Math 254A, UC Berkeley, Fall 2001 (Kedlaya) Solutions for Problem Sets 1 and 2

1 Gaussian Integers

1. Pair each of 1, …, p − 1 with its inverse mod p. Then only 1 and p − 1 are paired with themselves. Thus 1 ⋯ (p − 1) can be rewritten as 1 · (p − 1) times the product of pairs xx^{−1}, and so is congruent to −1 modulo p.

2. In fact, every euclidean ring is a principal ideal ring, since any nonzero element of an ideal of minimal norm generates the ideal. In fact, we will show that every principal ideal ring R is factorial. To prove existence of prime factorization, pick α a nonzero element of R. If α generates a prime ideal, then α is irreducible and we are done. Otherwise, the ideal (α) is contained in a maximal ideal (β_{1}); write α = β_{1}α_{1}. If α_{1} is irreducible, we are done again. Otherwise, (α_{1}) is contained in a maximal ideal (β_{2}), and we may put α_{1} = β_{2}α_{2}. If this process stops at some point, we obtain a factorization of α into irreducibles. Otherwise, we obtain an increasing sequence (α) ⊂ (α_{1}) ⊆ ⋯ of ideals which does not terminate, contradicting the fact that every principal ideal ring is noetherian. Thus the process must stop somewhere, and prime factorizations exist. As for uniqueness, if α_{1} ⋯ α_{m} = β_{1} ⋯ β_{m} is an equality of two products of irreducibles, then β_{1} ⋯ β_{m} belongs to the prime ideal (α_{1}), so one of the β_{i} must be a multiple of α_{1}, say β_{1}. In fact, β_{1} and α_{1} must generate the same ideal, since both generate maximal ideals. Thus we can factor off α_{1} from both sides and keep going.

3. One method is to start with x such that x^{2} + 1 is divisible by p (which can be found by exhaustive search, or by taking x ≡ ((p − 1)/2)! (mod p)), then compute a generator of (p, x+i) by imitating the proof from lecture that the Gaussian integers form a euclidean ring. That is, given generators (α, β), replace them by (β, γ), where γ − α is a multiple of β and |γ| < |β|. If γ = 0 at some point, then β generates the ideal (p, x + i), and if β = a + bi, then a^{2} + b^{2} = p.

4. I know, I know, you can do this without the Gaussian integers. So sue me. Anyway, x and y can’t both be odd by mod 4 considerations, so assume x is odd and y is even. Factor z^{2} = (x + yi)(x − yi). Then the GCD of x + yi and x − yi divides both 2x and 2yi. If x and y have a common divisor, so does z, contrary to assumption. Thus the GCD of x + yi and x − yi also divides 2, but since x is odd and y is even, x + yi is coprime to 2. Thus x + yi and x − yi are relatively prime. That means, up to units, each one is a perfect square. That is, there exist u, v such that x + yi = (u + vi)^{2}c for some unit c. Since (u + vi)^{2} = u^{2} − v^{2} + 2uv, x is odd and y is even, c must be 1 or −1. If c = −1, we can rewrite −(u + vi)^{2} as (−v + ui)^{2}. So without loss of generality, assume c = 1; then x = u^{2} − v^{2}, y = 2uv, and z = u^{2} + v^{2}, as desired.

5. We use the norm f(a + b\sqrt{2}) = |a^{2} − 2b^{2}|. (Note that the absolute value is needed in this case!) Clearly this norm is multiplicative. Now given α = a + b\sqrt{2} and β = c + d\sqrt{2}

6. Let s(n) be the number of ways to write n as a sum of two squares (by which I’ll mean the number of solutions of x^{2}+y^{2} = n over the rational integers); then s(n) is equal to the number of Gaussian integers of norm n. In particular, if m and n are relatively prime, α runs over the Gaussian integers of norm m, and β runs over the Gaussian integers of norm n, then αβ runs over the divisors of mn, but each divisor of mn shows up four times. (If a divisor shows up once as αβ, it shows up as (αi^{m})(βi^{−m}) for m = 0, 1, 2, 3.) Thus 4s(mn) = s(m)s(n). Suppose n = p^{e} for p a prime congruent to 3 modulo 4. Then p is a prime in the Gaussian integers, so if n = α\overline{α}, both of α and \overline{α} are divisible by p^{e/2}. This is impossible if e is odd, so in that case s(p^{e}) = 0; if e is even, then the only possible α are p^{e/2}i^{m} for m = 0, 1, 2, 3. Thus in this case s(p^{e}) = 4. Next, suppose n = p^{e} for p a prime congruent to 1 modulo 4. Then p factors as α\overline{α} for some prime α. If n = β\overline{β}, then β must be of the form α^{l}\overline{α}^{e−l}i^{m} for l ∈ \{0, …, e\} and m ∈ \{0, 1, 2, 3\}. In this case, s(p^{e}) = 4(e + 1). Putting it all together, we get for n = ∏_{i} p_{i}^{e_{i}},

