Brian Bi

## Section 11.3. Homomorphisms and Ideals

Exercise 11.3.1 Let $$I$$ be an ideal of $$R$$. By definition, $$I$$ is closed under addition. Also, either $$0 \in I$$ or $$I$$ contains some nonzero element $$x$$, whence $$0x = 0$$ is in $$I$$, so either way, $$0 \in I$$; $$I$$ contains the identity of $$R^+$$. Finally, let $$x \in I$$. Then $$(-1)x \in I$$, which is the inverse of $$x$$ in $$R^+$$. Therefore $$I$$ is a subgroup of $$R^+$$.

Exercise 11.3.2 Let $$I$$ be a nonzero ideal of $$\mathbb{Z}[i]$$ and let $$a + bi \in I$$ be nonzero. Then $$a^2 + b^2 = (a - bi)(a + bi) \in I$$, which is nonzero.

Exercise 11.3.3

1. This map sends all polynomials to their constant terms. It is therefore the set of polynomials with zero constant term. It is easy to see that this ideal is generated by $$x$$ and $$y$$.
2. Let $$I$$ denote the ideal so defined. $$I$$ is the set of real polynomials that have $$2 + i$$ as a zero. Since $$\mathbb{R}$$ is a field, Proposition 11.3.22 applies, and $$I$$ is generated by the monic polynomial of lowest degree that it contains. Clearly $$I$$ doesn't contain any linear polynomials, but it does contain the quadratic polynomial $$(x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5$$. Therefore $$x^2 - 4x + 5$$ generates $$I$$.
3. Denote the given ideal by $$I$$. $$I$$ is the set of all integer polynomials that have $$1 + \sqrt{2}$$ as a zero. Let $$I_{\mathbb{Q}}$$ be the ideal of rational polynomials that have $$1 + \sqrt{2}$$ as a zero. Evidently, $$I \subseteq I_{\mathbb{Q}}$$. By Proposition 11.3.22, $$I_{\mathbb{Q}}$$ is generated by the monic polynomial of lowest degree that it contains. Clearly $$I_{\mathbb{Q}}$$ doesn't contain any linear polynomials, but it does contain the quadratic polynomial $$(x - (1 + \sqrt{2}))(x - (1 - \sqrt{2})) = x^2 - 2x - 1$$. Therefore every element $$f \in I_{\mathbb{Q}}$$ can be written as $$f = (x^2 - 2x - 1)g$$ where $$g \in \mathbb{Q}[x]$$.

Let $$I'$$ denote the ideal of $$\mathbb{Z}[x]$$ generated by $$x^2 - 2x - 1$$. Clearly $$I' \subseteq I$$.

Suppose $$f \in I$$. Then $$f \in I_{\mathbb{Q}}$$, so we can write $$f = (x^2 - 2x - 1)g$$, that is, $$x^2 - 2x - 1$$ divides $$f$$ in $$\mathbb{Q}[x]$$. Since $$f$$ has integer coefficients and $$x^2 - 2x - 1$$ is monic, Lemma 11.3.24 applies, and $$x^2 - 2x - 1$$ divides $$f$$ in $$\mathbb{Z}[x]$$, that is, $$g \in \mathbb{Z}[x]$$. This establishes that $$f \in I'$$. Since this holds for all $$f \in I$$, we see that $$I \subseteq I'$$.

We conclude that $$I = I'$$, that is, $$I$$ is generated by $$x^2 - 2x - 1$$.

4. By Proposition 11.3.10, this homomorphism must send the constant polynomials to themselves. Proposition 11.3.4 then applies, and the map in question is the map that sends the polynomial $$f$$ to the value $$f(\sqrt{2} + \sqrt{3})$$. Thus, if we can find the monic polynomial $$P \in \mathbb{Q}[x]$$ of minimum degree that has $$\sqrt{2} + \sqrt{3}$$ as a zero, and $$P$$ turns out to have integer coefficients, then $$P$$ generates the ideal in question, by the reasoning applied in part (c).

