## 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.