Brian Bi
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Return to table of contents for Brian's unofficial solutions to Artin's 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:

  1. \(\varphi\) is surjective;
  2. \(\ker \varphi\) 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.