Monthly Archives: May 2011

Compute the degree of a field extension and find a basis

Let K = \mathbb{Q}(\sqrt[3]{2}, \sqrt{7}, i\sqrt{5}). Compute the degree of K over \mathbb{Q} and find a basis.


Note that \sqrt{7} is a root of a(x) = x^2 - 7, which is irreducible over \mathbb{Q} by Eisenstein and thus is the minimal polynomial of \sqrt{7} over \mathbb{Q}. So \mathbb{Q}(\sqrt{7}) has degree 2 over \mathbb{Q}, with \{1, \sqrt{7}\} a basis. Suppose there exist \alpha, \beta \in \mathbb{Q} such that (\alpha + \beta\sqrt{7})^2 = -5; comparing coefficients, we see that \alpha^2 + 7\beta^2 = -5. However, the left hand side of this equation is positive, while the right hand side is negative- a contradiction. In particular, b(x) = x^2 + 5 is irreducible over \mathbb{Q}(\sqrt{7}), and hence is the minimal polynomial of i\sqrt{5} over \mathbb{Q}(\sqrt{7}). So \mathbb{Q}(\sqrt{7}, i\sqrt{5}) has degree 4. We may take as a basis the set \{1, \sqrt{7}, i\sqrt{5}, i\sqrt{35} \}.

Note that the minimal polynomial of \sqrt[3]{2} over \mathbb{Q} is c(x) = x^3 - 2. (Use Eisenstein to show irreducibility.) By this previous exercise, c(x) is also irreducible over \mathbb{Q}(\sqrt{7}, i\sqrt{5}). Thus \mathbb{Q}(\sqrt[3]{2}, \sqrt{7}, i\sqrt{5}) has degree 12 over \mathbb{Q}. We may take as a basis the set \{ 1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt{7}, \sqrt{7}\sqrt[3]{2}, \sqrt{7}\sqrt[3]{4}, i\sqrt{5}, i\sqrt{5}\sqrt[3]{2}, i\sqrt{5}\sqrt[3]{4}, i\sqrt{35}, i\sqrt{35}\sqrt[3]{2}, i\sqrt{35}\sqrt[3]{4} \}.

Compute the degree of QQ(i + √2)

Compute the degree of K = \mathbb{Q}(i + \sqrt{2}) over \mathbb{Q}. Exhibit three distinct subfields of K. Deduce that the polynomial found in this previous exercise is irreducible over \mathbb{Q}.


Since we know \theta = i + \sqrt{2} is a root of p(x) = x^4 - 2x^2 + 9, the degree of \theta over \mathbb{Q} is at most 4. In this previous exercise, we showed that K contains both i and \sqrt{2}, so that \mathbb{Q}(i), \mathbb{Q}(\sqrt{2}) \subseteq K. Now i\sqrt{2} \in K, so we also have \mathbb{Q}(i\sqrt{2}) \subseteq K. Note that the minimal polynomial of i over \mathbb{Q} is a(x) = x^2 + 1 (irreducible because a(x+1) is Eisenstein at 2), the minimal polynomial of \sqrt{2} is b(x) = x^2 - 2 (Eisenstein at 2) and the minimal polynomial of i\sqrt{2} is c(x) = x^2 + 2 (Eisenstein at 2). So these subfields all have degree 2 over \mathbb{Q}.

Suppose \alpha,\beta \in \mathbb{Q} and (\alpha + i\beta)^2 = 2. Comparing coefficients, we see that \alpha^2 - \beta^2 = 2 and \alpha\beta = 0. If \alpha = 0, then \beta^2 = -2, a contradiction since \beta^2 \geq 0. If \beta = 0, then \alpha^2 = 2, a contradiction since \sqrt{2} is not rational. In particular, \sqrt{2} \notin \mathbb{Q}(i), so that \mathbb{Q}(i) and \mathbb{Q}(\sqrt{2}) are distinct.

Similarly, if (\alpha + \beta i\sqrt{2})^2 = 2, then either \alpha^2 = 2 or \beta^2 = -1. So \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(i\sqrt{2}) are distinct.

Finally, if (\alpha + i\beta\sqrt{2})^2 = -1, then either \alpha^2 = -1 or \beta^2 = 1/2, again a contradiction. So \mathbb{Q}(i) and \mathbb{Q}(i\sqrt{2}) are distinct.

Now we claim that the degree of \sqrt{2} over \mathbb{Q}(i) is 2. To see this, note that (since \mathbb{Q}(i) and \mathbb{Q}(\sqrt{2}) are distinct) x^2-2 is irreducible over \mathbb{Q}(i), and so is the minimal polynomial of \sqrt{2} over \mathbb{Q}(i). So \mathbb{Q}(i,\sqrt{2}) = \mathbb{Q}(i+\sqrt{2}) has degree 2 over \mathbb{Q}(i); hence K has degree 4 over \mathbb{Q}.

