## Tag Archives: degree

### Every degree n extension of a field F is embedded in the set of nxn matrices over F

Let $F$ be a field, and let $K$ be an extension of $F$ of finite degree.

1. Fix $\alpha \in K$. Prove that the mapping ‘multiplication by $\alpha$‘ is an $F$-linear transformation on $K$. (In fact an automorphism for $\alpha \neq 0$.)
2. Deduce that $K$ is isomorphically embedded in $\mathsf{Mat}_n(F)$.

Let $\varphi_\alpha(x) = \alpha x$. Certainly then we have $\varphi_\alpha(x+ry) = \alpha(x+ry) = \alpha x + r \alpha y$ $= \varphi_\alpha(x) + r \varphi_\alpha(y)$ for all $x,y \in K$ and $r \in F$; so $\varphi_\alpha$ is an $F$-linear transformation. If $\alpha \neq 0$, then evidently $\varphi_{\alpha^{-1}} \circ \varphi_\alpha = \varphi_\alpha \circ \varphi_{\alpha^{-1}} = 1$.

Fix a basis for $K$ over $F$; this yields a ring homomorphism $\Psi : K \rightarrow \mathsf{Mat}_n(F)$ which takes $\alpha$ and returns the matrix of $\varphi_\alpha$ with respect to the chosen basis. Suppose $\alpha \in \mathsf{ker}\ \Psi$; then $\varphi_\alpha(x) = 0$ for all $x \in K$, and thus $\alpha = 0$. So $\Psi$ is injective as desired.

### On the degree of an extension of F(x)

Let $F$ be a field, and consider the field $F(x)$ of rational functions over $F$ (that is, the field of fractions of the domain $F[x]$). Let $t(x) = p(x)/q(x)$, with $p,q \in F[x]$ and $q \neq 0$ such that the degree of $q$ is strictly larger than the degree of $p$. In this exercise, we will compute the degree of $F(x)$ over $F(t(x))$. (Note that if $p$ has degree larger than or equal to the degree of $q$, then we can use the division algorithm to find $p(x) = q(x)b(x) + r(x)$, and then $F(t(x)) = F(r(x)/q(x))$.)

1. Show that $p(y) - t(x)q(y) \in F(t(x))[y]$ is irreducible over $F(t(x))$ and has $x$ as a root (in the extension $F(x)$).
2. Show that the degree of $p(y) - t(x)q(y)$ as a polynomial in $y$ is the maximum of the degrees of $p$ and $q$.
3. Conclude that $[F(x):F(t(x))] = \mathsf{max}(\mathsf{deg}\ p(x)), \mathsf{deg}\ q(x))$.

(Fields of rational functions were introduced in Example 4 on page 264 of D&F.)

We claim that $t(x)$ is indeterminate over $F$– that is, that $t(x)$ does not satisfy any polynomial over $F$. Indeed, if $\sum_{i=0}^n c_i t(x)^i = \sum c_i \dfrac{p(x)^i}{q(x)^i} = 0$, then we have $\sum c_i p(x)^iq(x)^{n-i} = 0$. Let $d_p$ and $d_q$ denote the degrees of $p$ and $q$, respectively. Since $F$ has no zero divisors, the degree of the $i$th summand is $d_pi + d_q(n-i)$. Suppose two summands have the same degree; then $d_pj + d_q(n-i) = d_pj + d_q(n-j)$ for some $i$ and $j$, which reduces to $(d_p-d_q)i = (d_p-d_q)j$. Since (as we assume) $d_p \neq d_q$, we have $i = j$. In this case, we can pick out the summand with the highest degree, and be guaranteed that no other summands contribute to its term of highest degree. This gives us that $c_i = 0$ for the highest degree summand; by induction we have $c_i = 0$ for all the coefficients $c_i$, a contradiction. (In short, no two summands have the same highest degree term. Starting from the highest of the highest degree terms and working down, we show that each $c_i$ is 0.)

So $t(x)$ is indeterminate over $F$, and in fact $F[t(x)]$ is essentially a polynomial ring whose field of fractions is $F(t(x))$. By Gauss’ Lemma, the polynomial $p(y) - t(x)q(y)$ is irreducible over $F(t(x))$ if and only if it is irreducible over $F[t(x)]$. Now $F[t(x)][y] = F[y][t(x)]$, and our polynomial is irreducible in this ring since it is linear in $t(x)$. So in fact $p(y) - t(x)q(y)$ is irreducible over $F(t(x))$, and moreover has $x$ as a root.

The degree of $p(y) - t(x)q(y)$ in $y$ is the maximum of the degrees of $p$ and $q$ because the coefficient of each term (in $y$) is a linear polynomial in $t(x)$, which is nonzero precisely when one of the corresponding terms in $p$ or $q$ is nonzero. That is, we cannot have nonzero terms in $p(y)$ and $-t(x)q(y)$ adding to give a zero term.

