Tag Archives: prime

Field extensions of prime degree have no intermediate subfields

Let K/F be a field extension with prime degree p. Show that any subfield of K containing F is either K or F.


Let F \subseteq E \subseteq K. Then [K:E][E:F] = [K:F] = p by Theorem 14 in D&F. Since the degree of a (finite) field extension is an integer, either [K:E] = 1 (so E = K) or [E:F] = 1 (so E = F).

Compute a quotient module

Let R be a principal ideal domain, let a \in R be nonzero, and let M = R/(a). Given a prime p \in R, say a = p^nq, where p does not divide q. Prove that (p^k)M/(p^{k+1})M is module-isomorphic to R/(p) if k < n and to 0 if k \geq n.


We begin with some lemmas.

Lemma 1: Let C \subseteq A,B be ideals of a ring R, and consider A/C as an R-module. Then B(A/C) = (BA)/C. Proof: (\subseteq) If x \in B(A/C), then x = \sum b_i(a_i + C) = \sum (b_ia_i + C) = (\sum b_ia_i) + C \in (BA)/C. (\supseteq) If x \in (BA)/C, then x = (\sum b_ia_i) + C = \sum (b_ia_i + C) = \sum b_i(a_i+C). Thus x \in B(A/C). \square

Lemma 2: Let R be a principal ideal domain and let a,b,c \in R be nonzero with b|c. Note that (ac) \subseteq (ab). Prove that (ab)/(ac) \cong_R (b)/(c). Proof: Let \psi: (b) \rightarrow (ab)/(ac) be given by bx \mapsto \overline{abx}. (This is well-defined since R is a domain, and is clearly an R-module homomorphism.) Certainly \psi is surjective. Now if bx \in \mathsf{ker}\ \psi, then abx \in (ac), so that bx \in (c). Conversely, if bx \in (c), then bx = cy for some y, and so \overline{abx} = \overline{acy} = 0. By the First Isomorphism Theorem, (b)/(c) \cong_R (ab)/(ac). \square

Lemma 3: Let R be a principal ideal domain and a,b,c \in R such that b|c. If (a,c/b) = (1), then (a)((b)/(c)) = (b)/(c). Proof: Say c = bd. Note that ax+dy = 1 for some x,y \in R, so that abx + cy = b. Now b+(c) = abx + (c) = a(bx+(c)) \in (a)((b)/(c)), so that (a)((b)/(c)) = (b)/(c). \square

If k < n, then p^{k+1}|a, so that (a) \subseteq (p^k), (p^{k+1}). By Lemma 1, we have (p^k)(R/(a))/(p^{k+1})(R/(a)) = ((p^k)/(a))/((p^{k+1})/(a)) \cong_R (p^k)/(p^{k+1}) by the Third Isomorphism Theorem. Using Lemma 2, this is isomorphic to (1)/(p) \cong_R R/(p).

If k \geq n with (say) k = n+t, then ((p^k)(R/(a)))/((p^{k+1})/(R/(a)) = ((p^t)((p^n)/(p^nq)))/((p^{t+1})((p^n)/(p^nq))) = ((p^n)/(p^nq))/((p^n)/(p^nq)) = 0, using Lemma 3 and the Third Isomorphism Theorem.

A fact about the annihilators of torsion modules over a PID

Let R be a principal ideal domain, let B be a torsion (left, unital) R-module, and let p \in R be prime. Prove that if pb = 0 for some nonzero b \in B, then \mathsf{Ann}(B) \subseteq (p).


[Note: I’m not sure why D&F assume that B is torsion. This proof doesn’t use that fact as far as I can tell. Read with caution.]

Recall that \mathsf{Ann}(B) \subseteq R is an ideal (by this previous exercise). Since R is a principal ideal domain, we have \mathsf{Ann}(B) = (t) for some t \in R.