We can summarize this information using the following labeled lattice diagram.

A lattice of fields.

Cubic irreducibles over QQ are irreducible over extensions of degree 4

Let p(x) be an irreducible cubic polynomial over \mathbb{Q}. Suppose K is an extension of \mathbb{Q} of degree 4; prove that p(x) is irreducible over K.


We may assume without loss of generality that p(x) is monic.

Now suppose to the contrary that p(x) is reducible over K. Since p(x) is cubic, it must have a linear factor, and thus must have a root \alpha in K. Since \alpha is a root of the monic irreducible p(x) over \mathbb{Q}, p(x) is in fact the minimal polynomial of \alpha over \mathbb{Q}, and thus \mathsf{dim}_\mathbb{Q} \mathbb{Q}(\alpha) = 3. We thus have the following chain of fields: \mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq K. By Theorem 5.7 in TAN, we have that 3 divides 4, a contradiction.

So p(x) is irreducible over K.

Schanuel’s Lemma

Let R be a ring with 1.

  1. If 0 \rightarrow K_1 \stackrel{\alpha_1}{\rightarrow} P_1 \stackrel{\beta_1}{\rightarrow} M \rightarrow 0 and 0 \rightarrow K_2 \stackrel{\alpha_2}{\rightarrow} P_2 \stackrel{\beta_2}{\rightarrow} M \rightarrow 0 are short exact sequences of (left unital) modules and P_1 and P_2 are projective, show that P_1 \oplus K_2 \cong_R P_2 \oplus K_1.
  2. If 0 \rightarrow M \stackrel{\alpha_1}{\rightarrow} Q_1 \stackrel{\beta_1}{\rightarrow} L_1 \rightarrow 0 and 0 \rightarrow M \stackrel{\alpha_2}{\rightarrow} Q_2 \stackrel{\beta_2}{\rightarrow} L_2 \rightarrow 0 are short exact sequences of (left unital) modules and Q_1 and Q_2 are injective, show that Q_1 \oplus L_2 \cong_R Q_2 \oplus L_1.

(Side note: two modules A and B are called projectively equivalent if there exist projective modules P_1 and P_2 such that A \oplus P_1 \cong_R B \oplus P_2. Likewise, A and B are injectively equivalent if there are injective modules Q_1 and Q_2 such that A \oplus Q_1 \cong_R B \oplus Q_2.)


  1. Let X denote the pullback of \beta_1 and \beta_2. (See this previous exercise about pullbacks.) We claim that the pullback morphisms \lambda_1 and \lambda_2 are surjective. To see this, suppose a \in P_1 and consider \beta_1(a). Since \beta_2 is surjective, there exists b \in P_2 such that \beta_2(b) = \beta_1(a). Thus (a,b) \in X, and so \lambda_1(a,b) = a. Thus \lambda_1 is surjective; a similar argument shows that \lambda_2 is surjective. Thus 0 \rightarrow \mathsf{ker}\ \lambda_1 \stackrel{\iota}{\rightarrow} X \stackrel{\lambda_1}{\rightarrow} P_1 \rightarrow 0 and 0 \rightarrow \mathsf{ker}\ \lambda_2 \stackrel{\iota}{\rightarrow} X \stackrel{\lambda_2}{\rightarrow} P_2 \rightarrow 0 are short exact sequences. We now claim that there exist module homomorphisms \theta_1 : \mathsf{ker}\ \lambda_2 \rightarrow K_1 and \theta_2 : \mathsf{ker}\ \lambda_1 \rightarrow K_2 such that the following diagram of modules commutes.

    A diagram

    To this end, we prove a lemma.

    Lemma 1: Let f : A \rightarrow C and g : B \rightarrow C be injective module homomorphisms. If \mathsf{im} f \subseteq \mathsf{im}\ g, then there exists a unique module homomorphism h : A \rightarrow B such that f = g \circ h. Moreover, h is injective. Proof: Let a \in A. Since \mathsf{im}\ f \subseteq \mathsf{im}\ g, there exists b \in B such that f(a) = g(b). Since g is injective, this b is unique. Define h(a) = b. Certainly h is well defined. Moreover, if h(a_1) = b_1 and h(a_2) = b_2 and r \in R, then we have f(a_1+ra_2) = f(a_1)+rf(a_2) = g(b_1) + rg(b_2) = g(b_1+rb_2), so that h(a_1+ra_2) = b_1+rb_2 = h(a_1) + rh(a_2). So h is a module homomorphism. Moreover, (g \circ h)(a) = g(h(a)) = f(a), so that f = g \circ h. To see uniqueness, if k : A \rightarrow B is a module homomorphism such that f = g \circ k, then if k(a) = b, we have g(b) = f(a), so that b = h(a). Thus h = k. Finally, if h(x) = 0, then f(x) = 0, so that x = 0. Thus h is injective. \square

