Brian Bi
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Miscellaneous problems for Chapter 11

Problem 11.M.1 Suppose $$R$$ is a nonzero ring and $$a^2 = a$$ for all $$a \in R$$. Let $$a = 1 + 1$$. Then $$1 + 1 + 1 + 1 = a^2 = a = 1 + 1$$, so $$1 + 1 = 0$$. We conclude that $$R$$ has characteristic 2.

Problem 11.M.2 We may form fractions of $$S$$. Namely, let $$G$$ be the set of equivalence classes of $$S \times S$$ where the elements $$(a, b), (c, d)$$ are considered equivalent iff $$ad = bc$$. We may write $$a/b$$ for the equivalence class of $$(a, b)$$.

It is clear that this relation is symmetric since $$S$$ is commutative. To verify that it's really an equivalence relation, we need to show that it's also transitive. Suppose $$a/b = c/d$$ and $$c/d = e/f$$. We have $$ad = bc$$, so $$adef = bcef$$. But $$cf = de$$, so $$acff = bcef$$, and we can cancel $$cf$$ from both sides to give $$af = be$$. This shows that $$a/b = e/f$$.

We can endow $$G$$ with a law of composition by defining $$(a/b)(c/d) = ac/bd$$. To see that this is well-defined, suppose $$a'/b' = a/b, c'/d' = c/d$$; then $$(a'/b')(c'/d') = a'c'/b'd'$$, and $$ac/bd = a'c'/b'd'$$ since $$acb'd' = (ab')(cd') = (a'b)(c'd) = bda'c'$$. It is easy to see that the law of composition of $$G$$ is also commutative and associative.

The element $$1/1$$ serves as the identity of $$G$$. The element $$a/b$$ has inverse $$b/a$$. So $$G$$ is a group.

We can embed $$S$$ into $$G$$ by identifying $$a \in S$$ with the element $$a/1 \in G$$. Notice that if $$a/1 = b/1$$ then $$a1 = 1b$$, or $$a = b$$, so this mapping is injective as desired.

Problem 11.M.3 The set $$R$$ contains the sequences $$(0, 0, 0, \ldots)$$ and $$(1, 1, 1, \ldots)$$, which respectively serve as the additive identity and the multiplicative identity. It is closed under additive inverses since if $$a$$ is eventually constant then $$-a$$ is also eventually constant. Also, it is closed under addition since if $$a$$ is constant after term $$m$$ and $$b$$ is constant after term $$n$$ then $$a+b$$ is constant after term $$\max(m, n)$$. A similar argument applies for multiplication. It is obvious enough that the other ring axioms hold, as the addition and multiplication operations are those inherited from the superset consisting of all real sequences. So $$R$$ is a ring.

Consider the following subsets of $$R$$:

1. $$R_1, R_2, R_3, \ldots$$, where $$R_i = \{a \mid a \in R \wedge a_i = 0\}$$,
2. $$Z$$, consisting of all elements of $$R$$ that tend to zero.

It is easy to see that the $$R_i$$'s and $$Z$$ are ideals. It's also easy to see that the cosets of $$R_i$$ are just equivalence classes of sequences that are equal at the $$i$$th element, so $$R/R_i \cong \R$$. Likewise, the cosets of $$Z$$ are equivalence classes of sequences with the same limit, so $$R/Z \cong \R$$. As $$\R$$ is a field, this establishes that the ideals $$R_i$$ and $$Z$$ are maximal.

Now let $$I$$ be an arbitrary proper ideal of $$R$$. If all elements of $$I$$ tend to zero, then $$I \subseteq Z$$. Assume going forward that this is not the case. Choose $$a \in I$$ that doesn't tend to zero. Suppose that $$a \to x$$, so that there is some $$n$$ such that for all $$i \ge n$$, $$a_i = x$$. For each $$i = 1, 2, \ldots, n - 1$$, let $$b^i$$ be an element of $$I$$ such that $$b^i_i \ne 0$$. If all such $$b^i$$ exist, then the element $$a^2 + (b^1)^2 + \ldots + (b^{i-1})^2$$ is an element of $$I$$ with all entries nonzero, which is a unit. But we assumed $$I$$ to be proper, so this cannot be. Therefore, there is some $$i$$ such that $$b^i$$ doesn't exist; that is, all elements of $$I$$ have $$i$$th element equal to 0. Therefore, $$I \subseteq R_i$$. That is, every proper ideal of $$R$$ is contained in either $$Z$$ or one of the $$R_i$$'s, so these are all the maximal ideals of $$R$$.

Problem 11.M.4 I will assume that the statement $$R$$ contains $$\C$$ means there is an injective ring homomorphism from $$\C$$ to $$R$$.