In order to find $$P$$, we first observe that $$\mathbb{Q}[\sqrt{2}, \sqrt{3}]$$ has the following endomorphisms:

• $$\varphi_2(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}$$
• $$\varphi_3(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}$$
• $$\varphi_{23}(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a - b\sqrt{2} - c\sqrt{3} + d\sqrt{6}$$

If $$\varphi$$ is an endomorphism of a ring $$R$$, it is easy to see that for all $$f \in R[x], x \in R$$, we have $$\varphi(f(x)) = \varphi(f)(\varphi(x))$$, where $$\varphi(f)$$ is defined as the polynomial obtained by replacing each coefficient $$c_i$$ of $$f$$ by $$\varphi(c_i)$$.

Suppose $$f \in \mathbb{Q}[x]$$ and $$f(\sqrt{2} + \sqrt{3}) = 0$$. If we regard $$f$$ as an element of $$\mathbb{Q}[\sqrt{2}, \sqrt{3}][x]$$, then $$0 = \varphi_2(f(\sqrt{2} + \sqrt{3})) = \varphi_2(f)(\varphi_2(\sqrt{2} + \sqrt{3})) = f(-\sqrt{2} + \sqrt{3})$$. Similar logic using $$\varphi_3$$ and $$\varphi_{23}$$ shows that $$f(\sqrt{2} - \sqrt{3}) = f(-\sqrt{2} -\sqrt{3}) = 0$$. Therefore $$P$$ must have degree at least 4. In fact, the product of the four monomials $$x - (\pm \sqrt{2} \pm \sqrt{3})$$ is $$x^4 - 10x^2 + 1$$, which is monic and has integer coefficients; therefore this is the polynomial we are looking for. We conclude that the ideal is generated by $$x^4 - 10x^2 + 1$$.

5. Although this isn't explicitly stated, it seems we're supposed to assume that the map in question sends the constant polynomials to themselves, so that we can apply Proposition 11.3.4 as in part (d). Let $$I$$ be the ideal in question, and let $$I'$$ be the ideal of $$\mathbb{C}[x, y, z]$$ generated by $$y - x^2$$ and $$z - x^3$$. It is easy to see that $$I' \subseteq I$$.

Suppose $$f \in I$$. We can do division in the variable $$z$$ to write $$f(x, y, z) = (z - x^3)g(x, y, z) + h(x, y, z)$$ where the remainder $$h$$ must have degree 0 in $$z$$, that is, it is a polynomial of $$x$$ and $$y$$ only. We can then do division in the variable $$y$$ to write $$h(x, y) = (y - x^2)i(x, y) + j(x, y)$$ where $$j$$ must have degree 0 in $$y$$, that is, it is a polynomial in $$x$$ only. So $$f(x, y, z) = (z - x^3)g(x, y, z) + (y - x^2)i(x, y) + j(x)$$. Evidently $$f(t, t^2, t^3) = (t^3 - t^3)g + (t^2 - t^2)i + j(t)$$, and since $$f = 0$$, it follows that $$j = 0$$. Therefore $$f = (z - x^3)g + (y - x^2)i$$ for some $$g \in \mathbb{C}[x, y, z], h \in \mathbb{C}[x, y]$$. Therefore $$f \in I'$$. Since this holds for all $$f \in I$$, we have $$I \subseteq I'$$

Since $$I' \subseteq I$$ and $$I \subseteq I'$$, we conclude $$I = I'$$, so $$I$$ is generated by $$z - x^3$$ and $$y - x^2$$.

Exercise 11.3.4 Again, although this isn't explicitly stated, I assume the intended map is the one that sends the constant polynomials to themselves, so that Proposition 11.3.4 applies, and the map in question is unique and given by substitution. Let $$P(x, y) = y - (x - 1)^3 + 1$$. Then $$\varphi(P) = t^3 - 1 - ((t + 1) - 1)^3 + 1 = 0$$. Let $$I = (P)$$. It is clear that $$I \subseteq \ker K$$. On the other hand, let $$f \in \ker K$$. Using polynomial division in $$y$$, write $$f(x, y) = P(x, y) g(x, y) + r(x, y)$$. Since $$P$$ has degree 1 in $$y$$, the remainder $$r$$ must have degree 0 in $$y$$, that is, $$r$$ is a polynomial in $$x$$ alone. Then $$0 = \varphi(f) = \varphi(P)\varphi(g) + r(t + 1) = r(t + 1)$$, so the polynomial $$r(t + 1)$$ is identically zero. Since the field is $$\mathbb{C}$$, this implies that $$r$$ is the zero polynomial. Therefore $$f = Pg$$. We conclude that $$\ker K = \{Pg \mid g \in \mathbb{C}[x, y]\}$$.