    Now we claim that \lambda_1 \circ \iota : \mathsf{ker}\ \lambda_2 \rightarrow P_1 is injective. To see this, suppose (a,b) \in \mathsf{ker}\ (\lambda_1 \circ \iota). Then 0 = (\lambda_1 \circ \iota)(a,b) = \lambda_1(a,b) = a. Moreover, since (a,b) \in \mathsf{ker}\ \lambda_2, we have b = 0. Thus \mathsf{ker}\ (\lambda_1 \circ \iota) = 0, and so \lambda_1 \circ \iota is injective. Moreover, suppose a \in \mathsf{im}\ (\lambda_1 \circ \iota) with a = (\lambda_1 \circ \iota)(a,0). Then \beta_1(a) = \beta_2(0) = 0, so that a \in \mathsf{ker}\ \beta_2 = \mathsf{im}\ \alpha_1. Thus we have \mathsf{im}\ (\lambda_1 \circ \iota) \subseteq \mathsf{im}\ \alpha_1. By Lemma 1, there exists a unique module homomorphism \theta_1 : \mathsf{ker}\ \lambda_2 \rightarrow K_1 such that \lambda_1 \circ \iota = \alpha_1 \circ \theta_1. By a similar argument, \theta_2 : \mathsf{ker}\ \lambda_1 \rightarrow K_2 exists such that \lambda_2 \circ \iota = \alpha_2 \circ \theta_2. Moreover, \theta_1 and \theta_2 are injective.

    We claim now that \theta_1 and \theta_2 are surjective. To see this, let k \in K_1. Now k \in \mathsf{ker}\ \beta_1, so that \beta_1(k) = 0 = \beta_2(0). Now (k,0) \in X, and in fact (k,0) \in \mathsf{ker}\ \lambda_2. Clearly \theta_1(k,0) = k, so that \theta_1 is surjective. A similar argument shows that \theta_2 is surjective. In fact, we have \mathsf{ker}\ \lambda_1 \cong_R K_2 and \mathsf{ker}\ \lambda_2 \cong_R K_1. Because P_1 and P_2 are projective, the sequences centered at X are split. Thus we have K_1 \oplus P_2 \cong_R K_2 \oplus P_1.

  2. Let X denote the pushout of \alpha_1 and \alpha_2. (See this previous exercise about pushouts.) We claim that the pushout morphisms \sigma_1 and \sigma_2 are injective. To see this, suppose \sigma_1(x) = 0. Then (x,0) = (\alpha_1(m),-\alpha_2(m) for some m \in M. Since \alpha_2 is injective, we have m = 0, so that x = 0. Thus \sigma_1 is injective. Similarly, \sigma_2 is injective. This yields the two short exact sequences 0 \rightarrow Q_1 \stackrel{\sigma_1}{\rightarrow} X \stackrel{\pi}{\rightarrow} X/\mathsf{im}\ \sigma_1 \rightarrow 0 and 0 \rightarrow Q_2 \stackrel{\sigma_2}{\rightarrow} X \stackrel{\pi}{\rightarrow} X/\mathsf{im}\ \sigma_2 \rightarrow 0. We claim that there exist module homomorphisms \theta_1 : L_1 \rightarrow X/\mathsf{im}\ \sigma_2 and \theta_2 : L_2 \rightarrow X/\mathsf{im}\ \sigma_1 such that the following diagram of modules commutes.

    Another diagram

    To this end, we prove a lemma.

    Lemma 2: If f : A \rightarrow B and g : A \rightarrow C are surjective module homomorphisms and \mathsf{ker}\ f \subseteq \mathsf{ker}\ g, then there exists a unique module homomorphism h : B \rightarrow C such that g = h \circ f. Proof: Define h as follows: given b \in B, h(b) = g(a) where f(a) = b (a exists since f is surjective). Note that if f(a_1) = f(a_2) = b, then a_1 - a_2 \in \mathsf{ker}\ f \subseteq \mathsf{ker}\ g, so that g(a_1) = g(a_2). Thus h is well defined. Moreover, if b_1,b_2 \in B and r \in R, and f(a_1) = b_1 and f(a_2) = b_2, then f(a_1+ra_2) = b_1+rb_2. Thus h(b_1 + rb_2) = g(a_1+ra_2) = g(a_1) + rg(a_2) = h(b_1) + rh(b_2), so that h is a module homomorphism. That g = h \circ f is clear. To see uniqueness, suppose k : B \rightarrow C is a homomorphism such that g = k \circ f. Given b \in B, there exists a \in A such that b = f(a). Now k(b) = (k \circ f)(a) = g(a) = (h \circ f)(a) = h(b), so that k = h. Finally, to see that h is surjective, note that if c \in C then since g is surjective there exists a \in A with g(a) = c. Let b = f(a); then h(b) = c. \square