1. Let $$\alpha \in R$$ be such that $$\{1, \alpha\}$$ is linearly independent, i.e., $$\{1, \alpha\}$$ is a basis of $$R$$. Therefore $$\alpha^2 = p\alpha + q$$ for some $$p, q \in \C$$. Let $$\beta = \alpha - p/2$$. Then $$\beta^2 = \alpha^2 - p\alpha + p^2/4 = q + p^2/4$$. If this is zero, then the ring $$R$$ is isomorphic to $$\C[x]/(x^2)$$ where $$\beta$$ is the residue of $$x$$. If this is nonzero, let $$\gamma = \beta/\sqrt{q + p^2/4}$$ so that $$\gamma^2 = 1$$. Then the ring $$R$$ is isomorphic to $$\C[x]/(x^2 - 1)$$ where $$\gamma$$ is the residue of $$x$$.

So in general $$R$$ is always isomorphic to either $$\C[x]/(x^2)$$ or $$\C[x]/(x^2 - 1)$$. But are these rings isomorphic to each other? The answer is no, because $$\C[x]/(x^2 - 1)$$ contains a nontrivial idempotent element, namely $$\frac{1}{2}(1 + x)$$ (which induces a decomposition $$\C[x]/(x^2 - 1) \cong \C \times \C$$). On the other hand, we can easily check that $$\C[x]/(x^2)$$ contains no nontrivial idempotents (or, equivalently, is not a nontrivial product of rings). So there are exactly two isomorphism classes.

2. We consider a few cases, which might have some overlap, but we will address this at the end.

• Case 1: $$R$$ contains a nontrivial idempotent element, $$e \in R \setminus \{0, 1\}, e^2 = e$$. The ring homomorphism $$\varphi : R \to eR$$ with $$\varphi(x) = ex$$ is also a homomorphism of $$\C$$-vector spaces, and Proposition 11.6.2(c) tells us that $$eR$$ is not the entire ring $$R$$. Nor can $$eR$$ be the zero ring since $$e = e1 \in eR$$. Therefore, $$eR$$ is a subspace of $$R$$ of dimension 1 or 2, and $$R$$ decomposes as a nontrivial direct product $$R \cong eR \times (1-e)R$$. If $$eR$$ has dimension 2, then replace $$e$$ with $$1-e$$, so that $$eR$$ has dimension 1. Now we have the decomposition $$R \cong \C \times R'$$ where $$R'$$ has dimension 2. By part (a), we conclude that either $$R \cong \C \times \C[x]/(x^2)$$ or $$R \cong \C^3$$.

• Case 2: There exists $$\alpha \in R$$ such that $$\{1, \alpha, \alpha^2\}$$ is a basis for $$R$$. Therefore, $$R \cong \C[x]/(P(x))$$ where $$P$$ is some monic polynomial of degree 3. Factorize $$P(x)$$ as $$(x-a)(x-b)(x-c)$$. If $$P(x)$$ is a perfect cube, then $$R \cong \C[x]/(x^3)$$. Otherwise, without loss of generality, suppose $$a$$ is unequal to both $$b$$ and $$c$$. Write $$\alpha$$ for the residue of $$x$$ in $$\C[x]/(P(x))$$. Let the ideals $$I$$ and $$J$$ of $$\C[x]/(P(x))$$ be respectively generated by $$(\alpha-a)$$ and $$(\alpha-b)(\alpha-c)$$. It's not hard to see that $$I + J = R$$ and that $$IJ = 0$$, so by the Chinese Remainder Theorem, we obtain $$R \cong (R/(\alpha-a)) \times (R/((\alpha-b)(\alpha-c)))$$. The factor $$R/(\alpha-a)$$ is isomorphic to $$\C$$, so as in case 1, we must have $$R \cong \C \times \C[x]/(x^2)$$ or $$\R \cong \C^3$$.

• Case 3: Neither case 1 nor case 2 holds. Let $$\{1, \alpha, \beta\}$$ be a basis of $$R$$. By the argument in part (a), we can always choose $$\alpha$$ and $$\beta$$ such that $$\alpha^2$$ and $$\beta^2$$ are each either 0 or 1. But if $$\alpha^2 = 0$$ or $$\beta^2 = 0$$ then either $$\frac{1}{2}(1+\alpha)$$ or $$\frac{1}{2}(1+\beta)$$ would be a nontrivial idempotent. This was already covered in case 1, so we proceed assuming that $$\alpha^2 = \beta^2 = 0$$. Write $$\alpha\beta = p\alpha +q\beta + r$$ with $$p, q, r \in \C$$. By multiplying both sides by $$\alpha$$ we obtain $$q\alpha\beta + r\alpha = 0$$. Likewise if we chose to multiply both sides by $$\beta$$ we would obtain $$p\alpha\beta + r\beta = 0$$. We can now consider a few subcases:

