# Exam Review **Sections for the exam:** - 2.1-2.4 ($\mathbf R, \mathbf Q, \mathbf C$) and quadratic/cubic formulas - 3.1-3.3 (Euclidean algorithm, roots, $\mathbf Z[x]$) - 5.1 (Vector Spaces, toward field extensions) **Some unsolicited advice**: - First and foremost, review homeworks and problem sets. These are the most likely topics to show up on an exam. You'll want to know the statements of any results proved in exercises, along with the details of how you proved or computed things. - For every definition or theorem, have an example and a counterexample in mind! E.g. when defining a ring, know by heart a good example of a set that *is* a ring and a set that is *not* a ring. - Whenever possible, ask yourself "What's the picture?" I.e., is there some sketch/doodle/cartoon you can associate to a definition, theorem, or example that helps intuitively capture what's going on? Or if nothing else, you can use such pictures as mnemonic devices to help trigger recall later. - Know the definitions by heart! You'll want to have them memorized. - When reviewing proofs, try to remember a "sketch" of the idea first, i.e. a high-level understanding of the various steps or tricks used to move through the proof. A good example is exercise 3.3.8 on proving "Wilson's theorem" using field theory -- a sketch might look like - Trick: consider this very special polynomial $f(x) = x^p - x$. - Show it has $p$ distinct roots in $\mathbf{Z}/p\mathbf Z$ using the fundamental theorem of algebra and Fermat's Little Theorem - Factor out $f$ and remove the $(x-0)$ term to get $g(x) = {f(x) \over (x-0)} = x^{p-1} - 1 = (x-1)(x-2)\cdots$. - Plug in $g(p)$ and see what happens. - Some of these computations can be done by online calculators like WolframAlpha, Symbolab, Sagemath, etc. Do use these to check your work -- but also make sure you have tricks and techniques for doing it if you were stranded on a desert island with no electronics! (I.e. in an exam setting). **Notation**: - $R$ is usually a ring, and $A, B ,S$ etc are subsets of $R$. - $\mathbf{Z, Q, R, C}$ denote the fields of integers, rationals, reals, and complex numbers respectively. Sometimes these are written $\Z, \Q, \R, \C$ as well. - The notation $:=$ means "this is a definition". - $\mathbf{Z}/m\mathbf{Z}$ is the ring of integers modulo $m$, and its elements are often denoted $\left\{\bar 0, \bar 1,\cdots, \overline{m-1}\right\}$. - $K, L$ are abstract fields (which could for example be $\mathbf{Q}$) and $\bar K, \bar L$ denote algebraic closures. - $K[\alpha]$ is the field extension of $K$ obtained by adjoining an element $\alpha\in \bar K$ to the base field $K$. - $R[x]$ is the ring of polynomials with coefficients in $R$, so finite degree polynomials of the form $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ where $a_0, \cdots, a_n \in R$. - $i$ always means the purely imaginary complex number $e^{i\pi\over 2}$, which is a solution to the equation $x^2=-1$. - $\zeta_n := e^{2 i \pi \over n}$ is a *primitive $n$th root of unity*, one of the $n$ complex solutions to the equation $x^n=1$. - $\mathrm{M}_n(R), \mathrm{Mat}_n(R)$ denotes $n\times n$ matrices over a ring $R$. Note that such matrices may be singular. - $\mathrm{GL}_n(R)$ denotes the set of $n\times n$ invertible matrices with entries from $R$, i.e. matrices $M$ with $\det M \neq 0$. ## Rings and fields - Define what it means for a set $R$ to be a **ring**. > A set $R$ is a ring iff $(R, +)$ is an abelian group (associative, commutative, identities, inverses), $(R,\cdot)$ is a monoid (associative, identities), and these are compatible via the distributive law: $a(b+c) = ab+ac, (a+b)c = ac + bc$. - Define what it means for a subset $S \subseteq R$ to be a **subring**. > $S$ is a subring iff $(S, +, \cdot)$ forms a ring with the operations restricted from $R$, i.e. $S$ is closed with respect to all of these operations, and the multiplicative identity $1\in S$. > Mnemonic: a subset that itself forms a ring. - Let $A$ and $B$ be subrings of a ring $R$. Is $A \cap B$ a subring? Explain. - *Hint: true. How do you prove it?* > Idea: $A$ is closed under all operations, as is $B$, so any operation on $x,y\in A\cap B$ will again be in $A$ and in $B$ and thus back in $A\cap B$. > > An easier trick: the one-step subring test. Let $S = A\cap B$, check > > - $S\neq \emptyset$, need to assume this here > - $x,y\in S\implies x - y\in S$. True since $x,y\in S\implies x,y\in A, B \implies x-y\in A, B$. > - $x,y\in S\implies xy\in S$: True since $x,y\in S\implies x,y\in A, B\implies xy\in A, B$. - Let $A$ and $B$ be subrings of a ring $R$. Is $A \cup B$ a subring? Explain. - *Hint: false. Consider $a+b$ where $a\in A, b\in B$.