    Now we claim that \pi \circ \sigma_1 : Q_1 \rightarrow X/\mathsf{im}\ \sigma_2 is surjective. To see this, let ((a,b)+Y) + \mathsf{im}\ \sigma_2 \in X/\mathsf{im}\ \sigma_2. (Recall that X = (Q_1 \oplus Q_2)/Y, where Y = \{(\alpha_1(m),-\alpha_2(m)) \ |\ m \in M\}.) Note that ((a,b)+Y)+\mathsf{im}\ \sigma_2 = ((a,0)+Y)+\mathsf{im}\ \sigma_2 + ((0,b)+Y)+\mathsf{im}\ \sigma_2 = ((a,0)+Y)+\mathsf{im}\ \sigma_2 = (\pi \circ \sigma_1)(a). So \pi \circ \sigma_1 is surjective; likewise \pi \circ \sigma_2 : Q_2 \rightarrow X/\mathsf{im}\ \sigma_1 is surjective. By Lemma 2, there exist unique surjective module homomorphisms \theta_1 and \theta_2 making the diagram commute.

    Finally, we claim that \theta_1 and \theta_2 are injective. To see this, let x \in \mathsf{ker}\ \theta_1. Since \beta_1 is surjective, there exists an element y \in Q_1 such that \beta_1(y) = x. Now 0 = (\theta_1 \circ \beta_1)(y) = (\pi \circ \sigma_1)(y), so that \pi(\sigma_1(y)) = 0. Thus \sigma_1(y) \in \mathsf{ker}\ \pi = \mathsf{im}\ \sigma_2. That is, \sigma_1(y) = \sigma_2(z) for some z \in Q_2. Now \sigma_1(y) - \sigma_2(z) = (y,-z)+Y = 0 in X, so that (y,-z) = (\alpha_1(m),-\alpha_2(m)) for some m \in M. In particular, y \in \mathsf{im}\ \alpha_1, so that x = \beta_1(y) = 0. Thus \mathsf{ker}\ \theta_1 = 0, and so \theta_1 is injective. Similarly, \theta_2 is injective. So L_1 \cong_R X/\mathsf{im}\ \sigma_2 and L_2 \cong_R X/\mathsf{im}\ \sigma_1. Since Q_1 and Q_2 are injective, the middle row and column split, and we have L_1 \oplus Q_2 \cong_R L_2 \oplus Q_1.

Pullbacks and pushouts of modules

Let R be a ring with 1 and let M, A, and B be left unital R-modules.

  1. Let \varphi : A \rightarrow M and \psi : B \rightarrow M be module homomorphisms. Prove that there exists a module X and two module homomorphisms \lambda_1 : X \rightarrow A and \lambda_2 : X \rightarrow B such that \varphi \circ \lambda_1 = \psi \circ \lambda_2 and which have the following property: if Z is a module and \alpha_1 : Z \rightarrow A and \alpha_2 : Z \rightarrow B are module homomorphisms such that \varphi \circ \alpha_1 = \psi \circ \alpha_2, then there exists a unique module homomorphism \theta : Z \rightarrow X such that \alpha_1 = \lambda_1 \circ \theta and \alpha_2 = \lambda_2 \circ \theta. That is, given the following diagram of modules,

    The pullback of two modules

    there exists a unique \theta which makes the diagram commute. Deduce that X is unique up to isomorphism. We will call this X the pullback or fiber product of \varphi and \psi, and sometimes denote it by \mathsf{pb}(\varphi,\psi).

  2. Let \varphi : M \rightarrow A and \psi : M \rightarrow B be module homomorphisms. Prove that there exists a module X and two module homomorphisms \sigma_1 : A \rightarrow X and \sigma_2 : B \rightarrow X such that \sigma_1 \circ \varphi = \sigma_2 \circ \psi and which have the following property: If Z is a module and \alpha_1 : A \rightarrow Z and \alpha_2 : B \rightarrow Z are module homomorphisms such that \alpha_1 \circ \varphi = \alpha_2 \circ \psi, then there exists a unique module homomorphism \theta : X \rightarrow Z such that \theta \circ \sigma_1 = \alpha_1 and \theta \circ \sigma_2 = \alpha_2. That is, given the following diagram of modules,

    The pushout of two module homomorphisms

    there exists a unique \theta which makes the diagram commute. Deduce that X is unique up to isomorphism. We will call this X the pushout or fiber sum of \varphi and \varphi and \psi, and sometimes denote it by \mathsf{po}(\varphi,\psi).