To summarize, $p(y) - t(x)q(y)$ is an irreducible polynomial over $t(x)$ with $x$ as a root, and so must be (essentially) the minimal polynomial of $F(x)$ over $F(t(x))$. By the preceding paragraph, the degree of this extension is the larger of the degrees of $p(x)$ and $q(x)$.

### On the degrees of the divisors of a composite polynomial

Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$, and let $g(x)$ be any polynomial in $F[x]$. Prove that every irreducible factor of the composite $(f \circ g)(x)$ has degree divisible by $n$.

Let $\beta$ be a root of $f \circ g$, and let $m$ be the degree of $\beta$ over $F$. Now $g(\beta) \in F(\beta)$, and moreover $g(\beta)$ is a root of the irreducible polynomial $f(x)$. So $F(g(\beta))$ has degree $n$ over $F$. Graphically, we have the following diagram of fields.

A field diagram

By Theorem 14 in D&F, $n|m$.

### Odd degree extensions of formally real fields are formally real

A field $F$ is called formally real if -1 cannot be expressed as a sum of squares in $F$. Let $F$ be a formally real field and let $p(x)$ be irreducible over $F$ of odd degree with a root $\alpha$. Show that $F(\alpha)$ is formally real.

Suppose to the contrary that there exist formally real fields $F$ and irreducible, odd degree polynomials $p$ over $F$ with roots $\alpha$ such that $F(\alpha)$ is not formally real. Choose $F$ and $p$ from among these fields such that the degree of $p$ is minimal, let $\alpha$ be a root of $p$, and consider (1) $F(\alpha) \cong F[x]/(p(x))$.

Since (as we assume) $F(\alpha)$ is not formally real, we have $-1 = \sum \theta_i$ for some $\theta_i \in F(\alpha)$. Let $p_i$ be the residue in $F[x]$ corresponding to $\theta_i$ under the isomorphism (1); then we have $-1 \equiv \sum p_i(x)^2$ in $F[x]/(p(x))$, so that $-1 + p(x)q(x) = \sum p_i(x)^2$ in $F[x]$, for some $q(x)$.

Note that, since each $p_i$ is a residue mod $p$ (using the division algorithm), each $p_i$ has degree strictly less than the degree of $p$. Since $F$ is a domain, $\sum p_i(x)^2$ has even degree less than $2 \mathsf{deg}\ p$. So that $pq$, has even degree, and since $p$ has odd degree, $q$ has odd degree, and in fact the degree of $q$ is strictly less than the degree of $p$. Moreover, since the degree of $q$ is the sum of the degrees of its irreducible factors, some irreducible factor of $q$, say $t$, must also have odd degree (less than that of $p$).

Let $\beta$ be a root of $t(x)$. In $F(\beta) = F[x]/(t(x))$, we have that $-1 \equiv \sum p_i(x)^2$. So $F$ is formally real, $t$ irreducible over $F$, $\beta$ a root of $t$, and $F(\beta)$ is not formally real – but $F(\beta)$ has degree (over $F$) strictly less than the degree of $F(\alpha)$, a contradiction.

So no such fields and polynomials exist, and in fact every odd extension of a formally real field is formally real.

### If [F(α):F] is odd, then F(α²) = F(α)

Let $F$ be a field, and let $\alpha$ be algebraic over $F$. Prove that if $[F(\alpha):F]$ is odd, then $F(\alpha^2) = F(\alpha)$.

The inclusion $F(\alpha^2) \subseteq F(\alpha)$ is immediate.

Let $p(x) = \sum c_ix^i$ be the minimal polynomial of $\alpha$ over $F$. Now write $p(x) = \sum c_{2i}x^{2i} + \sum c_{2i+1}x^{2i+1}$, separating the even and odd terms. Now $p(x) = \sum c_{2i}(x^2)^i + \sum c_{2i+1}x(x^2)^i$ $= r(x^2) + xs(x^2)$, and we have $r(\alpha^2) + \alpha s(\alpha^2) = 0$. Since $p(x)$ has odd degree and the powers of $\alpha$ are linearly independent over $F$, $s(\alpha^2) \neq 0$. Thus $\alpha = -r(\alpha^2)/s(\alpha^2) \in F(\alpha^2)$, and we have $F(\alpha) \subseteq F(\alpha^2)$.

Note that our proof only needs some nonzero odd-degree term in the minimal polynomial of $\alpha$, so in some cases the proof holds if $\alpha$ has even degree.

### 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 the degree of a given extension of the rationals

Compute the degrees of $\mathbb{Q}(\sqrt{3+4i} + \sqrt{3-4i})$ and $\mathbb{Q}(\sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}})$ over $\mathbb{Q}$.