Let $$I'$$ be an ideal that contains $$\ker K$$. If $$I' = \ker K$$, then as we have established, $$I'$$ is generated by the element $$P = y - (x - 1)^3 + 1$$. Suppose on the other hand that $$I'$$ is strictly larger than $$\ker K$$. If $$f \in I' \setminus \ker K$$, then division by $$P$$ yields $$f = Pg + r$$ where $$r$$ is a nonzero polynomial of $$x$$ alone. Since $$Pg \in I'$$, it follows that $$r \in I'$$ also. In general, if $$S$$ is the set of remainders obtained by dividing elements of $$I'$$ by $$P$$, then this reasoning shows that $$S \subseteq I'$$. Observe that:

• If $$r \in S, f \in \mathbb{C}[x]$$, that is, there exists some $$g \in \mathbb{C}[x, y]$$ such that $$Pg + r \in I'$$, then since $$f\cdot(Pg + r) \in I'$$, we have $$P\cdot(fg) + fr \in I'$$, which establishes $$fr \in S$$.
• If $$r_1, r_2 \in S$$, that is, there exist $$g_1, g_2 \in \mathbb{C}[x, y]$$ such that $$Pg_1 + r_1, Pg_2 + r_2 \in I'$$, then since $$Pg_1 + r_1 + Pg_2 + r_2 \in I'$$, we have $$P\cdot(g_1 + g_2) + (r_1 + r_2) \in I'$$, which establishes $$r_1 + r_2 \in S$$.

The above results imply that $$S$$ is an ideal of $$\mathbb{C}[x]$$. By Proposition 11.3.22, there is some polynomial $$Q(x)$$ such that $$S = \{Qh \mid h \in \mathbb{C}[x]\}$$. Obviously $$Q \in S \subseteq I'$$, so $$(P, Q) \subseteq I'$$. Conversely, let $$f \in I'$$; write $$f = Pg + r$$, and since $$r \in S$$, we can write $$r = Qh$$, so $$f = Pg + Qh$$. Therefore $$I' \subseteq (P, Q)$$. We conclude that $$I' = (P, Q)$$, so at most two elements are needed to generate $$I'$$.

Exercise 11.3.5 Consider $$F[x]$$ as a vector space over $$F$$. One important property that the derivative operation is a linear endomorphism. We will not prove this here since the proof is easy, but we will use it in what follows.

1. Let $$f = x^a, g = x^b$$. Then $$(fg)' = (a+b)x^{a+b-1}$$, and $$f'g + fg' = ax^{a-1} x^b + x^a bx^{b-1} = (a+b)x^{a+b-1}$$, therefore $$(fg)' = f'g + fg'$$. Now let $$f$$ and $$g$$ be arbitrary elements of $$\mathbb{C}[x]$$ and let $$h(f, g) = (fg)' - f'g - fg'$$. By linearity of differentiation, it follows that $$h$$ is bilinear in its two arguments. We already established that $$h(x^a, x^b) = 0$$ for all monomials $$x^a, x^b$$, so $$h$$ vanishes on a basis. Therefore $$h$$ vanishes identically.
2. We'll first prove the claim for a restricted class of pairs of functions, those for which $$f$$ is of the form $$f(x) = x^a$$ while $$g \in \mathbb{C}[x]$$ is arbitrary. We'll proceed by induction on the degree $$a$$.

Base case: $$a = 0$$. Then $$(f \circ g)' = (1 \circ g)' = 1' = 0$$ while $$(f' \circ g)g' = (0 \circ g)g' = 0$$ so the two are equal.