  1. Let X = \{ (a,b) \in A \times B \ |\ \varphi(a) = \psi(b) \}, and let \lambda_1(a,b) = a and \lambda_2(a,b) = b. Note that if (a_1,b_1), (a_2,b_2) \in X and r \in R, then \varphi(a_1+ra_2) = \psi(b_1+rb_2), so that (a_1+ra_2,b_1+rb_2) \in X. Since (0,0) \in X, by the submodule criterion, X is a submodule of A \times B. Certainly if (a,b) \in X, then \varphi(a) = \psi(b), so that (\varphi \circ \lambda_1)(a,b) = (\psi \circ \lambda_2)(a,b). Thus \varphi \circ \lambda_1 = \psi \circ \lambda_2. Now suppose we have Z, \alpha_1, and \alpha_2. Note that for all z \in Z, we have \varphi(\alpha_1(z)) = \psi(\alpha_2(z)), so that (\alpha_1(z), \alpha_2(z)) \in X. Define \theta : Z \rightarrow X by \theta(z) = (\alpha_1(z), \alpha_2(z)). Clearly \theta is a module homomorphism. Moreover, \lambda_1 \circ \theta = \alpha_1 and \lambda_2 \circ \theta = \alpha_2. The uniqueness of \theta follows easily.
  2. Define Y = \{ (\varphi(m), -\psi(m)) \ |\ m \in M \} \subseteq A \oplus B. Note that (0,0) \in Y, and if (\varphi(m_1),-\psi(m_1)), (\varphi(m_2), -\psi(m_2)) \in Y and r \in R, then (\varphi(m_1),-\psi(m_1)) + r(\varphi(m_2), -\psi(m_2)) = (\varphi(m_1+rm_2), -\psi(m_1+rm_2)) \in Y. By the submodule criterion, Y \subseteq A \oplus B is a submodule. Let X = (A \oplus B)/Y, and define \sigma_1 : A \rightarrow X and \sigma_2 : B \rightarrow X by \sigma_1(a) = (a,0) + Y and \sigma_2(b) = (0,b) + Y. Note that for all m \in M, we have (\varphi(m),0) + Y = (\varphi(m),0) - (\varphi(m),-\psi(m)) + Y = (0,\psi(m)) + Y. Thus \sigma_1 \circ \varphi = \sigma_2 \circ \psi. Now suppose we have Z, \alpha_1 : A \rightarrow Z, and \alpha_2 : B \rightarrow Z. Define \theta : X \rightarrow Z by \theta((a,b)+Y) = \alpha_1(a) + \alpha_2(b). To see that \theta is well-defined, suppose (a_1,b_1) - (a_2,b_2) \in Y. Then we have a_1-a_2 = \varphi(m) and b_1-b_2 = \psi(m) for some m \in M. Applying \alpha_1 and \alpha_2, we see that \alpha_1(a_1) - \alpha_1(a_2) = (\alpha_1 \circ \varphi)(m) = (\alpha_2 \circ \psi)(m) = \alpha_2(b_1) - \alpha_2(b_2). Thus \alpha_1(a_1) + \alpha_2(b_1) = \alpha_1(a_2) + \alpha_2(b_2). Clearly \theta is a module homomorphism. Moreover, we see that \theta \circ \sigma_1 = \alpha_1 and \theta \circ \sigma_2 = \alpha_2. To see uniqueness, suppose there exists \zeta : X \rightarrow Z such that \zeta \circ \sigma_1 = \alpha_1 and \zeta \circ \sigma_2 = \alpha_2. Now \zeta((a,b)+Y) = \zeta((a,0)+Y) + \zeta((0,b)+Y) = (\zeta \circ \sigma_1)(a) + (\zeta \circ \sigma_2)(b) = \alpha_1(a) + \alpha_2(b) = \theta((a,b)+Y), so that \zeta = \theta.

There are no fields between QQ and QQ(¹³√2)

Let \alpha be a root of p(x) = x^{13} - 2. Does there exist a field K such that \mathbb{Q} \subsetneq K \subsetneq \mathbb{Q}(\alpha)?


Note that p(x) is irreducible by Eisenstein’s criterion; hence \mathsf{dim}_\mathbb{Q} \mathbb{Q}(\alpha) = 13. Suppose K is a field in between \mathbb{Q} and \mathbb{Q}(\alpha). By Theorem 5.7 in TAN, we have \mathsf{dim}_\mathbb{Q} \mathbb{Q}(\alpha) = \mathsf{dim}_\mathbb{Q} F \cdot \mathsf{dim}_F \mathbb{Q}(\alpha). Since 13 is prime, either \mathsf{dim}_\mathbb{Q} F = 1 or \mathsf{dim}_F \mathbb{Q}(\alpha) = 1; so F = \mathbb{Q} or F = \mathbb{Q}(\alpha).

Thus no fields intermediate to \mathbb{Q} and \mathbb{Q}(\alpha) exist.