* > Idea: not closed under addition. Take $R=\Z, A=2\Z, B=3\Z$, then $A\cup B$ consists of exactly multiples of $2$ or $3$. But let $x= 2, y=3$, then $x+y=5$ is not a multiple of $2$ or $3$. - Let $A$ and $B$ be subrings of a ring $R$. Is $A+B := \{ a+b\mid a\in A, b\in B\}$ a subring? Explain. - *Hint: false, consider $(a_1 + b_1)(a_2 + b_2)$*. > Idea: a counterexample is $A = R[x], B = R[y]$, then $(1+x)(1+y) = 1 + x + y + xy$ can not be written as $f(x) + g(y)$. - Prove that $M_2(\mathbf Z)$, the set of 2 by 2 matrices with integer coefficients, is a ring. > Idea: Matrix addition $A+B$ is commutative, associative, has identity the zero matrix, inverses $A \leftrightarrow -A$. Matrix multiplication is associative, has identity $I$, may or may not have inverses. The distributive law holds. - Prove that $GL_2(\mathbf{Z}) := \left\{ M\in M_2(\mathbf Z) \mid \det M \neq 0 \right\}$ is not a ring. > If $R$ is a ring then $R$ contains $I$, so it must contain $-I$, and thus it must contain $I + (-I) = 0$. But $\det 0 = 0$. - Define an **integral domain**. > A nonzero commutative ring with no nonzero zero divisors. - Define a **zero divisor**. > $r\in R$ is a zero divisor iff $\exists x\in R \text{ such that } xr = 0$. - Prove that $\mathbf{Z}/5\mathbf{Z}$ is an integral domain. > Check the multiplication table and note that there are no zeros in any nonzero row/column: > > | | $\bar 1$ | $\bar 2$ | $\bar 3$ | $\bar 4$ | $\bar 5 = \bar 0$ | > | ----------------- | -------- | ----------------- | ------------------ | ------------------ | -------------------- | > | $\bar 1$ | $\bar 1$ | $\bar 2$ | $\bar 3$ | $\bar 4$ | $\bar 0$ | > | $\bar 2$ | $\bar 2$ | $\bar 4$ | $\bar 6 = \bar 1$ | $\bar 8 = \bar 3$ | $\bar {10} = \bar 0$ | > | $\bar 3$ | $\bar 3$ | $\bar 6 = \bar 1$ | $\bar 9 = \bar 4$ | $\bar 12 = \bar 2$ | $\bar {15} = \bar 0$ | > | $\bar 4$ | $\bar 4$ | $\bar 8 = \bar 3$ | $\bar 12 = \bar 2$ | $\bar 16 = \bar 1$ | $\bar {20} = \bar 0$ | > | $\bar 5 = \bar 0$ | $\bar 0$ | $\bar 0$ | $\bar 0$ | $\bar 0$ | $\bar {25} = \bar 0$ | > > Note that this also proves this ring is a field, since there is exactly one copy of the multiplicative identity $\bar 1$ in each nonzero row/column. This also gives you the multiplicative inverses: $(\bar 3)^{-1} = \bar 2$ since $\bar 3\cdot \bar 2 = \bar 1$, and $(\bar 4)^{-1} = \bar {4}$ by looking for the $\bar 1$ in the $\bar 4$ row. - Prove that $\mathbf{Z}/6\mathbf{Z}$ is not an integral domain. > Do the multiplication table and find $\bar 2 \cdot \bar 3 = \bar 6 = \bar 0$ but $\bar 2\neq \bar 0, \bar 3\neq \bar 0$. So $\bar 2$ and $\bar 3$ are nonzero zero divisors. - Prove that if $R$ is an integral domain, then its polynomial ring $R[x]$ is also an integral domain. (See p. 84) > Proof: use degrees. Let $f, g\in R[x]$ with $\deg f = m, \deg g = n$, and consider $\deg ( f(x)\cdot g(x))$. This is surely $m+n$, *unless* somehow the leading term is zero! Write $f(x) = \sum_{i=1}^m a_i x^i, g(x) = \sum_{i=0}^n b_i x^i$, then their product is $f(x)\cdot g(x) = a_m b_n x^n + \cdots + a_0 b_0$, and $\deg f=m \implies a_m\neq 0, \deg g = n\implies b_n\neq 0$, and thus $a_m g_n\neq 0$ since $R$ has no nonzero zero divisors. - Prove that $\mathbf{Z}/n\mathbf{Z}$ is an integral domain $\iff n$ is prime. > $\not\impliedby$: if $n$ is composite, factor as $n=ab$, then take mod $n$ to get $\bar 0 = \bar {n} = \bar{ab} = \bar{a}\bar {b}$, but $\bar a, \bar b\neq \bar 0$. so these are zero divisors. > > $\impliedby$: let $n$ be prime, and toward a contradiction suppose $\bar b$ is a nonzero zero divisor. Then $x\bar b = 0$ has a nonzero solution, say $x=\bar a$. > Since $\bar 0 = \bar{a}\bar b = \bar{ab}$ we have $ab \equiv 0\mod n$, where e.g. we can pick $a,b\in \{1,2,\cdots,n-1\}$. > > Note that $a,b < n$ and $n$ is prime, so $\gcd(a, n) = \gcd(b, n) = 1$ and thus $\gcd(ab, n) = 1$, but this contradicts the fact that $ab\equiv 0\mod n\iff n\mid ab$. - Let $R$ be an integral domain and assume that $a b=a c$, where $a, b, c \in R$ and $a \neq 0$. Prove that $b=c$. > $ab=ac \iff ab-ac = 0\iff a(b-c) = 0$, and since $R$ is an integral domain, this forces $a=0$ or $b-c=0$. > Since $a\neq 0$ by assumption, $b-c = 0\iff b=c$. - Define a **field**. > A field is a commutative ring in which every nonzero element is a unit. - Define a **unit** in a ring. > A unit is an element with a multiplicative inverse, i.e. elements $r\in R$ such that ${1\over r} := r^{-1}\in R$. > > Note that in general, some inverse exists in some much bigger quotient field of $R$, but it may not exist in $R$ itself. > For example, $R=\Z$ is not a field because $2\in \Z$ but its multiplicative inverse $(2)^{-1} = {1\over 2}\in \Q$ is not contained in $\Z$. - Prove that if $R$ is a field then $R$ is an integral domain. > Suppose $R$ is a field, let $a,b\in R$ and suppose $ab=0$ where wlog $b\neq 0$. We want to show $b$ must be zero to conclude there are no nonzero zero divisors. Every element is a unit, so $a^{-1}$ exists in the ring, and we get $a^{-1}ab = a^{-1}0 \implies 1b = 0 \implies b=0$. - Prove that $\mathbf{Z}/5\mathbf{Z}$ is a field by showing the multiplicative inverse of every element exists in the ring. > Make the multiplication table (you don't need a row/column for $\bar 0$) and check that every row/column has exactly one copy of $\bar 1$. - Prove that $\mathbf{Z}/6\mathbf{Z}$ is not a field by finding an element without a multiplicative inverse. > An easier proof: $R$ a field $\implies R$ an integral domain, and $A\implies B$ is the same as $\not B\implies \not A$, so $R$ not an integral domain $\implies R$ is not a field. > More directly, pick any nonzero zero divisor, like $\bar 2$ or $\bar 3$, and notice that these rows do not contain a copy of $\bar 1$: > ![abstract algebra - Tables of $\mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z }/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$? - Mathematics Stack Exchange](https://i.stack.imgur.com/zRE8a.png) > > In particular, this ring is not a field since e.g. $\bar 2$ does not have an inverse in the ring. - Let $R = \mathbf{Z}/m\mathbf{Z}$ for $m$ arbitrary. When is an element $\bar a\in R$ a unit? > Claim: $\bar a$ is a unit $\iff \gcd(a, m) = 1$, i.e. the units are exactly the elements coprime to $m$. > $\implies$: suppose $\bar a$ is a unit, let $\bar b = (\bar a)^{-1}$, then $\bar a \bar b = \bar 1 \iff \bar a \bar b -\bar 1 = \bar 0 \iff m \mid \bar a \bar b - \bar 1 \iff \exists t \text{ such that } tm = ab-1 \iff ab-tm = 1 \iff \gcd(a, m) = 1$ since every expression of the form $t_1 a + t_2m = t_3$ is a multiple of an expression of the form $s_1 a + s_2m = d := \gcd(a, m)$. > $\impliedby$: Suppose $\gcd(a, m) = 1$, then $t_1 a + t_2 m = 1 \implies t_1 a \equiv 1 \mod m$, so $t_1 = (\bar a)^{-1}$. - Define an **ordered field** (see p. 47). > A field $F$ with distinguished set $F^+$ such that $\forall x\in F$, one of $-x\in F^{+}, x=0, x\in F^+$ holds, and $F^{+}$ is closed under $+, \cdot$. - Prove that $\mathbf{Q}$ is an ordered field (see p.48) > Write $r={a\over b}$ with $b$ positive, define $\mathbf{Q}^+$ by $r\in \mathbf{Q}^+$ when $a\in \mathbf{Z}^+$. > Show it's closed under addition and multiplication: ${a\over b} {c\over d} = {ac\over bd}$ where $a,c\in \Z^+ \implies ac\in \Z^+$, and similarly ${a\over b} + {c\over d} ={ ad+ cb \over db}$ and $ad+cb$ is positive since $d,b$ are assumed positive and $a,c\in \Z^+$. - Prove that $\sqrt 2$ is irrational. > $\sqrt 2 = {a\over b} \implies 2b^2 = a^2 \implies 2\mid a^2 \implies 2\mid a$, so write $a=2s$. > Then $2b^2 = a^2 = (2s)^2 = 4s^2 \implies b^2=2s^2 \implies 2\mid b^2\implies 2\mid b$, a contradiction. - Define the **least upper bound** or **supremum** of an ordered set (see p. 52). > $\alpha = \sup S \iff$ $\alpha$ is an upper bound, so $\alpha\geq s$ for all $s\in S$, and it is the least such: if $\beta$ is another upper bound, then $\alpha\leq \beta$. - (Slightly hard) Sketch a proof that $\sqrt 2 \in \mathbf{R}$ using the least upper bound property of the real numbers. > Idea: define $S = \{x\in \mathbf R \mid x^2< 2 \}$, show that $S$ is a bounded set (e.g. every $s\in S$ is nor more than $2$), and although it does not have a LUB in $\mathbf Q$, it does in $\mathbf R$ since this has the UB property. Such a LUB $\alpha$ must just be equal to $\sqrt 2$, since $|\alpha - \sqrt 2| < \varepsilon$ for any $\varepsilon$, since there are elements $s\in S$ arbitrarily close to $\sqrt 2$ and each satisfies $s < \alpha < \sqrt 2$. - Prove that $\mathbf{Q}[\sqrt 2] := \left\{ a_0 \cdot 1 + a_1 \cdot \sqrt{2} \mid a_0, a_1\in \mathbf{Q} \right\}$ is a field (see p.54). *Hint: it suffices to find the multiplicative inverse of $a_0 + a_1\sqrt{2}$ and show it is still an element of $\mathbf{Q}[\sqrt 2]$*. > Let $R = \mathbf Q[\sqrt 2] \ni r$, so $r=a+b\sqrt 2$. It suffices to find $r^{-1}$ and show it is in $R$. > Guess: conjugate, set $s = a-b\sqrt{2}$ and multiply to get $rs = (a+b\sqrt 2)(a-b\sqrt 2) = a^2 - 2b^2 \in \mathbf Q$, which is something we can put in a denominator. Setting $r^{-1} = {s\over a^2-2b^2}= {a-b\sqrt 2\over a^2-2b^2}$ yields $rr^{-1} = {rs\over a^2-2b^2} = {a^2-2b^2\over a^2-2b^2}=1$. > Finally, check that $r^{-1}\in \mathbf Q[\sqrt 2]$: write $r^{-1} = {a-b\sqrt 2\over a^2-2b^2} = {a\over a^2-2b^2} - {b\over a^2-2b^2}\sqrt 2 = a' + b'\sqrt 2$ where $a', b'\in \mathbf Q$, so it works. - Prove that $\mathbf{Q}[2^{1\over 3}]$ is a field. *Hint: find a basis and write it as a set as above.