Inductive case: $$a \ge 1$$. By the induction hypothesis, $$(g^{a-1})' = ((x \mapsto x^{a-1}) \circ g)' = ((x \mapsto (a-1)x^{a-2}) \circ g)g' = (a-1)g^{a-2}g'$$. Then, using this together with the product rule: \begin{align*} (f \circ g)' &= (g^a)' \\ &= (g g^{a-1})' \\ &= g' g^{a-1} + g (g^{a-1})' \\ &= g^{a-1} g' + g (a-1) g^{a-2} g' \\ &= a g^{a-1} g' \\ &= ((x \mapsto ax^{a-1}) \circ g) g' \\ &= (f' \circ g) g' \end{align*} as required.

Now suppose $$f, g \in \mathbb{C}[x]$$ are arbitrary. Define $$h(f, g) = (f \circ g)' - (f' \circ g)g'$$. We have already established that $$h$$ vanishes whenever $$f$$ is of the form $$x^a$$. Also, $$h$$ is linear in its first argument. Since $$\{x^a \mid a \in \mathbb{N}\}$$ is a basis for $$\mathbb{C}[x]$$, it follows that $$h$$ vanishes identically.

Exercise 11.3.6 Again, although it isn't explicitly stated, I assume the map in question is intended to send the constant polynomials to themselves, so that it acts by substitution. That is, for a given $$f$$, we are dealing with the unique endomorphism $$\varphi_f$$ that is the ideneity on $$R$$ and that sends $$x$$ to $$x + f(y)$$ and $$y$$ to $$y$$. We simply need to show that the map $$\varphi_f$$ is bijective.

We first show that $$\varphi_f$$ is injective. Suppose $$\varphi_f(g_1) = \varphi_f(g_2)$$. Since the map $$\varphi_f$$ is a ring endomorphism, we have $$\varphi_f(g_1 - g_2) = 0$$. Denote $$g_1 - g_2$$ by $$h$$. Suppose $$h = \sum_{i=0}^m \sum_{j=0}^n c_{ij} x^i y^j = \sum_{i=0}^m x^i P_i(y)$$, where we have defined $$P_i(y) = \sum_{j=0}^n c_{ij} y^j$$. Then $0 = \varphi_f(h) = \sum_{i=0}^m (x + f(y))^i P_i(y)$ If we group terms in the RHS by powers of $$x$$, we see that the coefficient of the $$x^m$$ term is $$P_m(y)$$. Therefore $$P_m(y) = 0$$, implying $$c_{mj} = 0$$ for all $$j$$, so that $$0 = \sum_{i=0}^{m-1} (x + f(y))^i P_i(y)$$. Repeating this for $$m-1, m-2, \ldots$$, we find that all $$P_i$$'s vanish, so $$h$$ vanishes. Therefore $$g_1 = g_2$$. This completes the proof that $$\varphi_f$$ is injective.

We now show that $$\varphi_f$$ is surjective. Let $$g(x, y)$$ be given and let $$h(x, y) = g(x - f(y), y)$$. Then $$\varphi_f(h) = h(x + f(y), y) = g((x + f(y)) - f(y), y) = g(x, y)$$.

We have shown that $$\varphi_f$$ is both surjective and injective, so we are done.

Exercise 11.3.7 By Proposition 11.3.10, any endomorphism of $$\mathbb{Z}[x]$$ must send the constant polynomials to themselves. By the subtitution principle, for each $$f \in \mathbb{Z}[x]$$, there is a unique endomorphism $$\varphi_f$$ on $$\mathbb{Z}[x]$$ that sends the constant polynomials to themselves and that sends $$x$$ to $$f(x)$$, which is given by substituting $$f(x)$$ for $$x$$; that is, $$\varphi_f(g) = g \circ f$$. It is clear that all endomorphisms of $$\mathbb{Z}[x]$$ are of this form, since each endomorphism must send $$x$$ to some polynomial. So we need to determine which $$\varphi_f$$ are bijective.

We can rule out all constant $$f$$, since the image of $$\varphi_f$$ would contain only constants.