• Case 3a: $$r \ne 0$$. Then $$\alpha = -\frac{q}{r}\alpha\beta$$ and $$\beta = -\frac{p}{r}\alpha\beta$$. This contradicts the assumption of linear independence.
• Case 3b: $$r = 0$$ and $$p = q$$ with $$p \ne 0$$. Then $$\alpha\beta = p(\alpha+\beta)$$. Then $$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 = 2p(\alpha+\beta)$$. By appropriately rescaling $$\alpha$$ and $$\beta$$ we can force $$\alpha +\beta$$ to be a nontrivial idempotent. However, this was considered in case 1.
• Case 3c: Otherwise, either we have $$r = 0$$ and $$p \ne q$$, in which case $$(p-q)\alpha\beta = 0$$ implying $$\alpha\beta = 0$$, or else $$p = q = r = 0$$ and again $$\alpha\beta = 0$$. This completely determines the operations of $$R$$. In particular, $$R \cong \C[x,y]/(x^2,y^2,xy)$$.

There are therefore up to four isomorphism classes that $$R$$ might belong to, namely $$R_1 = \C^3$$, $$R_2 = \C \times \C[x]/(x^2)$$, $$R_3 = \C[x]/(x^3)$$, and $$R_4 = \C[x,y]/(x^2,y^2,xy)$$. We can verify by direct calculation that $$R_3$$ and $$R_4$$ don't contain any nontrivial idempotents (the details are omitted here), so $$R_3$$ and $$R_4$$ are not isomorphic to $$R_1$$ and $$R_2$$. Since $$\C[x]/(x^2)$$ contains no nontrivial idempotents, $$R_2$$ has a total of four idempotent elements, namely $$(0, 0), (0, 1), (1, 0), (1, 1)$$, while $$R_1$$ has eight, so $$R_1$$ and $$R_2$$ aren't isomorphic. Finally, by direct calculation, we can verify that $$R_3$$ contains only one linearly independent element that squares to zero, namely $$x^2$$, so $$R_3$$ and $$R_4$$ aren't isomorphic. We conclude that there are exactly four isomorphism classes of these rings (which are better known as commutative algebras of dimension 3 over $$\C$$).

(To contribute a better solution to this problem, email me.)

Problem 11.M.5 Let $$P \in \C[x,y]$$ be given by $$\sum_{i,j} c_{ij}x^i y^j$$. We then have $P(x, 0) = \sum_i c_{i0} x^i; \quad P(0, y) = \sum_j c_{0j} y^j; \quad P(t, t) = \sum_k \sum_{i+j=k} c_{ij} t^k$ Clearly all three of $$f(x, 0), f(0, y), f(t, t)$$ have the same constant term, namely $$c_{00}$$. Also, the linear coefficient of $$f(t, t)$$ is $$c_{01} + c_{10}$$, which is the sum of the linear coefficients of $$f(x, 0)$$ and $$f(0, y)$$.

We now claim that $$(Q, R, S) \in \im \varphi$$ for all $$Q \in \C[x], R \in \C[y], S \in \C[t]$$ as long as the aforementioned conditions are satisfied. Let $$Q, R, S$$ be given with $Q(x) = \sum_i q_i x^i; \quad R(y) = \sum_j r_j y^j; \quad S(t) = \sum_k s_k z^k;$ where $$q_0 = r_0 = s_0$$ and $$s_1 = q_1 + r_1$$. Choose $$P_0 \in \C[x, y]$$ as follows: $P_0(x, y) = q_0 + \sum_{i=1}^\infty q_i x^i + \sum_{j=1}^\infty r_j x^j$ It is easy to see that $$P_0(x, 0) = Q(x)$$ and $$P_0(0, y) = R(y)$$. Also, $$P_0(t, t) = q_0 + (q_1 + r_1)t + \ldots$$, so $$P_0(t, t)$$ already agrees with $$S(t)$$ at the constant and linear terms. Now let $P(x, y) = P_0(x, y) + \sum_{k=2}^\infty (s_k - q_k - r_k)xy^{k-1}$ Then $$P(x, 0) = P_0(x, 0) = Q(x); P(0, y) = P_0(0, y) = R(x)$$; and $$P(t, t) = P_0(t, t) + \sum_{k=2}^\infty (s_k - q_k - r_k)t^k = q_0 + (q_1 + r_1)t + \sum_{k=2}^\infty s_k t_k = S(t)$$. So our claim is proven.

