MHB Prime Field of a Field k - Rotman Theorem 3.110

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.8 Quotient Rings and Finite Fields ...

I need help with an aspect of the proof of Theorem 3.110.

Theorem 3.110 and its proof read as follows:View attachment 4693In the above proof, we read the following ...

" ... ... Since $$k$$ is a field, $$\text{ I am } \chi$$ is a domain ... ..."Can someone please explain why/how it follows that $$\text{ I am } \chi$$ is a domain ... ... Hope someone can help ...

Peter

 

[Deveno]

$\text{im }\chi$ is a subring of a field. Since the field it is contained in has no zero-divisors (except $0$, which doesn't count as a zero-divisor), any subring of it must not have any zero-divisors, either.

Since the multiplicative group (of $k^{\ast}= U(k)$) is abelian, this subring is a commutative ring without zero-divisors, that is, an integral domain.

Note that the ring-homomorphism:

$\chi: \Bbb Z \to R$ given by $\chi(n) = \sum\limits_{i = 1}^n 1_R$ is uniquely determined for any ring with unity $R$, another way of expressing this is that $\Bbb Z$ is initial in the category of rings with unity (with morphisms ring-homomorphisms that preserve unity).

This is a fancy way of saying $\Bbb Z$ is the "most basic ring with unity", and all other rings with unity depend on its structure, in some way.

I find the following proof of this theorem more "intuitive":

Let $c$ be the characteristic of a field $F$. Then either $c = 0$, or $c = p$, a prime integer.

If $c = 0$ (that is, all sums of $1_F$ are *distinct*), then there is nothing to prove. So suppose $c = n > 0$.

If $n = ab$ with $1 < a,b < n$, we have:

$(a\cdot 1_F)(b\cdot 1_F) = \left(\sum\limits_{i = 1}^a 1_F\right)\left(\sum\limits_{j= 1}^b 1_F\right)$

$= \sum\limits_{k = 1}^{ab} 1_F = \sum\limits_{k = 1}^n 1_F = 0$ (by the definition of characteristic).

Thus $a\cdot1_F$ is a zero-divisor in the field (which is also an integral domain) $F$, a contradiction.

Hence $n$ cannot be composite. Since $0_F \neq 1_F$ in a field, $n$ is prime.

Now the prime field of a field is the smallest field contained within the original field (the field *generated* by $1_F$).

If $c = 0$, then we have an infinite cyclic subgroup of $(F,+)$, which can be made into a ring by using the laws of exponents (written as "multiples" when we have an abelian group) for groups. It is not hard to check that this must agree with the multiplication on $F$ (use the field axioms).

(An aside: given any cyclic group, we have a natural $\Bbb Z$-action: $n\cdot g = g^n$, which turns our (cyclic, thus abelian) group into a $\Bbb Z$-module. This works for direct products of cyclic groups, as well, and so can be extended to any finitely-generated abelian group. In fact, it turns out that this works for any abelian group, since $g \mapsto g^n$ is an abelian group endomorphism. This lies behind the statement: "abelian groups are just $\Bbb Z$-modules").

This ring (generated by $1_F$) is clearly the smallest subring of $F$ containing $1_F$, which any subfield must, and is also thus contained in the smallest subfield of $F$. It is not hard to show that if $c = 0$, that $\langle 1_F\rangle \cong \Bbb Z$ *as rings* (and not merely abelian groups).

In the $c = 0$ case, we have that $Q(\langle 1_F\rangle)$, the field of fractions of the ring $\langle 1_F\rangle$ is the smallest field contained in $F$.

On the other hand, if $c = p$, a prime, then $\langle 1_F \rangle$ is already a field, and since it is minimal as a ring containing $1_F$, it is minimal as a subfield (the subfields of a field are a subset of the set of subrings of a field, which respects the partial order given by set-inclusion).

 

[Math Amateur]

$\text{im }\chi$ is a subring of a field. Since the field it is contained in has no zero-divisors (except $0$, which doesn't count as a zero-divisor), any subring of it must not have any zero-divisors, either.

Since the multiplicative group (of $k^{\ast}= U(k)$) is abelian, this subring is a commutative ring without zero-divisors, that is, an integral domain.

Thanks Deveno ...

Just a clarification ... I can see that the multiplicative group of the non-zero elements of the field k is abelian ... but how are we sure that the sub-ring is therefore abelian ... how exactly does this follow ...

Can you help further regarding this point ...

PeterEDIT ... Oh ... just thought ... ... any two elements of the sub-ring are elements of the field and hence commute ... is that correct?

 

[Deveno]

Thanks Deveno ...

Just a clarification ... I can see that the multiplicative group of the non-zero elements of the field k is abelian ... but how are we sure that the sub-ring is therefore abelian ... how exactly does this follow ...

Can you help further regarding this point ...

PeterEDIT ... Oh ... just thought ... ... any two elements of the sub-ring are elements of the field and hence commute ... is that correct?

Yes, a sub-monoid of an abelian group is always commutative.

 

[Euge]

$\chi: \Bbb Z \to R$ given by $\chi(n) = \sum\limits_{i = 1}^n 1_R$ is uniquely determined for any ring with unity $R$, another way of expressing this is that $\Bbb Z$ is initial in the category of rings with unity (with morphisms ring-homomorphisms that preserve unity).

This is a small technicality, but although the equation for $\chi(n)$ is true for $n \ge 0$, $\chi(n) = \sum\limits_{i = 1}^{-n} (-1_R)$ if $n \le 0$.

 

[Deveno]

This is a small technicality, but although the equation for $\chi(n)$ is true for $n \ge 0$, $\chi(n) = \sum\limits_{i = 1}^{-n} (-1_R)$ if $n \le 0$.

Of course, how silly of me.