* > A basis is $1, 2^{1\over 3}, 2^{2\over 3}$, and this is enough because $2^{1\over 3}$ satisfies $x^3-2$ which is degree 2, irreducible, and monic. > So $\mathbf{Q}[2^{1\over 3}] = \left\{ a_0 \cdot 1 + a_1 \cdot 2^{1\over 3} +a_2 \cdot 2^{2\over 3}\mid a_1,a_2,a_3\in \mathbf Q \right\}$. How to find multiplicative inverses? This is harder than I initially thought! But a general fact helps: theorem 2.3, page 129 in Shifrin: if $F$ is a field and $f\in F[x]$ is irreducible, then $F[x]/(f)$ is always a field. But $\mathbf{Q}[\alpha] = \mathbf{Q}[x]/(f)$ where $f$ is the minimal polynomial of $\alpha$, which is always irreducible. > > If you want a direct proof, $\{1, \alpha,\alpha^2,\cdots, \alpha^{n-1} \}$ is a basis for $F[\alpha]$, so any $\beta \in \mathbf{Q}[\alpha]$ can be written as $\beta = \sum_{i=0}^{n-1} b_i \alpha^i$. So set $p(x) = \sum_{i=0}^{n-1} b_i x^i$, so $p(\alpha)=\beta$. Let $f(x)$ be the minimal polynomial of $\alpha$. Then one can write > $$ > a_1(x) p(x) + a_2(x) f(x) = 1 > $$ > for some $a_1, a_2\in F[x]$ since $\gcd(p, f) = 1$ (using that $f$ must be irreducible and hence has no nontrivial divisors to share with $p$). Now plug in $\alpha$ to get > $$ > a_1(\alpha)p(\alpha) + a_2(\alpha)f(\alpha) = 1 \implies a_1(\alpha)\beta + a_2(\alpha)\cdot 0 =1\implies a_1(\alpha)\beta = 1 \implies \beta^{-1} = a_1(\alpha). > $$ > - Prove that $\mathbf{Q}[\sqrt d]$ is a field for any $d\in\mathbf{Z}$. > The above proof shows that $\mathbf{Q}[\alpha]$ is always a field when $\alpha$ is algebraic, i.e. satisfies a monic irreducible polynomial $f\in \mathbf{Q}[x]$, e.g. $\alpha = i \leadsto x^2+1, \alpha = \sqrt{2}\leadsto x^2-2, \alpha=\sqrt{d}\leadsto x^2-d$. > But you can also just construct inverses directly: if $r = a+b\sqrt d$, let $s=a-b\sqrt d$ to get $rs = a^2 -db^2$ and take $r^{-1} := {a-b\sqrt d\over a^2-db^2} = {a\over a^2-db^2} + {b\over a^2-db^2}\sqrt d = a' + b'\sqrt d$. - Write down all of the elements in $$R=\left\{\left(\begin{array}{ll}a & b \\ 0 & a\end{array}\right) \mid a, b \in \mathbf{Z}/2\mathbf Z\right\} \subset M_2\left(\mathbf{Z}/2\mathbf{Z}\right)$$. Is $R$ commutative? What are the additive and multiplicative inverses of all elements? Is $R$ an integral domain? Is $R$ a field? > There are two choices for $a$ and 2 for $b$, so 4 total possibilities > $$ > (a,b) = (0,0): A_{0, 0} = \begin{bmatrix} > 0 & 0 \\ > 0 & 0 > \end{bmatrix},\quad > (a,b) = (0,1): A_{0, 1} = > \begin{bmatrix} > 0 & 1 \\ > 0 & 0 > \end{bmatrix}\, \\ > (a,b) = (1,0): A_{1, 0} = > \begin{bmatrix} > 1 & 0 \\ > 0 & 1 > \end{bmatrix},\quad > (a,b) = (1,1): A_{1, 1} = > \begin{bmatrix} > 1 & 1 \\ > 0 & 1 > \end{bmatrix} > $$ > This contains the identity as $A_{1,0}$, but $A_{0, 1}$ isn't invertible set $\det A_{0, 1} = 0$, so if this is a ring then it can't be a field. > Check that > $$ > \left(\begin{array}{ll}a_1 & b_1 \\ 0 & a_1\end{array}\right) + \left(\begin{array}{ll}a_2 & b_2 \\ 0 & a_2\end{array}\right) = \left(\begin{array}{ll}a_1 + a_2 & b_1 + b_2 \\ 0 & a_1 + a_2\end{array}\right) \in R \\ > \left(\begin{array}{ll}a_1 & b_1 \\ 0 & a_1\end{array}\right) \cdot \left(\begin{array}{ll}a_2 & b_2 \\ 0 & a_2\end{array}\right) = \left(\begin{array}{ll}a_1a_2 & a_1b_2 + b_1a_2 \\ 0 & a_1a_2\end{array}\right) \in R > $$ > so $R$ is closed under addition and multiplication. It contains $-A_{i, j}$ for all $i,j$ since $-1 \equiv 1 \mod 2$. > To check if it's a domain, check the multiplication table, where we can omit zero. > I won't write it here, but you'll find that $A_{0, 1}^2 = 0$, so $A_{0, 1}$ is a nonzero zero divisor. > The general trick is that if $r\in R$ is **nilpotent**, i.e. $r^n = 0$ for some $n$, then $r$ must be a zero divisor because $r^n = r^{n-1}\cdot r =0$ so some nonzero element (namely $r^{n-1}$) kills $r$. > Since $R$ is not a domain, it can not possibly be a field. ## Complex numbers and field extensions - Let $R$ be a ring and $f\in R[x]$ a polynomial. Define what it means for $f$ to be **irreducible** in $R[x]$. > $f$ is reducible if it factors as $f=gh$ with $g,h$ non-constant polynomials, and irreducible if it is not reducible. - State the **rational root test**. > If $f(x) := a_n x^n + \cdots + a_1 x + a_0$ has a rational root $r = {p\over q}$, then $p\mid a_0$ and $q\mid a_n$, so that ${p\over q} = \pm{\text{factors of } a_0 \over \text{factors of } a_n}$. - State **Eisenstein's criterion**. > If $f$ as above and there exists a $p$ such that $p\mid a_{n-1},\cdots, a_n$, $p\not \mid a_n$, and $p^2\not\mid a_0$, then $f$ is irreducible. - Let $f\in \mathbf{Z}[x]$ be a polynomial with integer coefficients. How does the irreducibility of $f\in \mathbf{Z}[x]$ relate to the reducibility of $\bar f\in {\mathbf{Z} \over p\mathbf{Z}}[x]$? How can you use this to prove or disprove irreducibility of a polynomial? *Hint: there is a statement like "if $f$ is reducible over $\mathbf{Z}$, then $\bar f$ is ??? when reduced mod $p$".