s(n) = 4 ∏_{i} \frac{s(p_{i}^{e_{i}})}{4} = 4 ∏_{i} \begin{cases}e_{i} + 1 & p_{i} ≡ 1 (mod 4) \\ 0 & p_{i} ≡ 3 (mod 4), e_{i} ≡ 1 (mod 2) \\ 1 & p_{i} ≡ 3 (mod 4), e_{i} ≡ 0 (mod 2). \end{cases}

2 Number Fields

1. We start with the basis 1 and \sqrt{D}, which may or may not be integral. Note that Trace(1) = 1 + 1 = 2 and Trace(\sqrt{D}) = \sqrt{D} + (−\sqrt{D} = 0. Thus

Disc(1, \sqrt{D}) = det (\begin{matrix}Trace(1 · 1) & Trace(1 · \sqrt{D}) \\ Trace(\sqrt{D} · 1) & Trace\sqrt{D} · \sqrt{D} \end{matrix}) = det (\begin{matrix}2 & 0 \\ 0 & 2D \end{matrix}) = 4D.

Recall that if m is the index, in the ring of integers, of the ℤ-submodule generated by this basis, then m^{2} divides the discriminant of the basis. Since we assumed D is squarefree, either the basis 1, \sqrt{D} is integral, or it generates a ℤ-submodule of the ring of integers of index 2. First suppose D ≡ 1 (mod 4). In that case, (1 + \sqrt{D})/2 has minimal polynomial x^{2} + x + (1 − D)/4, and so is an integer. The basis 1, (1 + \sqrt{D})/2 generates a ℤ-submodule between the ring of integers and ℤ + ℤ\sqrt{D}, but the index between these is 2 and ℤ + ℤ\sqrt{D} has index 2 in ℤ + ℤ(1 + \sqrt{D})/2. Thus 1, (1 + \sqrt{D})/2 is an integral basis. Next, suppose D ≢ 1 (mod 4). In this case, we claim 1, \sqrt{D} is already an integral basis; since the index of its ℤ-span in the ring of integers is at most 2, it suffices to check that if a, b are not both even, then (a + b\sqrt{D})/2 is not an algebraic integer. In fact, it suffices to check this for (a, b) = (1, 0), (0, 1), (1, 1). In these cases, the minimal polynomials of these are x − 1/2, x^{2} − D/4, and x^{2} + x + (1 − D)/4, none of which have integer coefficients. Thus 1, \sqrt{D} is an integral basis. In summary: an integral basis is given by

\begin{cases}1, ^{\frac{1+\sqrt{D}}{2}} & D ≡ 1 (mod 4) \\ 1, \sqrt{D} & D ≢ 1 (mod 4). \end{cases}

2. First we compute Trace(1) = 3, Trace(\sqrt[3]{2}) = 0, Trace(\sqrt[3]{4}) = 0. (Either add up the conjugates in ℂ or write out the matrices by which these act on ℚQ(\sqrt{3}) as a ℚQ-vector space.) Thus

Disc(1, \sqrt[3]{2}, \sqrt[3]{2^{2}}) = det (\begin{matrix}Trace(1 · 1) & Trace(1 · \sqrt[3]{2}) & Trace(1 · \sqrt[3]{2^{2}} \\ Trace(\sqrt[3]{2} · 1) & Trace(\sqrt[3]{2} · \sqrt[3]{2}) & Trace(\sqrt[3]{2} · \sqrt[3]{2^{2}} \\ Trace(\sqrt[3]{2^{2}} · 1) & Trace(\sqrt[3]{2^{2}} · \sqrt[3]{2}) & Trace(\sqrt[3]{2^{2}} · \sqrt[3]{2^{2}}) \end{matrix}) = det (\begin{matrix}3 & 0 & 0 \\ 0 & 0 & 6 \\ 0 & 6 & 0 \end{matrix}) = −2^{2}·3^{3}.