In general, for nonzero $$f, g$$ the degree of $$g \circ f$$ is the product of the degrees of $$f$$ and $$g$$. Therefore if $$f$$ has degree 2 or greater, the image of $$\varphi_f$$ doesn't contain any linear polynomials. So we can rule out all such $$f$$ as well.

If $$f$$ is linear, $$f = ax + b$$, then the leading coefficient of $$g \circ f$$ will always be a multiple of $$a$$. So $$\varphi_f$$ will not be surjective unless $$a$$ is a unit. Thus, if $$\varphi_f$$ is to be an automorphism, we must have $$a = \pm 1$$.

Suppose $$f$$ is indeed of this form. If $$h \in \mathbb{Z}[x]$$ is given, then let $$g(x) = h(x/a - b/a)$$. Then $$g \in \mathbb{Z}$$ and $$\varphi_f(g) = g(ax + b) = h((ax + b)/a - b/a) = h(x)$$. This establishes that $$\varphi_f$$ is surjective.

Suppose $$g, h \in \mathbb{Z}$$ such that $$\varphi_f(g) = \varphi_f(h)$$. Then $$\varphi_f(g - h) = 0$$. Let $$i = g - h$$. Then we have that $$i(ax + b) = 0$$. If we consider $$i$$ as a polynomial over $$\mathbb{R}$$, then since $$i(ax + b) = 0$$ for all $$x \in \mathbb{Z}$$, either $$i$$ is identically zero or else $$i$$ has infinitely many zeroes, namely $$(n-b)/a$$ for each integer $$n$$. The latter is impossible, so $$i = 0$$. Therefore $$g = h$$, that is, $$\varphi_f$$ is injective. This completes the proof that $$\varphi_f$$ is bijective whenever $$f(x) = \pm x + b$$.

Thus, the endomorphisms of $$\mathbb{Z}[x]$$ are given by $$g \mapsto g \circ f$$ where $$f = \pm x + b$$ for some $$b \in \mathbb{Z}$$.

Exercise 11.3.8 Let $$\varphi$$ denote the map in question. We have the following:

• $$\varphi(1) = 1^p = 1$$
• $$\varphi(ab) = (ab)^p = a^p b^p = \varphi(a)\varphi(b)$$, where we have used the fact that multiplication in a ring is commutative
• $$\varphi(a + b) = (a + b)^p = \sum_{i=0}^p \binom{p}{i} a^i b^{p-i} = a^p + b^p = \varphi(a) + \varphi(b)$$ where we have used the fact that the binomial coefficients $$\binom{p}{i}$$ are divisible by $$p$$ whenever $$1 \le i \le p-1$$.
We conclude that $$\varphi$$ is a ring endomorphism.

Exercise 11.3.9

1. Suppose $$x^k = 0$$. Observe that $$(1 + x)(1 - x + x^2 - x^3 + \ldots + (-1)^{k-1} x^{k-1}) = 1 + (-1)^{k-1} x^k = 1$$. Therefore $$1 + x$$ is a unit.
2. Suppose $$a^k = 0$$. Let $$p^e$$ be a power of $$p$$ that is greater than or equal to $$k$$, so that $$a^{p^e} = 0$$. Observe that $$(1 + a)^p = \sum_{i=0}^p \binom{p}{i} a^i = 1 + a^p$$, where we have used the fact that the binomial coefficients $$\binom{p}{i}$$ are divisible by $$p$$ whenever $$1 \le i \le m - 1$$. Iterating this $$e$$ times, we obtain that $$(1 + a)^{p^e} = 1 + a^{p^e} = 1$$.

Exercise 11.3.10 Let $$I$$ be an ideal of $$F[[t]]$$. If $$I$$ is not the zero ideal, then choose $$P \in I \setminus \{0\}$$ such that $$P$$ has a term with nonzero coefficient with degree $$d$$ as small as possible. Thus, $$I \subseteq (x^d)$$. Observe that $$P/x^d \in F[[t]]$$ and has nonzero constant term, so according to Exercise 11.2.2, there exists $$Q \in F[[t]]$$ with $$(P/x^d)Q = 1$$. Therefore $$x^d = PQ \in I$$, so $$(x^d) \subseteq I$$. This establishes that $$I = (x^d)$$. Therefore, $$F[[t]]$$ has one ideal for each $$d$$, namely $$(x^d)$$, together with the zero ideal.