* > If $f$ is reducible over $\mathbf Z$, then it is reducible over every $\mathbf Z/p\mathbf Z$. So toward a contrapositive, if $f$ is irreducible over **any** $\mathbf Z/p\mathbf Z$, then $f$ must be irreducible over $\mathbf Z$. > Note that this is delicate, since being reducible $\mod p$ tells you nothing! There are polynomials reducible $\mod p$ for every $p$ which are not irreducible over $\mathbf Z$, - Let $K$ be a field and $a \in K$ by any element. Show that $f(x)\in K[x]$ is irreducible iff $g(x) := f(x+a)\in K[x]$ is irreducible. > Suppose $f$ is reducible, so $f(x) = a(x)b(x)$, then $g(x) = f(x+a) = a(x+a)b(x+a)$ is a reduction of $g$. > Conversely, if $g(x)$ is reducible, then $a(x) b(x) = g(x) = f(x+a) \implies f(x) = a(x-a)b(x-a)$ is a reduction of $f$. - Use this to conclude that $f(x)\in \mathbf{Q}[x]$ is irreducible iff $f(x+1) \in \mathbf{Q}[x]$ is irreducible. > Use $K=\mathbf Q$ and $a=1$ above. Useful test!! - Checking irreducibility: - Prove that $f(x) = x^2-2$ is irreducible in $\mathbf{Q}[x]$. > Trick: factor over the algebraic closure $\mathbf C$ and check that the factors aren't in $\mathbf Q[x]$. Here we can $x^2-2 = (x-\sqrt 2)(x+\sqrt 2)$ (it's a difference of squares) and neither factor is in $\mathbf Q[x]$ since $\sqrt 2$ is irrational. - Prove that if $f\in R[x]$ has a root $\alpha\in R$, then $f$ is reducible in $R[x]$. > If $f$ has a root $\alpha\in R$, then $x-\alpha$ divides $f$ as a polynomial. This means $f(x) = (x-\alpha)g(x)$ where $\deg g < \deg f$, so $f$ has been reduced into smaller nonconstant factors. - Prove that $x^4-5x^2+6$ has no rational roots $\alpha\in \mathbf{Q}$, but is still reducible in $\mathbf{Q}[x]$. > Use the RRT: rational roots must be of the form $\alpha = \pm{\text{factors of } a_0 \over \text{factors of } a_n} = \pm{1\over 1, 2,3, 6}$, but plugging in all 8 possibilities for $\alpha$ yields $f(\alpha)\neq 0$, so there are no rational roots. It is still reducible though, since it breaks into two quadratic factors: let $g(z) = z^2-5z+6$, then $g$ factors as $g(z) = (z-3)(z-2)$ and thus $f(x) = (x^2-3)(x^2+2)$, making it reducible. - Show that $p(x)=x^3+x+1$ is irreducible in ${\mathbf Z \over 2\mathbf Z}[x]$. > A degree 3 polynomial is reducible over $R$ iff it has a root in $R$, so plug in all of the elements of $R=\mathbf Z/2\mathbf Z$, i.e. check $f(\bar 0) = \bar 1 \neq \bar 0, f(\bar 1) = \bar 3 =\bar 1 \neq \bar 0$. - Let $\mathbf{F}=\mathbf{Z}/2\mathbf{Z}$ and factor the polynomial $x^8+x \in \mathbf{F}[x]$ into irreducible factors. > Write $f(x) = x^8+x$, factor as $f(x) = x(x^7+1) = xg(x)$ where $g(x) = x^7+1$, check $g(\bar 0) = \bar 1\neq \bar 0$ and $g(\bar 1) = \bar 2=\bar 0$ so $\bar 1$ is a root of $g$. Do polynomial long division ${x^7+1\over x-1} = h(x)$ where $h(x) = x^6-x^5+x^4-x^3+x^2-x+1$. Check that $h(\bar 0) = h(\bar 1) = \bar 1$, so it has no root in $\mathbf F$, so it has to break up as polynomials of degrees $(2,4)$ or $(3,3)$. Use the list of irreducible degree 2 and 3 polynomials from an old homework, and check that none of them evenly divide $h(x)$ to conclude it is irreducible. So $f(x) = x(x-1)(x^6-x^5+x^4-x^3+x^2-x+1)$. - Prove that $h(x)=x^4+2 x^2+5 x+1$ is irreducible over $\mathbf{Q}$ by reducing mod $p$ for some prime $p$. > Check mod $2$: we get $f(x) = x^4+x+1$ and check $f(\bar 0)\neq 0, f(\bar 1)\neq 0$, so no root. You can also try dividing it by all other irreducible quadratic polynomials in $\mathbf{Z}/2\mathbf{Z}[x]$ (from a previous HW) and see none divide it. So it's irreducible mod 2, and thus irreducible in $\mathbf Q[x]$ using proposition 3.4 (p. 108 in Shifrin). - Prove that the polynomial $f(x)=1+x+x^3+x^4$ is reducible over *any* field $\mathbf{F}$. > Check that $f(\overline{-1}) = \bar 1- \bar 1-\bar 1+\bar 1 = \bar 0$, so $\overline{-1}$ is always a root (and $-1$ is in every field). - Let $R$ be a ring and $f, g\in R[x]$ be two polynomials. Define what the **greatest common divisor (GCD)** of $f$ and $g$ is, usually written $d(x) = \gcd(f(x), g(x))$. > $d$ is the GCD of $f$ and $g$ if (1) $d$ divides $f$ and $d$ divides $g$, and (2) it is the greatest such: if $d'$ also divides $f$ and $g$, then $d'$ must divide $d$ (in particular, forcing $d' \leq d$). - Suppose that $f(x)=x^{99}-5 x^{49}+1 \in \mathbf{Q}[x]$. Compute the remainder when $f(x)$ is divided by - $g_1(x) = x$; - $g_2(x) = x-1$; - $g_3(x) = x+1$. > These are all just polynomial long division. - State **Euler's formula for complex numbers**. *Hint: this is the one that converts $e^{i\theta}$ to cosines and sines.