Thus the index of ℤ + ℤ\sqrt[3]{2} + ℤ\sqrt[3]{2^{2}} in the full ring of integers is either 1, 2, 3 or 6. We next check that the index is not divisible by 2. That is, we check that if a, b, c are rational integers, not all even, then α = (a + b\sqrt[3]{2} + c\sqrt[3]{2^{2}})/2 is not an algebraic integer. One could do this by computing the minimal polynomials for all possible choices of a, b, c ∈ {0, 1}, but in this case there is a shortcut. If α is an algebraic integer, so is α\sqrt[3]{2^{2}} = b + c\sqrt[3]{2} + (a/2)\sqrt[3]{2^{2}}, as then is (a/2)\sqrt[3]{2^{2}}. But the latter has minimal polynomial x^{3} − a^{3}/2, which has integer coefficients only if a is even. Thus (b\sqrt[3]{2} + c\sqrt[3]{2^{2}})/2 must also be an integer; multiplying by \sqrt[3]{2}, we find that b must be even, then c. Finally, we check that the index is not divisible by 3. Again, we must check that if a, b, c are rational integers, not all divisible by 3, then α = (a+b\sqrt[3]{2}+c\sqrt[3]{2^{2}})/3 is not an algebraic integer. It suffices to check this for a, b, c ∈ {0, 1, 2}. Again, one can list all of the minimal polynomials of these elements, but one can also do a bit better. Note that if we put β = 1 + \sqrt[3]{2} and γ = 1−\sqrt[3]{2}+\sqrt[3]{2^{2}}, then βγ = 3. Now αγ = (a+2b−2c)/3+(−a+b+2c)\sqrt[3]{2}/3+(a−b+c)\sqrt[3]{2^{2}}/3 is equal to an element of ℤ + ℤ\sqrt[3]{2} + ℤ\sqrt[3]{2^{2}} plus (a − b + c)(1 + \sqrt[3]{2} + \sqrt[3]{2^{2}})/3. This can only be integral if a−b+c is divisible by 3, since 1+\sqrt[3]{2}+\sqrt[3]{2^{2}} has minimal polynomial x^{3}−3x^{2}−3x+1 whose constant coefficient is not divisible by any cube. We now know that if α is integral, then a − b + c is divisible by 3. Next, suppose c = 0. Then Then α(1 + \sqrt[3]{2} + \sqrt[3]{2^{2}}) = (a + 2b)/3 + (b + a)\sqrt[3]{2}/3 + (b + a)\sqrt[3]{2^{2}}/3 is integral, so a + 2b is divisible by 3, which now implies a = b = 0. A similar argument applies if a = 0, or if b = 0. The only remaining cases are a = c = 1, b = 2 or a = c = 2, b = 1. These are equivalent: in the former, the minimal polynomial of α is x^{3} − x^{2} − x − 1/3, which is not integral. We conclude 1, \sqrt[3]{2}, \sqrt[3]{2^{2}} is an integral basis.

3. From what we have shown in class, it suffices to check that the discriminant of the polynomial P (x) equals the discriminant of the basis 1, x, …, x^{n−1}. Let r_{1}, …, r_{n} be the roots of P (x) in ℂ, which are precisely the conjugates of x in ℂ. The discriminant of the basis 1, x, …, x^{n−1} is

(det(r_{i}^{j−1})_{i,j=1}^{n})^{2} = ∏_{i<j}(r_{i} − r_{j})^{2}

by the properties of the Vandermonde determinant. (See Neukirch, p. 11. One proof of the Vandermonde formula: view both sides as polynomials in the r_{i}, note that the determinant vanishes whenever r_{i} = r_{j}, then check that the coefficient of some monomial matches up.) The latter is precisely the discriminant of P.

4. One example is P (x) = x^{3} + 2x + 2, which has discriminant −140. The proof that 1, x, x^{2} is an integral basis is similar to the argument given in problem 2.

5. Note 2 cos(2π/n) = ζ_{n} + ζ_{n}^{−1} and 2 sin(2π/n) = −i(ζ_{n} + ζ_{n}^{−1}; since i, ζ_{n} and ζ_{n}^{−1} are algebraic integers, these combinations are too. From this representation, we also see that the conjugates of cos(2π/n) are all of the form cos(2πk/n). If cos(2π/n) were an algebraic integer, its norm would be an integer; but the absolute value of the product of the conjugates is less than 1 for n ≥ 5, contradiction. The argument for sin(2π/n) is similar.

