Gallian (9th ed), Chapter 20

5

Take $\mathbb{C}$ an extension of $\mathbb{Q}$. Interpret the polynomial over it.

The roots of $x^2 + x + 1$ over $\mathbb{C}$, using the quadratic formula, are

$$x = \frac{-1 \pm \sqrt{-3}}{2}$$

Similarly, the roots of $x^2 - x + 1$ over $\mathbb{C}$ are

$$x = \frac{1 \pm \sqrt{-3}}{2}$$

It follows a splitting field is $\mathbb{Q}(\frac{-1 \pm \sqrt{-3}}{2},\frac{1 \pm \sqrt{-3}}{2}) = \mathbb{Q}(\frac{1 \pm \sqrt{-3}}{2})$, eliminating algebraic dependency over $\mathbb{C}$.

By the interpretation of vector spaces:

$$\mathbb{Q} \left ( \frac{1 \pm \sqrt{-3}}{2} \right ) = \left \{a + b \left ( \frac{1 + \sqrt{-3}}{2} \right ) + c \left ( \frac{1 - \sqrt{-3}}{2} \right ) + d \left ( \frac{1 + \sqrt{-3}}{2} \right ) \left ( \frac{1 - \sqrt{-3}}{2} \right ) \mid a,b,c,d \in \mathbb{Q} \right \}$$

12

Take $p(x) = x^3 - \pi^3$. Then $p(\pi) = \pi^3 - \pi^3 = 0$. Hence the extension is algebraic. In order to conclude the basis is $\{1, \pi, \pi^2\}$, we show $p$ is a minimal degree polynomial with root $\pi$.

Assume towards a contradiction $p'(\pi) = a\pi + b = 0$ for some $p'$. It follows $\pi = -b/a$, implying $\pi \in \mathbb{Q}(\pi^3)$. Hence $\pi = P_0(\pi^3)/Q_0(\pi^3)$ for some $P_0, Q_0 \in \mathbb{Q}[x]$. Observe:

$$Q_0(\pi^3) \pi = P_0(\pi^3)$$

Interpret that expression as polynomials over the single variable $\pi$. Clearly the degree of the L.H.S cannot match the degree of the R.H.S. Contradiction.

Assume towards a contradiction $p'(\pi) = a\pi^2 + b\pi + c = 0$ for some $p'$. Since by definition $a,b,c \in \mathbb{Q}(\pi^3)$:

$$p'(\pi) = \frac{P_0(\pi^3)}{Q_0(\pi^3)} \pi^2 + \frac{P_1(\pi^3)}{Q_1(\pi^3)} \pi + \frac{P_2(\pi^3)}{Q_2(\pi^3)} = 0$$

where $P_0, Q_0, P_1, Q_1, P_2, Q_2 \in \mathbb{Q}[x]$. By multiplying both sides by $Q_0(\pi^3) Q_1(\pi^3) Q_2(\pi^3)$:

$$\begin{aligned} Q_1(\pi^3) Q_2(\pi^3) P_0(\pi^3) \pi^2 + Q_0(\pi^3) Q_2(\pi^3) P_1(\pi^3) \pi + Q_0(\pi^3) Q_1(\pi^3) P_2(\pi^3) = 0 \\ Q_0(\pi^3) Q_1(\pi^3) P_2(\pi^3) = -Q_1(\pi^3) Q_2(\pi^3) P_0(\pi^3) \pi^2 - Q_0(\pi^3) Q_2(\pi^3) P_1(\pi^3) \pi \end{aligned}$$

Interpret that expression as polynomials over the single variable $\pi$. Clearly the degree of the L.H.S cannot match the degree of the R.H.S. Contradiction.

14

Fact. For an isomorphism $\phi: \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]$:

For (1), it suffices to show $\phi(1/n) \cdot n = 1$. Observe

$$\phi(1/n) \cdot n = \phi(1/n) \cdot \phi(n) = \phi(n/n) = \phi(1) = 1$$

For (2), let $k = a/b$ for integers $a,b \in \mathbb{Z}$. Then

$$\phi(k) = \phi(a/b) = \phi(1/b) \phi(a) = (1/b) a = a/b$$

For (3), it follows by definition since $\phi$ is a homomorphism, interpreting $k$ as a constant polynomial in $\mathbb{Q}[x]$. Then we use (3) to get $\phi(k)\phi(p) = k \phi(p)$.

