## 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 $t$th 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 $I \subseteq R$.