6. The root of a polynomial P with algebraic integer coefficients is an algebraic number; let K be the number field generated by the roots and coefficients of P. The integrality of the root now follows from the fact that 𝔬_{K} is integrally closed, which we checked in class.

3 Interlude on Symmetric Polynomials

1. Let d be a positive integer; we check that the homogeneous symmetric polynomial of degree d with integer coefficients can be written by integer polynomials in σ_{1}, …, σ_{n}. We order the monomials x_{1}^{e_{1}} ⋯ x_{n}^{e_{n}} in lexicographic order. That is, if two monomials have different e_{1}, whichever has the bigger e_{1} is the bigger of the two. If the e_{1}’s are equal, compare e_{2}’s, and so on. Now given a homogeneous symmetric polynomial P with integer coefficients, let cx_{1}^{e_{1}} ⋯ x_{n}^{e_{n}} be the largest monomial in lexicographic order. In particular, we must have e_{1} ≥ ⋯ ≥ e_{n}, otherwise the monomial with the exponents sorted in order also appears and is larger in lexicographic order. Then P − cσ_{1}^{e_{1}−e_{2}} ⋯ σ_{n−1}^{e_{n−1}−e_{n}}σ_{n}^{e_{n}} is symmetric with integer coefficients, and its largest monomial is strictly smaller than that of P. Thus this process eventually stops at the zero polynomial, and P has been expressed as an integer polynomial in the σ_{i}.

2. We just give the argument for α+β here, as the argument for αβ is similar. Let α_{1}, …, α_{m} be the conjugates of α in ℂ, and let β_{1}, …, β_{n} be the conjugates of β in ℂ. Let P (x) be the polynomial ∏_{i=1}^{m} ∏_{j=1}^{n}(x − α_{i} − β_{j}). The polynomial ∏_{i=1} m ∏_{j=1}^{n}(x − t_{i} − u_{j}) can be viewed as a symmetric polynomial in the u_{j} with coefficients in the ring ℤ[t_{1}, …, t_{m}], so by the previous problem is a polynomial in the elementary symmetric functions of the u_{i}. But the coefficients of this polynomial are themselves symmetric functions of the t_{i}, so can be written as integer polynomials in the elementary symmetric functions of the t_{i}. Substituting t_{i} = α_{i} and u_{j} = β_{j}, the elementary symmetric functions of the t_{i} and u_{j} become the coefficients of the minimal polynomial of α and β, which are integers. Thus P becomes a monic polynomial with integer coefficients, so its root α + β is an algebraic integer.

4 Conjugates, Trace and Norm

1. Let [ℚ(α) : ℚ] = n. Let

X^{n} + a_{1}X^{n−1} + … a_{n}

be the minimal polynomial of α over ℚ. Since α is algebraic we may assume that a_{i} ∈ ℤ. Let α_{i} be the conjugates of α, and let σ_{k} be the k^{th} symmetric polynomial. Using the triangle inequality, and the identity |α_{i}| = 1, we conclude that

∥a_{i}∥ = ∥σ_{k}(α_{1}, …, α_{n})∥ ≤ \binom{n}{k}.

Since this bound depends only on n, we conclude that there can only be finitely many algebraic integers of degree n with this property, since there are only finitely many polynomials for each n satisfying the above bound. On the other hand, if α has this property, then so does α^{k} for every k. Thus α^{i} = α^{j} for some i ≠ j and we conclude (since α ≠ 0) that α is a root of unity.

2. Let α be an algebraic number of degree d. Suppose that Trace(α^{n}) ∈ ℤ for all n. Let α_{i} for i = 1, …, d be the conjugates of α. Observe that symmetric polynomials (in α_{i}) may be expressed in terms of the “Newton polynomials”, which are symmetric polynomials given by taking the sum of the k^{th} powers (i.e. ∑ α_{i}^{k}). This follows from the formal identity:

exp (− ∑_{k=1}^{∞} \frac{t^{k}}{k} ∑_{i=1}^{d} α_{i}^{k}) = exp (∑_{i=1}^{d} log(1 − tα_{i})) = ∏_{i=1}^{d}(1 − tα_{i})

In particular, the minimal polynomial of α has denominators bounded purely in terms of d, and so c_{d} · α is an algebraic integer for some universal constant c_{d}. Applying the same argument to α^{n} for all n we conclude that c_{d} · α^{n} is also an algebraic integer for all n. One may infer that α itself is algebraic in various ways. One such way is as follows. Suppose that (α) = 𝔭_{1}^{a_{1}} … 𝔭_{n}^{a_{n}}. We wish to prove that a_{i} ≥ 0 for all i. If a_{i} < 0, then for c_{d} · α to be integral we must have c_{d} ∈ 𝔭_{i}^{−a_{i}}. Applying the same argument to α^{n} we must have c_{d} ∈ 𝔭_{i}^{−a_{i}·n} for all n, which is a contradiction. Alternatively, note that the ℤ-submodule generated by 1, α, α^{2}, … is finitely generated, since it’s contained in the finitely generated ℤ-module c_{d}^{−1}𝔬, where 𝔬 is the ring of integers of ℚQ(α). That implies α is algebraic by a theorem from lecture.