Fact. An isomorphism $\phi: \mathbb{Q}[x] \rightarrow \mathbb{Q}[x]$ is decided by $\phi(x)$.

It follows immediately by applying the shown linearity. For example, $ax^2 + bx + c \mapsto a[\phi(x)]^2 + b \phi(x) + c$.

Lemma. The only isomorphism $\phi: \mathbb{Q}[x] \mapsto \mathbb{Q}[x]$ is the identity.

If $\phi(x) = c$ then $\phi$ won’t be surjective.

If $\phi(x) = a_mx^{k_m} + \cdot + a_0x^{k_0}$ with degree at least $2$, then $\phi$ won’t be surjective. We know $\phi$ maps a constant to a constant. If $\phi$ is applied to a non-constant polynomial, it yields a polynomial of degree at least $2$. Thereby, polynomials of degree $1$ are missed.

Therefore, $\phi(x) = kx$. For some polynomials, by definition we know:

$$\begin{aligned} \phi(a_0x \cdot a_1x) &= \phi(a_0x) \cdot \phi(a_1x) \\ k a_0a_1x^2 &= ka_0x \cdot ka_1x \\ &= k^2a_0a_1x^2 \end{aligned}$$

It concludes $k = k^2$ which is true only if $k = 0$ or $k = 1$. The former maps all polynomials to constants violating the surjectivity of $\phi$.

Therefore, $\phi(x) = x$, i.e the only isomorphism is the identity.

Theorem. Main Problem.

Observe $p(x) = x^3 - 5$ satisfies $p(5^{1/3}) = 0$. By Eisenstein’s criteria with prime $5$, the polynomial $p$ is irreducible over $\mathbb{Q}$. By vector space interpretation:

$$\mathbb{Q}(5^{1/3}) = \{ a + b 5^{1/3} + c 5^{2/3} \mid a,b,c \in \mathbb{Q} \}$$

Considering the interpretation of polynomials over the variable $5^{1/3}$, the only isomorphism $\phi: \mathbb{Q}(5^{1/3}) \rightarrow \mathbb{Q}(5^{1/3})$ is the identity.

20

Partially Solved.

It suffices to find a polynomial $p(x) \in \mathbb{Q}[x]$ which has $\sqrt{1 + \sqrt{5}}$ as a root.

Take $p(x) = (x^2 - 1)^2 - 5 = x^4 - 2x^2 - 4$. Trivially, polynomial $p$ has $\sqrt{1 + \sqrt{5}}$ as a root. Thereby, there exists an irreducible minimal degree polynomial which has $\sqrt{1 + \sqrt{5}}$ as a root.

If $p$ is a minimal degree polynomial, then we are done. Maybe $p$ is reducible over $\mathbb{Q}$ and hence can be factored.

38

Fact. $(\mathbb{Z}/3\mathbb{Z})(i) = \mathbb{Z}_3(i)$ is a field.

Factorize $p(x) = x^4 - x^2 - 2 = (x^2 + 1)(x^2 - 2)$. It suffices to construct an extension where each factor has a root.

Observe $x^2 + 1$ is irreducible and has a root over $\mathbb{Z}_3(i)$.

However, $x^2 - 2$ has no roots over $\mathbb{Z}_3(i)$. Consider $\mathbb{Z}_3(i)/\langle x^2 - 2 \rangle$. Clearly $x^2 - 2$ is irreducible and has a root over it. The root is $x + \langle x^2 - 2 \rangle$; Call it $\sqrt{2}$. It follows

$$\mathbb{Z}(i, \sqrt{2}) = \mathbb{Z}(i)(\sqrt{2}) = \{ a + bi + c\sqrt{2} + di\sqrt{2} \mid a,b,c,d \in \mathbb{Z} \}$$

is a splitting field of $x^4 - x^2 - 2$ over $\mathbb{Z}_3$.