Consider the cyclic submodule (b)_R generated by b. Again, \mathsf{Ann}((b)_R) \subseteq R is an ideal, so \mathsf{Ann}((b)_R) = (s) for some s. By our assumption, p \in (s), so that p = sr for some r \in R. Since p is prime, either p|r or p|s. If p|r, then we have r = pq for some q \in R. Now pqs = p, so that qs = 1; in particular, s is a unit, and so \mathsf{Ann}((b)_R) = R. But then 0 = 1 \cdot b = b, a contradiction. So p|s, and thus (s) \subseteq (p).

Since \mathsf{Ann}(B) \subseteq \mathsf{Ann}((b)_R), we have (t) \subseteq (s), and so (t) \subseteq (p), as desired.

Factor the principal ideals in an algebraic integer ring which are generated by ramified rational primes

Let K = \mathbb{Q}(\sqrt{-5}) and let \mathcal{O} be the ring of integers in K. Factor (p) in \mathcal{O}, where p is a ramified rational prime.


By Theorem 9.6, if p does not divide the discriminant d = -20 of K (using Theorem 6.11), then p is not ramified. Now 20 = 2^2 \cdot 5, so that if p is ramified in K, it is either 2 or 5.

We claim that (2) = (2,1+\sqrt{-5})^2. Indeed, the (\supseteq) direction is clear, and we have 2 = -(1+\sqrt{-5})^2 + 2 \cdot (1+\sqrt{-2}) - 2 \cdot 2. We claim also that P = (2,1+\sqrt{-5}) is maximal. To this end, let a+b \sqrt{-5} \in \mathcal{O}, and say a - b \equiv c mod 2 where c \in \{0,1\}. Evidently, a+b\sqrt{-5} \equiv c mod P; if c \equiv 0 mod 2 then a+b\sqrt{-5} \equiv 0 mod P, and if c \equiv 1 mod 2 then a+b\sqrt{-5} \equiv 1 mod P. Now suppose 1 \in P; then 1 = 2(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5}). Comparing coefficients mod 2, we have 0 \equiv h+k \equiv 1 mod 2, a contradiction. So \mathcal{O}/(2,1+\sqrt{-5}) = \{ \overline{0}, \overline{1}\}, and thus \mathcal{O}/(2,1+\sqrt{-5}) \cong \mathbb{Z}/(2) is a field. Hence P is maximal, and (2) = (2,1+\sqrt{-5})^2 is the prime factorization of (2).

Certianly (5) = (\sqrt{-5})^2. We claim that Q = (\sqrt{-5}) is prime. To see this, note that 5 \in (\sqrt{-5}). If a+b\sqrt{-5} \in \mathcal{O}, then a+b\sqrt{-5} \equiv a \equiv a_0 mod Q, where a_0 \in \{0,1,2,3,4\}. Suppose t \in Q \cap \mathbb{Z}. Then t = (a+b\sqrt{-5})\sqrt{-5} for some a,b \in \mathbb{Z}. Comparing coefficients, we have t \equiv 0 mod 5. In particular, r \not\equiv s mod Q, where r,s \in \{0,1,2,3,4\} are distinct. Thus \mathcal{O}/(\sqrt{-5}) \cong \mathbb{Z}/(5) is a field, so that (\sqrt{-5}) is maximal. Thus (5) = (\sqrt{-5})^2 is the prime factorization of (5).

In a quadratic field, rational primes have at most two irreducible factors

Let K be a quadratic extension of \mathbb{Q} and let p be a rational prime. Prove that, as an algebraic integer in K, p has at most two irreducible factors.


Recall that the conjugates of p for K are p itself with multiplicity 2. So the norm of p over K is p^2. Since (by Lemma 7.1) the norm of an algebraic integer is a rational integer and the norm is multiplicative, p has at most two irreducible integer factors in K.

Every prime congruent to 1 mod 4 is a sum of squares

Prove that every prime integer p congruent to 1 mod 4 is a sum of two squares. Show that every product of two such primes is also a sum of two squares.