5 Ideals and Unique Factorization

1. Let R = ℤ[\sqrt{−7}]. Let η = (2, 1 + \sqrt{−7}). Then

η^{2} = (4, −6 + 2\sqrt{−7}, 2 + 2\sqrt{−7}) = (4, 2 + 2\sqrt{−7}) = (2)(2, 1 + \sqrt{−7}) = (2)η.

Assuming unique factorization for ideals in R we conclude that η = (2). Yet since the element (1 + \sqrt{−7})/2 ∉ R, (1 + \sqrt{−7}) cannot lie in (2), and we are done.

2(i). Suppose first of all that there exists a solution to x^{2} = d mod p. Clearly we may take 2x ≤ p, and thus write x^{2} − d = pm with m < p. Let 𝔭_{1} = (p, x + \sqrt{d}) and 𝔭_{2} = (p, x − \sqrt{d}). We observe that

𝔭_{1}𝔭_{2} = (p^{2}, p(x ± \sqrt{d}), pm)

since m < p, (m, p) = 1 and so (p^{2}, pm) = (p). Thus 𝔭_{1}𝔭_{2} = (p). Observe that 𝔭_{1} is not the unit ideal, since if 𝔭_{1} = (1) then by symmetry 𝔭_{2} = (1) and (p) = 𝔭_{1}𝔭_{2} = (1). On the other hand, 𝔭_{1} ≠ (p) since for p > 2, (x + \sqrt{d})/p is not an algebraic integer (see previous assignment).

2(ii). In the other direction, suppose that pO is not prime in O, and that p does not divide 2d. Then we may write (p) = 𝔞𝔟 with 𝔞 and 𝔟 neither (p) nor the unit ideal. Write p = ab with a ∈ 𝔞 and b ∈ 𝔟. Since N(p) := Norm(p) = p^{2}, we must have N(a)N(b) = p^{2}. If N(a) = 1, then a is a unit and 𝔞 is the unit ideal. Thus p|N(a), and similarly p|N(b). Thus N(a) = ±p. Write a = x + y\sqrt{d}, with 2x, 2y ∈ ℤ. Then

N(a) = x^{2} − dy^{2} = ±p

Suppose that p|y. Then p|x^{2}, and thus p|x. This would imply, however, that p^{2}|N(a) = ±p, which is clearly impossible. Thus p does not divide y. In particular, since p ≠ 2, y is invertible modulo p, and (xy^{−1})^{2} ≡ d mod p, which was to be shown.

3. Suppose 𝔞 is an integral ideal which is not principal. Let α be a nonzero element of 𝔞 of minimum norm. By assumption, there exists β ∈ 𝔞 \ (α). Choose γ ∈ ℤ[\sqrt{−5}] to minimize |δ|, where δ = β/α − γ; then in particular, ℜ(δ) ∈ [−1/2, 1/2] and ℑ(δ) ∈ [−\sqrt{5}/2, \sqrt{5}/2]. On the other hand, by assumption, δα = β − γα is a nonzero element of 𝔞, so |δα| ≥ |α|, which is to say |δ| ≥ 1. Since |ℜ(δ)| ≤ 1/2, it follows that |ℑ(δ)| ≥ \sqrt{3}/2. Without loss of generality, assume ℑ(δ) > 0; then \sqrt{3} − \sqrt{5} < ℑ(2δ − \sqrt{−5}) < 0. If c is the closest integer to ℜ(2δ), then put ϵ = 2δ − \sqrt{−5} − c; we have

|ϵ|^{2} ≤ ℜ(ϵ)^{2} + ℑ(ϵ)^{2} ≤ \frac{1}{4} + (\sqrt{3} − \sqrt{5})^{2} < 1.