Let $\zeta = \sqrt{3+4i} + \sqrt{3-4i}$. Evidently, $\zeta^2 = 16$. (WolframAlpha agrees.) That is, $\zeta$ is a root of $p(x) = x^2 - 16 = (x+4)(x-4)$. Thus the minimal polynomial of $\zeta$ over $\mathbb{Q}$ has degree 1, and the extension $\mathbb{Q}(\zeta)$ has degree 1 over $\mathbb{Q}$.

Similarly, let $\eta = \sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}}$. Evidently, $\eta^2 = 6$ (WolframAlpha agrees), so that $\eta$ is a root of $q(x) = x^2-6$. $q$ is irreducible by Eisenstein’s criterion, and so is the minimal polynomial of $\eta$ over $\mathbb{Q}$. The degree of $\mathbb{Q}(\eta)$ over $\mathbb{Q}$ is thus 2.

### Compute the degree of a given extension of the rationals

Find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$.

We have $\sqrt{3+2\sqrt{2}} = \sqrt{3 + \sqrt{8}}$, and $3^2 - 8 = 1^2$ is square over $\mathbb{Q}$. By this previous exercise, $\sqrt{3+2\sqrt{2}} = 1 + \sqrt{2}$. So $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}}) = \mathbb{Q}(1+\sqrt{2}) = \mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$.

### Compute the degree of a given extension of the rationals

Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Conclude that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ has degree 4 over $\mathbb{Q}$. Find an irreducible polynomial satisfied by $\sqrt{2} + \sqrt{3}$.

The $(\supseteq)$ inclusion is clear. To see the $(\subseteq)$ inclusion, let $\zeta = \sqrt{2} + \sqrt{3}$. Then evidently $\frac{1}{2}(\zeta^3 - 9\zeta) = \sqrt{2}$ and $-\frac{1}{2}(\zeta^3 - 11\zeta) = \sqrt{3}$. (WolframAlpha agrees; see here and here.) Thus the extensions are equal.

Recall that $\mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$. Now we claim that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree 2 over $\mathbb{Q}(\sqrt{2})$. To see this, suppose to the contrary that $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$. Say $(a+b\sqrt{2})^2 = 3$ for some rationals $a$ and $b$. Coparing coefficients, we have $a^2 + 2b^2 = 3$ and $2ab = 0$; if $a = 0$, then $b^2 = \frac{3}{2}$, and if $b = 0$, then $a^2 = 3$. Either case yields a contradiction.

So $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ has degree 4 over $\mathbb{Q}$.

The minimal polynomial of $\zeta$ has degree 4 over $\mathbb{Q}$. We can solve the system $\zeta^4 + a\zeta^3 + b\zeta^2 + c\zeta + d = 0$ over $\mathbb{Q}$, and in so doing we see that $p(x) = x^4-10x^2+1$ is the minimal polynomial of $\sqrt{2} + \sqrt{3}$.

### Compute the degree of a given extension of QQ

Compute the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$, where $\alpha$ is $2+\sqrt{3}$ or $1 + \sqrt[3]{2} + \sqrt[3]{4}$.

Since $2+\sqrt{3} \in \mathbb{Q}(\sqrt{3})$, the degree of $\mathbb{Q}(2+\sqrt{3})$ over $\mathbb{Q}$ is at most 2. We can solve the linear system $\alpha^2 + a\alpha + b = 0$ in $\mathbb{Q}(\sqrt{3})$ (as a vector space over $\mathbb{Q}$) to find a polynomial satisfied by $\alpha$; evidently $2+\sqrt{3}$ is a root of $p(x) = x^2 - 4x + 1$. (WolframAlpha agrees.) Evidently, $p(x+1) = x^2-2x-2$ (WolframAlpha agrees), which is irreducible by Eisenstein’s criterion; so $p(x)$ is irreducible. Thus $p(x)$ is the minimal polynomial of $2+\sqrt{3}$ over $\mathbb{Q}$, and so the degree of $\mathbb{Q}(2+\sqrt{3})$ over $\mathbb{Q}$ is 2.

In a similar fashion, $\beta = 1+\sqrt[3]{2} + \sqrt[3]{4}$, as an element of $\mathbb{Q}(\sqrt[3]{2})$, has degree at most 3 over $\mathbb{Q}$. Evidently, $\beta$ is a root of $q(x) = x^3 - 3x^2 - 3x - 1$ (WolframAlpha agrees). Evidently, $q(x+1) = x^3-6x-6$, which is irreducible by Eisenstein. So $q(x)$ is the minimal polynomial of $\beta$, and thus $\mathbb{Q}(\beta)$ has degree 3 over $\mathbb{Q}$.