Let p be such a prime. We know that p is not irreducible in \mathbb{Z}[i]; thus there exist nonunits \alpha,\beta such that p = \alpha\beta. Note that neither of \alpha and \beta can have norm 1. Since p^2 = N(p) = N(\alpha)N(\beta), letting \alpha = a+bi, we see that a^2 + b^2 = p as desired.

Now suppose p,q are integer primes congruent to 1 mod 4. As above, we have p = \alpha_1\beta_1 and q = \alpha_2\beta_2, where \alpha_i and \beta_i have norm p and q as needed. Now N(\alpha_1\beta_2) = pq. In particular, if \alpha_1\beta_2 = a+bi, then a^2+b^2 = pq.

A fact about irreducible Gaussian integers

Let \pi \in \mathbb{Z}[i] be irreducible which is not real and not an associate of 1+i. Prove that if \pi|\alpha and \pi|\overline{\alpha}, then N(\pi)|\alpha.


Since \pi|\overline{\alpha}, we have \pi\tau = \overline{\alpha}. Note that conjugation preserves products and has order 2; thus \overline{\pi}\overline{\tau} = \alpha. Since \pi|\alpha and \pi is irreducible, either \pi|\overline{\pi} or \pi|\overline{\tau}. In the first case, since \pi is not real, by this previous exercise, \pi is an associate of 1+i, a contradiction. So \pi|\overline{\tau}. Thus N(\pi) = \pi\overline{\pi} divides \alpha.

A fact about irreducibles in ZZ[i]

Let \pi = a+bi be an irreducible Gaussian integer with a,b \neq 0. Show that if \pi is a factor of its conjugate \overline{\pi} = a-bi, then \pi is an associate of 1+i.


Let \tau = c+di and suppose \pi\tau = \overline{\pi}. This equality yields the two equations ac-bd=a and ad+bc=-b, which can be rearranged as a(c-1) = bd and ad = -b(c+1). Now ad^2 = -bd(c+1) = -a(c+1)(c-1), so that $latex d^2 + c^2 = 1. Hence \tau \in \{ 1,-1,i,-i \}.

If \tau = 1, then a+bi = a-bi, so that b = 0, a contradiction. Similarly, if \tau = -1 then -a-bi = a-bi and we have a = 0. If \tau = i, then -b+ai = a-bi, so that a = -b. If \tau = -i, then b-ai = a-bi, and we have a = b. In either case, |a| = |b|. If |a| > 1, then \pi = a+bi has a nontrivial factorization (namely a(1+i)) and so is not irreducible. Thus |a| = 1, and so \pi \in \{1+i, 1-i, -1+i, -1-i\}. These are precisely the associates of 1+i.

In the Gaussian integers, the conjugate of a prime is prime

Prove that if a+bi is irreducible in \mathbb{Z}[i], then so is b+ai.


Note that conjugation preserves multiplication in \mathbb{Z}[i]; \overline{\alpha\beta} = \overline{\alpha} \overline{\beta} for all \alpha and \beta. As a consequence, if \alpha is irreducible then \overline{\alpha} is as well.

Note that b+ai = \overline{(-i)(a+bi)}. Thus, if a+bi is irreducible, then so is b+ai.

Half of Wilson’s Theorem

Let n be a natural number. Prove that if (n-1)! \equiv -1 \mod n, then n is prime.


Suppose n is composite; say n = p_1p_2 \cdots p_k, where the p_i are pairwise coprime and greater than 1. (Say each p_i is the highest power of some prime dividing n.) If k \geq 2, each of the p_i is represented in the set \{1,2,\ldots,n-1\}. Thus we have (n-1)! \equiv 0 \mod n. If on the other hand n = q^t is a prime power, and t \geq 3, then q and q^{t-1} are in the set \{1,2,\ldots,n-1\}, so that (n-1)! \equiv 0. If n = q^2 and q > 2, then q,2q \in \{1,2,\ldots,n-1\}, and (n-1)! \equiv 0 \mod n. If n = 4, then (4-1)! = 6 \equiv 2 \mod 4.

So n must be prime.