*Algebra*

## Section 11.7. Fractions

Exercise 11.7.1 Let \(R\) be an integral domain, and let \(a \in R\) be nonzero. Let \(\varphi_a : R \to R\) be the map such that \(\varphi_a(x) = ax\). The cancellation law (11.7.1) implies that \(\varphi_a\) is injective, but since \(R\) is finite, this implies that \(\varphi_a\) is also surjective. The element \(\varphi_a^{-1}(1)\) is then the inverse of \(a\). Since all nonzero elements have inverses by this argument, \(R\) is a field.

Exercise 11.7.2 Suppose \(R\) is an integral domain. Let \(f, g \in R[x]\) be nonzero. Write \(m = \deg f, n = \deg g\) so that \(f = a_0 + a_1 x + \ldots + a_m x^m\) and \(g = b_0 + b_1 x + \ldots + b_n x^n\) with \(a_m, b_n\) nonzero. Then the coefficient of the \(x^{m+n}\) term in \(fg\) is \(a_m b_n\), which is nonzero since \(R\) is an integral domain. That is \(\deg fg = \deg f + \deg g\). So \(fg\) is also nonzero. This establishes that \(R[x]\) is an integral domain.

Since \(\deg fg = \deg f + \deg g\), it follows that only the polynomials of degree 0 can be units in \(R[x]\). It is also clear that in order to get \(fg = 1\) where \(f\) has degree 0, \(g\) must also have degree 0. So \(fg = 1\) in \(R\), requiring that \(f\) is a unit in \(R\). This condition is also obviously sufficient. So the units of \(R[x]\) are precisely the constant polynomials where the constant term is a unit in \(R\).

Exercise 11.7.3 No. Exercise 11.7.1 implies
that an integral domain with 15 elements would be a field, but as 15 is not a
prime power, there is no field of order 15. *(Remark: The proof that a
finite field must have prime power order hasn't been covered at this point in
the text, but it's easy. A finite field \(F\), being an integral domain, must
have prime characteristic \(p\), so it contains \(\mathbb{F}_p\) as a
subfield and therefore has the structure of a vector space over
\(\mathbb{F}_p\). If the dimension is \(e\), then \(|F| = p^e\).)*

Exercise 11.7.4 Denote the fraction field of \(F[[x]]\) by \(G\). As \(F[[x]]\) embeds into \(G\) in the natural way, there is a unique ring homomorphism \(\varphi : F[[x]][t] \to G\) such that \(\varphi(f) = f/1\) whenever \(f \in F[[x]]\), and \(\varphi(t) = 1/x\). We would like to verify that:

- \(\varphi\) is surjective;
- \(\ker \varphi\) is the principal ideal generated by \(xt - 1\).

Consider an element of \(G\), which we can write in the form \(f/g\) with \(f, g \in F[[x]], g \ne 0\). Let \(i\) be the smallest number such that the coefficient of the \(x^i\) term in \(g\) is nonzero. Since \(g/x^i\) has a nonzero constant term and \(F\) is a field, by Exercise 11.2.2, \(g/x^i\) has an inverse in \(F[[x]]\), call it \(h\). Then \(f/g = f/(x^i (g/x^i)) = fh/(x^i h (g/x^i)) = fh/x^i\). Therefore \(f/g = \varphi(fht^i)\). This establishes that \(\varphi\) is surjective.

Now \(\varphi(xt - 1) = \varphi(x)\varphi(t) - 1 = (x/1)(1/x) - 1 = 0\), so \((xt - 1) \subseteq \ker \varphi\). Conversely, suppose we have \(f \in F[[x]][t]\) such that \(\varphi(f) = 0\). We will show that \(f \in (xt - 1)\). The proof is by induction in the \(t\)-degree of \(f\):

- Base case: \(f\) has degree 0 in \(t\). That is, \(f = f_0\) where \(f_0 \in F[[x]]\). Then \(\varphi(f) = f_0\), so \(f_0 = 0\), so \(f = 0\). Therefore \(f \in (xt - 1)\).
- Inductive case: Suppose the claim holds for all \(t\)-degrees less than \(n\) and suppose \(f = f_0 + f_1 t + \ldots + f_n t^n\) where \(f_i \in F[[x]]\) for each \(i\). Then, \(\varphi(f) = f_0 + f_1/x + \ldots + f_n/x^n = (x^n f_0 + x^{n-1} f_1 + \ldots + f_n)/x^n\). Since this is zero, it follows that \(x^n f_0 + x^{n-1} f_1 + \ldots + f_n = 0\). This implies that \(f_n = -xg\) where \(g = x^{n-1} f_0 + x^{n-2} f_1 + \ldots + f_{n-1}\). Let \(f' = f + (xt - 1)t^{n-1}g\). Then the coefficient of \(t^n\) in \(f'\) is \(f_n + xg = 0\), so \(f'\) has \(t\)-degree strictly less than \(n\). By the induction hypothesis, \(f' \in (xt - 1)\), therefore \(f = f' - (xt - 1)t^{n-1}g \in (xt - 1)\) also.

Having established that \(\varphi\) is surjective and that \(\ker \varphi = (xt - 1)\), we conclude that the ring obtained by adjoining an inverse to \(F[[x]]\), that is, \(F[[x]][t]/(xt - 1)\), is isomorphic to \(G\), the fraction field of \(F[[x]]\).

The field \(F[[x]][t]/(xt - 1)\) can alternatively be identified with the
field of *formal Laurent series* in \(x\), that is, power series in
\(x\) with at most finitely many terms of negative degree; let us call this
field \(L\). The map \(\gamma : F[[x]][t] \to L\) is defined very similarly to
\(\varphi\): let \(\gamma(f) = f\) for all \(f \in F[[x]]\) as \(F[[x]\) embeds
into \(L\) in a natural way, and let \(\gamma(t) = x^{-1}\); this uniquely
defines the ring homomorphism \(\gamma\). The proof that \(\gamma\) is
surjective is easy; the proof that \(\ker \gamma = (xt - 1)\) parallels the
argument given above. So \(F[[x]][t]/(tx - 1) \cong L\).

Exercise 11.7.5 This exercise appears to be tedious and mechanical, so I have skipped it.