phi(1_L) = 0_K$, so such an assumption rules out the triviality of $\phi$ since it implies $1_K = \phi(1_L) = \phi(ss^{-1}) = \phi(s) \phi(s^{-1})$, which implies $\phi(s) \ne 0$ for $s \ne 0$. But the amount of work is about the same . . . also, in the above, I used the property $0s = 0$ without proof, but I'll leave that one to my audience to figure out. End of Note.
Generally, when we present an algebraic theory, we do so by listing a number of constants and operations.
For example, the theory of a (multiplicative) monoid consists of:
and we further require these to satisfy the monoid axioms.
When we do this, a homomorphism of models of this theory are required to preserve the constants and operations:
A monoid homomorphism $M \to N$ is a function $f$ on the underlying sets of elements that satisfies
- $f(1) = 1$
- $f(x \cdot y) = f(x) \cdot f(y)$
(where, e.g., the $1$ on the left hand side refers to the unit of $M$, and on the right the unit of $N$)
You're being asked to go through this routine procedure yourself (rather than having the book spell it out for you).
Do note that you should generally be wary of taking groups as an example to model things after; they (and their homomorphisms) are often introduced in a way that fails to convey the flavor of how these sorts of things work.