Decide whether a given element is algebraic or transcendental over a field

Let F be a field. Suppose \alpha is algebraic over F, \gamma is transcendental over F, that \beta is algebraic over F(\alpha), and \delta is transcendental over F(\alpha). Decide whether the following elements are algebraic or transcendental over the given field.

  1. \alpha + \beta over F
  2. 1/\alpha over F
  3. \alpha+\delta transcendental F
  4. \gamma over F(\alpha)
  5. Is \gamma+\delta transcendental over F?

Note that F(\alpha,\beta) = F(\theta) is a finite extension of F. By Theorem 5.5 in TAN, \alpha+\beta is algebraic over F.

Since 1/\alpha \in F(\alpha) and F(\alpha) is an algebraic (hence finite) extension of F, by Theorem 5.5, 1/\alpha is algebraic over F.

Suppose \alpha+\delta is algebraic over F. Now F(\alpha + \delta) is an algebraic, hence finite, extension of F. Then F(\alpha+\delta,\alpha) is also a finite extension. But (\alpha+\delta) - \alpha = \delta \in F(\alpha+\delta,\alpha), so that (by Theorem 5.5) \delta is algebraic over F. So \delta is algebraic over F(\alpha), a contradiction. Thus \alpha+\delta is transcendental over F.

Suppose \gamma is algebraic over F(\alpha). Now F(\alpha,\gamma) is a finite extension of F, so that \gamma is algebraic over F– a contradiction. Thus \gamma is transcendental over F(\alpha).

Suppose \delta = -\gamma; then \gamma + \delta = 0 is (trivially) algebraic over F. So \gamma + \delta need not be transcendental over F. More generally, consider \alpha+\gamma and -\gamma.

If a and b are algebraic over F and F(a) = F(b), then a and b have the same degree over F

Let F be a field and E an extension of F. Suppose a,b \in E are algebraic over F and that F(a) = F(b). Prove that a and b have the same degree over F.


Note that b \in F(a). By Theorem 4.9 in TAN, \mathsf{deg}_F(b) \leq \mathsf{deg}_F(a). Similarly, \mathsf{deg}_F(a) \leq \mathsf{deg}_F(b). Hence \mathsf{deg}_F(a) = \mathsf{deg}_F(b).

Over a PID, flat and torsion free are equivalent

Let R be a principal ideal domain. Prove that a right unital R-module A is flat if and only if it is torsion free.


Suppose A is flat as a right R-module. Given r \in R nonzero, define \psi_R : R \rightarrow R by \psi_r(a) = ra. Because R is a commutative ring, \psi_r is a left module homomorphism. Because R is a domain, \psi_r is injective. Because A is flat, 1 \otimes \psi_r : A \otimes_R R \rightarrow A \otimes_R R is injective. Now suppose ar = 0, with r \neq 0. In A \otimes_R R, we have a \otimes r = 0 \otimes r, so that (1 \otimes \psi_r)(a \otimes 1) = (1 \otimes \psi_r)(0 \otimes 1). Since 1 \otimes \psi_r is injective, we have a \otimes 1 = 0 \otimes 1. Recall that A \otimes_R R \cong A via the multiplication map a \otimes r \mapsto ar. Thus we have a = 0, and so A is torsion free as a right R-module.

Conversely, suppose A is torsion free as a right R-module, and let I \subseteq R be a nonzero ideal. Since R is a PID, say I = (r) with r \in R nonzero. Note that the map \theta_r : A \rightarrow A given by \theta_r(a) = ar is a module homomorphism. Moreover, \theta_r has a trivial kernel since A is torsion free- so \theta_r is injective. Now consider the map \rho : R \rightarrow I given by \rho(x) = rx; this is a module homomorphism. Certainly \rho is surjective, and moreover since R is a domain, \rho is injective. Thus \rho is a module isomorphism. Now the homomorphism 1 \otimes \rho : A \otimes_R R \rightarrow A \otimes_R I is also an isomorphism since (1 \otimes \rho) \circ (1 \otimes \rho^{-1}) = 1 \otimes 1 = 1 and (1 \otimes \rho^{-1}) \circ (1 \otimes \rho) = 1 \otimes 1 = 1. Recall that A and A \otimes_R R are isomorphic as abelian groups via the mappings \iota : A \rightarrow A \otimes_R R and \tau : A \otimes_R R \rightarrow A given by \iota(a) = a \otimes 1 and \tau(a \otimes r) = ar. Consider the map 1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R.

Now (\tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota)(a) = \tau((1 \otimes \iota)((1 \otimes \rho)(a \otimes 1))) = \tau((1 \otimes \iota)(a \otimes r)) = \tau(a \otimes r) = ar = \theta_r(a). Thus we have \tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota = \theta_r, and so 1 \otimes \iota = \tau^{-1} \circ \theta_r \circ \iota^{-1} \circ (1 \otimes \rho). Thus 1 \otimes \iota is injective.