We further claim that $$\ker \varphi$$ is the principal ideal generated by $$xy(x-y)$$. It's easy to see that $$\varphi(xy(x-y)) = 0$$. Suppose $$\varphi(P) = 0$$. Then the constant and linear terms of $$P$$ must all vanish. The $$x^2$$ and $$y^2$$ terms must vanish too, and so $$P(x, y)$$ equals $$c_{11} xy$$ plus terms of degree at least 3. Since $$P(t, t) = 0$$, this implies that $$c_{11} = 0$$ also. So $$P$$ consists of terms of degree at least 3, which must be mixed, that is, not be of the form $$x^k$$ or $$y^k$$. Thus $$P' = P/(xy)$$ is a polynomial, and $$P'(t, t) = 0$$ for all $$t \ne 0$$. Perform division in $$x$$ of $$P'$$ by $$x - y$$ to obtain $$P'(x, y) = (x-y)Q(x, y) + R(x, y)$$, where $$R$$ has degree 0 in $$x$$, so $$R(x, y) = R(y)$$. Then for all $$t \ne 0$$, we have $$0 = P'(t, t) = R(t)$$, so $$R(y)$$ is the zero polynomial, and $$x - y \divides P'(x, y)$$. Therefore $$xy(x-y) \divides P(x, y)$$, and the claim is proven.

Problem 11.M.6 Suppose $$P(x, y) \in \C[x, y]$$ satisfies $$P(x, \sin x) = 0$$ for each $$x \in \C$$. For each $$a \in [-1, 1]$$, the curve $$y = \sin x$$ intersects the line $$y = a$$ an infinite number of times, so $$P$$ has an infinite number of common zeros with the polynomial $$y - a$$. By Theorem 11.9.10, $$y - a \divides P(x, y)$$ for all $$a \in [-1, 1]$$, which is only possible if $$P$$ is the zero polynomial.

Problem 11.M.7

1. Let $$f = f_1^2 + \ldots + f_n^2$$. Since $$f_1, \ldots, f_n$$ have no common zero, it follows that $$f$$ is everywhere nonvanishing, and is therefore a unit in $$R$$. This implies that $$f_1, \ldots, f_n$$ generate the unit ideal of $$R$$.
2. Let $$J_c \subseteq R$$ be the ideal consisting of the functions that vanish at $$c$$. Then $$J_c$$ contains some functions that vanish only at $$c$$, such as $$f(x) = x - c$$. Together with the result of part (a), this implies that $$J_c$$ is a maximal ideal.

Now let $$J \subseteq R$$ be an ideal that isn't contained in any of the $$J_c$$. That is, $$J$$ is nonzero and the elements of $$J$$ don't have any common zero. So for each $$x \in X$$, there is some $$f_x \in J$$ with $$f_x(x) \ne 0$$. Since $$f_x$$ is continuous, there is an open interval containing $$x$$, call it $$I_x$$, such that $$f_x$$ is nonvanishing on $$I_x$$. The collection $$\{I_x \mid x \in X\}$$ is an open cover of $$X$$, and since $$X$$ is compact, we may select finitely many of the $$I_x$$'s that cover $$X$$, say, $$I_{x_1}, \ldots, I_{x_n}$$. Then the finite subset of $$J$$ given by $$\{f_{x_1}, \ldots, f_{x_n}\}$$ has no common zero, so $$J$$ is the unit ideal.

Since every ideal of $$R$$ that isn't contained in any $$J_c$$ is the unit ideal, we conclude that the $$J_c$$'s ($$c \in X$$) are all the maximal ideals of $$R$$, or, as the problem statement suggests, there is a natural bijection between the maximal ideals of $$R$$ and the points of $$X$$.

Remark 1: This proof depends crucially on the compactness of $$X$$. It works if $$X$$ is replaced by any compact subset of $$\R$$, but the statement is false whenever $$X$$ fails to be compact. If $$X$$ is not closed, let $$p$$ be some limit point of $$X$$ that isn't contained in $$X$$; then then the proper ideal consisting of all $$f$$ such that $$\lim_{x\to p}f(x) = 0$$ is a proper ideal that isn't contained in any of the $$J_c$$. Or if $$X$$ is not bounded above, the proper ideal consisting of all $$f$$ with $$\lim_{x\to\infty}f(x) = 0$$ isn't contained in any of the $$J_c$$; a similar statement holds when $$X$$ isn't bounded below. For example, if $$X$$ is the open unit interval, then $$\{f \mid \lim_{x\to 0}f(x) = 0\}$$ is a maximal ideal that isn't equal to any of the $$J_c$$.

Remark 2: In order to form the collection $$\{I_x \mid x \in X\}$$ we are implicitly invoking the full axiom of choice. However, a proof that only uses countable choice can be found here.