* > $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. - Convert $5+5i$ to polar coordinates . > $z = 5+5i \implies z = \sqrt{5^2+5^2} e^{i\arctan(5/5)} = 5e^{i\pi\over 4}$. - $7e^{i\pi\over 3}$ to rectangular coordinates. $z = 7( \cos(\pi/3) + i\sin(\pi/3) ) = {7\over 2}(1 + i\sqrt 3)$. - Let $z = re^{i\theta}$ and $w = \rho e^{i\phi}$, and prove that $z w=r \rho(\cos (\theta+\phi)+i \sin (\theta+\phi))$. > $zw = r\rho e^{i(\theta + \phi)} = r\rho(\cos(\theta+\phi) + i\sin(\theta + \phi))$. - Prove de Moivre's theorem: $(\cos(x) +i \sin(x))^n=\cos(n x)+ i \sin (n x)$ > $(\cos(x) + i\sin(x))^n = (e^{ix})^n = e^{inx} = \cos(nx) + i\sin(nx)$. - Practice factoring: - Factor $f(x) = x^n-1$ into irreducible polynomials. > Slight problem: the word "irreducible" is always with respect to some ring! For example, consider $f(x) = x^2+1$. This is an irreducible quadratic if viewed as an element of $\mathbf{R}[x]$, but is reducible in $\mathbf{C}[x]$ since it factors as $f(x)(x+i)(x-i)$. > If working over $\mathbf{C}$, we have $x^n-1 = \prod_{i=1}^n (x-\zeta_n^i) = (x-\zeta_n)(x-\zeta_n^2)\cdots(x-1)$, recalling that $\zeta_n := e^{2\pi i\over n}$. > > If working over $\mathbf R$, note that $x=1$ is a real root, so $x^n-1 = (x-1)g(x)$ for some $g(x)$ of degree $n-1$. > Perform polynomial long division to find ${x^n-1\over x-1} = x^{n-1} + x^{n-2} + \cdots + x + 1$. > > However! $g(x)$ is only irreducible when $n$ is prime, by exercise 3.3.7 in Shifrin (the last homework). Recall how that proof went: write $g(x) = x^{p-1} + x^{p-2} + \cdots + x + 1$ and use some tricky sums to show $g(x+1) = x^{p-1} + {p\choose 1}x^{p-2} + \cdots + {p\choose p-2}x + {p\choose p-1}$, use that ${p\choose p-1} = p$, and apply Eisenstein with the prime $p$: since $p\mid {p\choose k}$ for every $1\leq k\leq p-1$, $p\not \mid a_{p-1} = 1$, and $p^2\not\mid a_0 = p$, it works. > I forgot that factoring $g(x)$ for non-prime $n$ is hard, sorry! > > Here's an example of with the various powers $\zeta_n^k$ look like on the unit circle in $\mathbf C$ for a few examples of $n$: > ![Example plots of the distinct Nth roots of unity | Download Scientific Diagram](https://www.researchgate.net/publication/263237071/figure/fig2/AS:614119153627153@1523428824479/Example-plots-of-the-distinct-Nth-roots-of-unity.png) - Factor $f(x) = x^n-a$. *Hint: you'll need a term like $a^{1\over n}$. Try some concrete examples like $a=2, 3,4,\cdots$.* > $$ > x^n-a = (x- a^{1\over n}\zeta_n)(x-a^{1\over n}\zeta_n^2)\cdots(x-a^{1\over n}\zeta_n^{n-1})(x-a^{1\over n}) \\ = \prod_{i=0}^n (x-a^{1\over n}\zeta_n^i) > $$ - Factor $f(x) = x^n+1$. > Over $\mathbf R$, note $x=-1$ is a root when $n$ is odd so $f(x) = (x+1)g(x)$ for some $g(x)$, and polynomial long division will show $g(x) = x^{n-1}-x^{n-2}+x^{n-3}-\cdots -x + 1$. Factoring $g$ over $\mathbf R$ is hard, sorry! > > > Over $\mathbf C$, in general find a "principal root" $w_0$, then this will factor as $\prod_{i=1}^n (x-w_0 \zeta_n^k)$. For $x^n-a$, we have $w_0 = a^{1\over n}$. For $x^n+1$ we have $a=-1$, so we need an $n$th root of $-1$ -- take $w_0 = e^{i\pi \over n}$, so $w_0^n = (e^{i\pi\over n})^n = e^{i\pi} = -1$, then > $$ > f(x) = (x - w_0\zeta_n)(x-w_0\zeta_n^2)\cdots(x-w_0\zeta_n^{n-1})(x-w_0) \\ = (x- e^{i\pi\over n} e^{2i\pi\over n})(x-e^{i\pi\over n} e^{2i\pi\cdot 2\over n})(x-e^{i\pi\over n} e^{2i\pi\cdot 3 \over n})\cdots (x-e^{i\pi\over n} e^{2i\pi\over n}) \\ > =(x-e^{3i\pi\over n})(x-e^{5i\pi \over n})\cdots (x-e^{(2n-1)i\pi\over n}))(x-e^{i\pi \over n}) \\ > = \prod_{k=0}^{n-1} (x-e^{(2k+1)i\pi \over n }) > $$ > - Factor $f(x) = x^n + a$. > $f(x) = \prod_{k=0}^{n-1}(x - a^{1\over n} e^{(2k+1)i\pi\over n})$ - Factoring over the complex numbers: - Plot the complex roots of $x^3-1$ in $\mathbf{C}$. > The roots are $1$ and ${1\over 2}(-1\pm i\sqrt 3)$. > > Check Wolfram: https://www.wolframalpha.com/input?i=x%5E3-1 > ![image-20221114182438988](/home/zack/.config/Typora/typora-user-images/image-20221114182438988.png) - Plot the complex roots of $x^3+1$ in $\mathbf{C}$. > The roots are $-1$ and ${1\over 2}(1\pm i\sqrt 3)$. > Check wolfram: https://www.wolframalpha.com/input?i=x%5E3%2B1 > ![image-20221114182423642](/home/zack/.config/Typora/typora-user-images/image-20221114182423642.png) - Plot the complex roots of $x^4-3$ in $\mathbf{C}$. > These are on a circle of radius $3^{1\over 4}$ and are of the form $\pm 1 \sqrt[4]3, \pm i\sqrt[4]3$: > > https://www.wolframalpha.com/input?i=x%5E4-3 > ![image-20221114182409917](/home/zack/.config/Typora/typora-user-images/image-20221114182409917.png) - Plot the complex roots of $x^4+3$ in $\mathbf{C}$ > Again on a circle of radius $3^{1\over 4}$, but a little different because of the positive sign. The roots are of the form $\sqrt[4]3\cdot e^{\pm i\pi \over 4}, \sqrt[4]3\cdot e^{\pm 3i\pi \over 4}$: > > https://www.wolframalpha.com/input?i=x%5E4%2B3 > ![image-20221114182347163](/home/zack/.config/Typora/typora-user-images/image-20221114182347163.png) - Perform the division algorithm to show the following (See