By the flatness criterion for modules (proved here), A is flat as a right R-module.

A flatness criterion for modules

Let R be a ring with 1, and let A be a right unital R-module.

  1. Prove that if A is flat then for every ideal I \subseteq R the mapping A \otimes_R I \rightarrow A \otimes_R R induced by the inclusion I \rightarrow R is injective.
  2. Suppose that for every finitely generated ideal I \subseteq R, the mapping A \otimes_R I \rightarrow A \otimes_R R induced by the inclusion map is injective. Prove that in fact the induced map A \otimes_R I \rightarrow A \otimes_R R is injective for every ideal I. Show further that if K is any submodule of a finitely generated free module F then A \otimes_R K \rightarrow A \otimes_R F is injective. Show that the same is true for any free module (not necessarily finitely generated).
  3. Under the hypothesis in part (2), suppose \psi : L \rightarrow M is an injective left R-module homomorphism. Prove that the induced map 1 \otimes \psi : A \otimes_R L \rightarrow A \otimes_R M is injective. Conclude that A is flat.
  4. Let F be a flat right unital R-module and K \subseteq F an R-submodule of F. Prove that F/K is flat if and only if FI \cap K = KI for every finitely generated ideal I \subseteq R.

  1. This is true by our definition of flat module.
  2. Suppose the induced map 1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R is injective for all finitely generated ideals I \subseteq R. Let I \subseteq R be an arbitrary ideal and consider the induced map 1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R. Suppose (1 \otimes \iota)(\sum a_i \otimes r_i) = 0. Now there is a finitely generated ideal J \subseteq R which contains all of the r_i. Moreover, 1 \otimes \iota restricted to A \otimes J is injective by our hypothesis. Thus \sum a_i \otimes r_i = 0, and so 1 \otimes \iota is injective.

    Now let F = \oplus_T R be a finitely generated free module, and let K \subseteq F be a submodule. Now K \subseteq \oplus_T I_t, where I_t \subseteq R is a submodule (i.e. a left ideal). (Specifically, I_t is the set of all tth coordinates of elements of K. This set is an R-submodule of R (an ideal) since K is an R-module.) Now A \otimes_R (\oplus_T I_t) \cong \oplus_T (A \otimes_R I_t) \hookrightarrow \oplus_T A \otimes_R R \cong A \otimes_R (\oplus_T R) = A \otimes_R F. Thus the induced map 1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F is injective.

    Finally, let F be an arbitrary free module and let K \subseteq F be an arbitrary submodule. Consider the induced map 1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F. If (1 \otimes \iota)(\sum a_i \otimes k_i) = 0, then there is a finitely generated submodule of F containing all the k_i. Moreover, 1 \otimes \iota restricted to this submodule is injective, so that \sum a_i \otimes k_i = 0. Thus 1 \otimes \iota is injective.

  3. Let \psi : L \rightarrow M be an injective left module homomorphism. Now every module (for instance M) is isomorphic to a quotient of a free module. Suppose F is a free module and K \subseteq F a submodule such that M = F/K. In particular, the sequence 0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0 is short exact. Let J \subseteq F be the subset consisting of precisely those elements x \in F such that \pi(x) \in \mathsf{im}\ \psi, where \pi : F \rightarrow M is the natural projection. That is, J = \{ x \in F \ |\ \pi(x) \in \mathsf{im}\ \psi \}. Define \theta : J \rightarrow L by \theta(x) = y, where \pi(x) = \psi(y). Now \theta is total by definition and is well-defined because \psi is injective. Suppose now that x,y \in J and r \in R. There exist a,b \in L such that \theta(x) = a and \theta(y) = b. So \pi(x) = \psi(a) and \pi(y) = \psi(b). Now \pi(x) + r\pi(y) = \psi(a) + r\psi(b), so that \pi(x+ry) = \psi(a+rb). Thus we have \theta(x+ry) = a+rb = \theta(x) + r\theta(y), and so \theta is a left R-module homomorphism. We claim also that K \subseteq J. It is certainly the case that K \subseteq J, since K = \mathsf{ker}\ \pi. This gives the following diagram of module homomorphisms.

    A diagram of modules

    We claim that this diagram commutes and that the rows are exact. First we show exactness; the bottom row is exact by definition. For the top row, certainly \mathsf{im}\ \iota \subseteq \mathsf{ker}\ \theta, since \psi is injective. Now suppose x \in \mathsf{ker}\ \theta. Then \pi(x) = 0, so that x \in \mathsf{ker}\ \pi = K = \mathsf{im}\ \iota. Thus the rows are exact. Now we show that the diagram commutes. The left square certainly commutes, since the maps are all simply inclusions. Now let x \in J. Note that (\psi \circ \theta)(x) = \psi(\theta(x)) = \psi(y), where \pi(x) = \psi(y). Thus \psi \circ \theta = \pi, and so the diagram commutes. In fact, the triple (\mathsf{id}, \iota, \psi) is a homomorphism of short exact sequences.