If ϵ ≠ 0, then ϵα is an element of 𝔞 of norm less than |α|, contradiction. Thus ϵ = 0 and so 2β ∈ (α). Since this holds for all β, we conclude [𝔞 : (α)] = 2 and that 2𝔞 ⊆ (α). Now let 𝔭 = (2, 1 + \sqrt{−5}) and note that 𝔭^{2} = (2). That means that (α)|2𝔞 = 𝔭^{2}𝔞 and 𝔞|(α). That means that 𝔞 is either equal to (α), 𝔭^{−1}(α) or 𝔭^{−2}(α) = (α/2). The first and third choices are principal ideals, so we must have 𝔞 = 𝔭^{−1}(α). Thus 𝔞 is in the ideal class of 𝔭, as desired.

4. Let I be an ideal in a Dedekind domain R. Let a ∈ I be a non zero element. Then (a) is a non-trivial ideal in R, and by the hint in the exercise, R/a is a PID. On the other hand, since (a) ⊆ I, there is a commutative diagram:

R ⟶ R/a ⟶ R/I.

Let J be the kernel of R/a → R/I. Since R/a is principal, we may write J = b + aR in R/a, with b ∈ R. Thus (a, b) = Ker(R → R/I) = I. (This last claim could be expanded upon briefly, but it is completely trivial).

5. Since the group of ideals is generated by prime ideals, it suffices to prove the following statement for Dedekind domains: for any (non-zero) prime ideal 𝔭, there exists another ideal 𝔮 such that 𝔮 represents the same element of the class group of 𝔭, and 𝔮 is prime to any particular choice of finitely many ideals: 𝔭_{1}, 𝔭_{2}, …, 𝔭_{n}. Since R is a Dedekind domain, 𝔭_{i} + 𝔭_{j} = R for all i, j. By the Chinese remainder theorem:

R/𝔭^{2}𝔭_{1}𝔭_{2} … 𝔭_{n} ≃ R/𝔭^{2} ⊕ R/𝔭_{1} ⊕ … ⊕ R/𝔭_{n}.

In particular, since 𝔭 ≠ 𝔭^{2}, there exists an element r ∈ R such that r ∈ 𝔭 \ 𝔭^{2}, and r ≠ 𝔭_{i} for any i. Let us consider the factorization of the ideal (r). By construction, the ideal 𝔭 occurs exactly once, and the ideals 𝔭_{i} do not occur. Write (r) = 𝔭𝔮. Then 𝔮 is prime to 𝔭 and 𝔭_{i} for all i. On the other hand, 𝔮 is the inverse element to 𝔭 in the class group, since their product is principal. One finishes the proof by either taking some appropriate power of 𝔮 (since the class group is finite), or more appropriately, taking N(𝔮)/𝔮, which is manifestly an integral ideal. (Or for a third argument, repeat the argument above with 𝔭 replaced by 𝔮, and 𝔭_{i} replaced by 𝔭_{i} ∪ 𝔭 to produce an ideal inverse to 𝔮 in the classgroup, and coprime to 𝔭 and 𝔭_{i}.)

6. We imitate the proof of Theorem I.3.1 in Neukirch; the key point is that K[x] is a principal ideal domain. (Every nonzero ideal in K[x] is generated by an element of minimum degree.) Let R be the integral closure of K[x] in L. By Proposition 2.10, every ideal of R is a finitely generated K[x]-module, and hence is a finitely generated ideal. Thus R is noetherian. Moreover, since R is the integral closure of K[x] in L, it is automatically integrally closed. (This was essentially a problem on the first homework; see also section 2 of Chapter 1 of Neukirch.) Now let 𝔭 be a nonzero prime ideal of R; it remains to show that 𝔭 must be maximal. For any y ∈ 𝔭 nonzero, there exists a polynomial y^{n} + a_{1}y^{n−1} + ⋯ + a_{n} = 0. Thus a_{n} ∈ 𝔭, so the intersection 𝔮 = 𝔭 ∩ K[x] is a nonzero ideal; it is also prime because K[x]/𝔮 injects into R/𝔭 and the latter is an integral domain. Thus 𝔮 is maximal, so K[x]/𝔮 is a field and R/𝔭 is an integral extension of K[x]/𝔮. That forces R/𝔭 to be a field: if z ∈ R/𝔭 has minimal polynomial z^{m} + b_{1}z^{m−1} + ⋯ + b_{m} over K[x]/gothq, then b_{m} ≠ 0, so z is invertible. We conclude every nonzero prime ideal of R is maximal, so R is a Dedekind domain.