p.85): $$ {x^3+2 \over 2x^2 + x + 1}= ({1\over 2}x - {1\over 4}) + {-{1\over 4}x + {9\over 4}\over 2x^2 + x + 1} \implies x^3 + 2 = ({1\over 2}x - {1\over 4})(2x^2+x+1) + (-{1\over 4}x + {9\over 4}) $$ > Be sure to practice by hand, but I used a calculator: https://www.snapxam.com/solver?p=%5Cfrac%7B%5Cleft(x%5E3%2B2%5Cright)%7D%7B2x%5E2%2Bx%2B1%7D&method=0 > ![image-20221114115806318](/home/zack/.config/Typora/typora-user-images/image-20221114115806318.png) - Define the gcd of two polynomials. How do you compute it? > The polynomial $d(x) = \gcd(f(x), g(x))$ is a polynomial such that $d$ divides both $f$ and $g$ (it's a common divisor) and if any $d'$ also divides both $f$ and $g$, $d'$ must also divide $d$ (it's the "greatest" such divisor). - Let $d(x) = \gcd(x^3-8, x^2-x-2)$ and show $d(x) = x-2$. Find polynomials $a(x), b(x) \in \mathbf{Q}[x]$ such that $$ d(x) = a(x)f(x) + b(x) g(x) $$ and by a computation, show that $$ d(x) = {1\over 3}f(x) + (-{1 \over 3 })(x+1)g(x) $$ > Important: the $\gcd$ depends on the ambient ring, which I didn't specify! Over $\mathbf R$ vs $\mathbf C$, they may differ. > > Easy solution: factor these over $\mathbf R$ as $x^3-8 =(x-2)\left(x^2+2 x+4\right)$ and $x^2-x-2 = (x+1)(x-2)$ and observe the common divisor $g(x) = x-2$. > The general solution: use the Extended Euclidean algorithm. > I checked my work with a calculator: https://mathsci2.appstate.edu/~cookwj/sage/algebra/Euclidean_algorithm-poly.html > > ![image-20221114120618894](/home/zack/.config/Typora/typora-user-images/image-20221114120618894.png) Find the greatest common divisor of the following polynomials over $\mathbf{Q}$. - - $x^2+x-2$ and $x^5-x^4-10 x^3+10 x^2+9 x-9$. > Answer: $x-1$. Find using the algorithm: > ![image-20221114120814926](/home/zack/.config/Typora/typora-user-images/image-20221114120814926.png) - $x^2+1$ and $x^6+x^3+x+1$. > Answer: $x^2+1$. Again use the algorithm: > ![image-20221114120847870](/home/zack/.config/Typora/typora-user-images/image-20221114120847870.png) ## Splitting Fields and Extensions - State the **fundamental theorem of algebra** over an arbitrary algebraically closed field $K$. > See Shifrin p. 96. If $K$ is algebraically closed field and $f\in K[x]$ is degree $n\geq 1$, then $f$ has a root in $K$, i.e. there exists some $\alpha \in K$ such that $f(\alpha) = 0$. - Prove **Wilson's theorem**: if $p$ is prime, then $(p-1)! \equiv -1 \mod p$. Hint: use the very special polynomial $f(x) := x^p - x \in {\mathbf Z\over p\mathbf Z}[x]$. > HW exercise. Check that every element of $\mathbf{Z}/p\mathbf Z$ is a root of $f$, and $f(x) = x(x^{p-1}-1) = x g(x)$ where $g(x) = x^{p-1}-1 = (x-1)(x-2)\cdots(x-(p-1))$, so $g(p) = p^{p-1}-1 = (p-1)(p-2)\cdots(2)(1)\implies -1\equiv(p-1)! \mod p $. - Let $K$ be a field and define what it means for a set $V$ to be a **vector space** over $K$. > See Shifrin p.150. Basically one needs $V$ to be closed under $\pm$, associative for $\pm$, needs a zero vector, negative vectors, and a notion of scalar multiplication $c\vec v$ satisfying distributivity and similar compatibility axioms. - Define what it means for a subset $S\subseteq V$ of a vector space to be **linearly independent**. > $S = \{\vec s_1,\cdots, \vec s_n\}$ is linearly independent iff $\sum_{i=1}^n a_i \vec s_i = 0 \implies a_i = 0 \,\,\forall i$, where the $a_i\in F$ are scalars. - Define what it means for $S$ to be a **spanning set**. > $S$ is a spanning set of $V$ if every $\vec v\in V$ can be written as $\vec v = \sum_{i=1}^n a_i \vec s_i$ for some collection of scalars $a_1,\cdots, a_n \in F$. - Define what it means for a subset $S \subseteq V$ of a vector space to be a **basis** of $V$. > $S$ is a basis iff $S$ is both linearly independent and spanning. - Define the **dimension** of a vector space $V$. > The dimension of $V$ is the cardinality of any basis of $V$. - Prove that $\mathbf{C}$ is a 2-dimensional vector space over the field $\mathbf{R}$. > Take as a basis $S = \{1, i\}$; this is linearly independent since $a+bi=0\iff a=b=0$ (where $a,b\in \mathbf R$) and spanning since every complex number is of the form $a+bi$ when written in rectangular coordinates. - Prove that $\mathbf{R}^n$ is an $n$-dimensional vector space over $\mathbf R$. > Take as a basis the vectors $\vec e_i = (0,0,\cdots, 0,1,0,\cdots,0,0)$ which has only a 1 in the $i$th spot. Every vector $\vec v = (a_1, a_2,\cdots, a_n)$ can thus be written as $\vec v = a_1 \vec e_1 + a_2 \vec e_2 + \cdots + a_n \vec e_n$. - Let $V = \left\{ a_0 + a_1 x + a_2 x^2 \mid a_i\in \mathbf R \right\} \subseteq \mathbf R[x]$ be the subset of all polynomials with real coefficients of degree at most 2. Prove that $V$ is a vector space over $\mathbf R$. What is its dimension? > Take as a basis $S = \{1,x,x^2\}$; linearly independent