    Recall that the functor A \otimes_R - is right exact. Thus the induced diagram of modules

    Another diagram of modules

    commutes and has exact rows. By part (2) above, 1 \otimes \iota : A \otimes_R J \rightarrow A \otimes_R F is injective, and certainly \mathsf{id} : A \otimes_R K \rightarrow A \otimes_R K and 1 \otimes \theta are surjective. By part 4 of this previous exercise, 1 \otimes \psi is injective.

    Thus A is flat as a right R-module.

  4. Suppose F is flat, K \subseteq F a submodule, and that for all finitely generated ideals I we have FI \cap K = KI. We wish to show that F/K is flat. Let I \subseteq R be an arbitrary finitely generated ideal. By the flatness criterion proved above, it suffices to show that 1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R is injective. To that end, consider the exact sequences shown in the rows of the following diagram.

    Yet another diagram of modules

    With t_K given by k \otimes a \mapsto ka (and likewise t_F) and \iota_K by k \mapsto k \otimes 1 (and likewise \iota_F), we claim that there exist suitable injective maps \alpha and \beta so that this diagram commutes.

    Define \alpha^\prime : F/K \times I \rightarrow FI/KI by \alpha^\prime(x+K, a) = xa+KI. Certainly if x-y \in K, then xa-ya \in KI, so that \alpha^\prime is well defined. Moreover, this map is clearly bilinear. Thus we have an R-module homomorphism \alpha : F/K \otimes_R I \rightarrow FI/KI given by (x+K) \otimes a \mapsto xa+KI. Clearly the upper right hand square commutes.

    Note that t_K is surjective (obviously). Moreover, since F is flat, the map 1 \otimes \iota : F \otimes_R I \rightarrow F \otimes_R R is injective (by part (1) above). Now t_F : F \otimes_R R \rightarrow F is an isomorphism and we have t_F = t_F \circ (1 \otimes \iota), so that t_F in our diagram is injective. By part 1 of this previous exercise, \alpha is injective.

    Now define \beta : FI/KI \rightarrow F/K \otimes_R R by x+KI \mapsto x \otimes 1. First, if x-y \in KI \subseteq K, then (x+K) \otimes 1 = (y+K) \otimes 1, so that \beta is well defined. It is clear that \beta is a homomorphism of modules, and that the lower right hand square commutes. We claim also that \beta is injective. To see this, suppose x+KI \in \mathsf{ker}\ \beta. Now (x+K) \otimes 1 = 0. Recall that F/K \otimes_R R \cong F/K via the “multiplication” homomorphism (x+K) \otimes r \mapsto xr+K; so in fact we have x+K = 0, so that x \in K. Now x \in FI and x \in K, so that (by our hypothesis) x \in KI. So x+KI = 0, and in fact \beta is injective.

    Finally, we claim that \beta \circ \alpha = 1 \otimes \iota. To see this, note that (\beta \circ \alpha)((x+k) \otimes a) = \beta(xa+KI) = (xa+K) \otimes 1 = (x+K) \otimes a.

    Thus 1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R is injective for all finitely generated ideals I \subseteq R. By the flatness criterion, F/K is flat as a right R-module.

    Conversely, suppose F and F/K are flat and let I \subseteq R be a finitely generated ideal. Note that KI \subseteq FI and KI \subseteq K, so that the inclusion KI \subseteq FI \cap K always holds. Now consider the following commutative diagram of modules.

    Last one, I promise!

    Again, t_K (and t_F and t_{F/K} denotes the isomorphism K \otimes_R R \rightarrow K given by k \otimes r \mapsto kr. Let \sum \zeta = x_ia_i \in FI \cap K. (Letting x_i \in F and a_i \in I.) We will chase \zeta around the diagram to see that \zeta \in KI. Note that 0 = \pi(\zeta) = \sum x_ia_i + K, so that \sum (x_i+K) \otimes a_i = 0 in F/K \otimes_R R. Note that (1 \otimes \iota)(\sum (x_i+K) \otimes a_i) = \sum (x_i+K) \otimes a_i, and because 1 \otimes \iota is injective (by our hypothesis that F/K is flat), \sum (x_i+K) \otimes a_i = 0 in F/K \otimes_R I. In particular, \sum x_i \otimes a_i \in \mathsf{ker}\ \pi \otimes 1. Since the top row is exact, we have \sum x_i \otimes a_i = \sum y_i \otimes b_i for some y_i \in K and b_i \in I. Following these elements down to F, we see that \zeta = \sum y_ib_i \in KI as desired.

    Thus if F and F/K are flat, then FI \cap K = KI for all finitely generated ideals I \subseteq R.