Exercise 11.3.11 This is clearly not true; in $$\mathbb{Z}[x]$$, the principal ideal generated by $$2x$$ has least degree 1, but it doesn't contain any monic polynomial of degree 1.

Exercise 11.3.12 Suppose $$z_1, z_2 \in I +J$$, that is, $$z_1 = x_1 + y_1, z_2 = x_2 + y_2$$, where $$x_1, x_2 \in I$$ and $$y_1, y_2 \in J$$. Then $$z_1 + z_2 = (x_1 + x_2) + (y_1 + y_2)$$, but $$x_1 + x_2 \in I$$ and $$y_1 + y_2 \in J$$, so $$z_1 + z_2 \in I + J$$. Now suppose $$z \in I + J$$, so that $$z = x + y$$ with $$x \in I, y \in J$$. If $$r \in R$$, then $$rz = rx + ry$$, but $$rx \in I$$ and $$ry \in J$$, so $$rz \in I + J$$. We conclude that $$I + J$$ is an ideal of $$R$$.

Exercise 11.3.13 Suppose $$z_1, z_2 \in I \cap J$$. Then $$z_1, z_2 \in I$$, so $$z_1 + z_2 \in I$$, and $$z_1, z_2 \in J$$, so $$z_1 + z_2 \in J$$. Therefore $$z_1 + z_2 \in I \cap J$$. Also, suppose $$z \in I \cap J, r \in R$$. Then $$z \in I$$ so $$rz \in I$$, and $$z \in J$$ so $$rz \in J$$, therefore $$rz \in I \cap J$$. This establishes that $$I \cap J$$ is an ideal of $$R$$.

Let $$I$$ be the ideal of $$\mathbb{R}[w, x, y, z]$$ generated by by the elements $$w, x$$ and let $$J$$ be the ideal generated by $$y, z$$. Then $$S = \{pq \mid p \in I, q \in J\}$$ contains the elements $$wy$$ and $$xz$$, but it doesn't contain their sum $$wy + xz$$, as the latter is irreducible while $$I$$ and $$J$$ don't contain any constant polynomials. So $$S$$ is not an ideal.

Suppose $$z, z' \in IJ$$. Write $$z = \sum_{i=1}^m x_i y_i, z' = \sum_{j=1}^n x_j' y_j'$$ where all $$x_i, x_j' \in I$$ and all $$y_i, y_j' \in J$$. Then $$z + z'$$ is a sum of $$m + n$$ terms that are each the product of an element of $$I$$ and an element of $$J$$, therefore $$z + z' \in IJ$$. Now suppose $$r \in R$$. Then $$rz = \sum_{i=1}^m r(x_i y_i) = \sum_{i=1}^m (rx_i) y_i$$. Since $$I$$ is an ideal, each $$rx_i$$ is in $$I$$. So $$rz$$ is a sum of $$m$$ terms that are each a product of an element of $$I$$ and an element of $$J$$, so $$rz \in IJ$$. We conclude that $$IJ$$ is an ideal of $$R$$.

In general, $$IJ \subseteq I \cap J$$; every element of $$IJ$$ is an element of $$I \cap J$$, but the converse isn't true. For example, if $$I = 4\mathbb{Z}, J = 6\mathbb{Z}$$, then $$IJ = 24\mathbb{Z}$$ and $$I \cap J = 12\mathbb{Z}$$, so $$IJ$$ is a strict subset of $$I \cap J$$. To see that $$IJ \subseteq I \cap J$$ in general, let $$z \in IJ$$ and write $$z = \sum_i x_i y_i$$ with $$x_i \in I, y_i \in J$$. Then $$z \in I$$ since $$z$$ is a combination of the $$x_i$$'s with the coefficients, namely $$y_i$$'s, being elements of $$R$$, and similar reasoning shows $$z \in J$$, therefore $$z \in I \cap J$$.