Brian Bi
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.