since $a_0+a_1x + a_2x^2=0 \iff a_0=a_1=a_2=0$ by equating coefficients of both sides, regarding $0$ as the polynomial $0 + 0x + 0x^2 + \cdots$, and spanning since $V$ was defined as exactly linear combinations of these. So $\dim V = 3$. Check that this satisfies all of the axioms on p.150 in Shifrin. - Let $S\subseteq \mathbf{R}^3$ be the plane defined by $S := \left\{ (x,y,z)\in \mathbf{R}^3 \mid x+y+z = 1\right\}$. Prove that $S$ is not a vector subspace of $\mathbf{R}^3$. > Note that $\vec 0\not\in S$, since $(x,y,z) = (0,0,0) \implies x+y+z=0\neq 1$. This is a plane that misses the origin. > However, any vector space (and any subspace) must contain $\vec 0$. - Prove that $\mathrm{GL}_2(\mathbf R)$ is not a vector space. > Note that $0\not \in \mathrm{GL}_2(R)$, since it would have to be the 2x2 matrix with all zeros, but this has determinant 0 and is thus singular (recalling that $M\in \mathrm{GL}_n(\mathbf R) \iff \det M\neq 0$). - Let $f\in \mathbf{Q}[x]$ and define what the **splitting field** of $f$ is. > See Shifrin p.99. $f$ splits in $K$ iff all of the roots of $f$ are in $K$, so $f(x)$ breaks into a product of linear factors over $K[x]$, and it is the smallest such field in the following sense: if $f$ also splits in $L$, then $K \subseteq L$. - Find the splitting field of $f(x) = x^3-1$. > Factor as $(x-1)(x-\zeta_3)(x-\zeta_3^2)$, so we only need to add in $\zeta_3$ to get all of the roots. So take $K = \mathbf Q[\zeta_3]$. > One can check that $\zeta_3$ satisfies the polynomial $x^2+x+1$, which is ${x^3-1\over x-1}$, so this is a degree 2 extension and thus a 2-dimension vector space. This has a basis $\{1,\zeta_3, \zeta_3^2\}$. - Find the splitting field of $f(x) = (x^2-2)(x^2+1)$. > The first factors as $(x-\sqrt 2)(x+\sqrt 2)$ and the second as $(x-i)(x+i)$, so we can take $\mathbf Q[\sqrt 2, i]$. > $x^2-2$ is the minimal polynomial of $\sqrt 2$ and $x^2+1$ is minimal for $i$, and we get lucky in this case: the total degree $[\mathbf Q[\sqrt 2, i]: \mathbf Q] = 2\cdot 2 = 4$. This has a basis $\{1, \sqrt 2, i, i\sqrt 2\}$. - Find the splitting field of $f(x) = x^2+x+1$. > Use the quadratic formula to find that the roots are $-{1\over 2}\pm i{\sqrt 3\over 2} = \zeta_3, \zeta_3^2$. > So you could take either - Let $L$ be a field extension of $K$, and define the **degree** of $L$ over $K$, denoted $[L : K]$. > Since $L$ is a vector space over $K$, the degree is the dimension of $L$ over $K$. - What are the degrees of $\mathbf{Q}[\sqrt d]$ over $\mathbf Q$ for $d=2,3,4,5,6,7$? > For $d=2,3,5,6,7$ this is degree 2. For $d=4$ we have $\mathbf{Q}[\sqrt 4] = \mathbf Q$ since $\sqrt 4\in \mathbf Q$ already (the issue is that 4 contained a perfect square, and the others were *squarefree*). - What is the degree of $\mathbf C$ over $\mathbf R$? > We have a basis $\{1, i\}$, so the dimension is 2 and thus so is the degree. - Let $\alpha \in \mathbf{C}$ and define what the **minimal polynomial** of $\alpha$ is. > Sorry, this may not be covered in the book!! It's a polynomial $f\in \mathbf R[x]$ such that $f$ is monic, irreducible, and $f(\alpha) = 0$ (i.e. $\alpha$ is a root of $f$). - How does the degree of the minimal polynomial of $\alpha$ relate to the dimension of $\mathbf{Q}[\alpha]$ as a vector space over $\mathbf{Q}$? > In good cases, the degree of $f$ equals the dimension of $\mathbf Q[\alpha]$! - Find the minimal polynomial of $\sqrt{d}$ for any $d\in \mathbf{Z}$ (positive or negative). > $f(x) = x^2-d$. This works if $d$ is negative, e.g. $\sqrt{-3}$ is a root of $x^2+3$. - Find the minimal polynomials of $\zeta_3, \zeta_4, \zeta_5$. *Hint: these are not $x^3-1, x^4-1, x^5-1$ respectively! There are even smaller polynomials they satisfy.* > $\zeta_n$ will always be a root of $x^n-1$, but may satisfy a smaller polynomial. The idea is to start taking powers of e.g. $\zeta_3$ and see if you can cook up a linear combination that is zero. E.g. $\zeta_3 = e^{2\pi i \over 3} = \cos(2\pi/3) + i\sin(2\pi /3) = -{1\over 2} + i{\sqrt 3\over 2}$ and $\zeta_3^2 = e^{4\pi i \over 3} = \cos(4\pi/3) + i\sin(4\pi/3) = -{1\over 2} - i{\sqrt 3\over 2}$, so $\zeta_3^2 + \zeta_3 = -{1\over 2} + -{1\over 2} = -1 \implies \zeta_3^2 +\zeta_3 - 1 = 0$. So the minimal polynomial is $x^2+x+1$. > Similarly, > > - $\zeta_3: x^2+x+1$ > - $\zeta_4: x^2+1$ > - $\zeta_5: x^4+x^3 +x^2+x+1$ (this one is hard without the theory of *cyclotomic polynomials*) - Show $\mathbf{Q}[\sqrt{3} + i] = \mathbf{Q}[\sqrt 3, i]$ by explicitly showing each field is contained in the other. *Hint: it's enough to show the generators of one are contained in the other and vice versa.* > $\subseteq: $ this is clear, since elements in the RHS are of the form $a\sqrt 3 + bi$, so take $a=1,b=1$. > $\supseteq:$ fiddle with $\alpha = \sqrt 3 + i$ and hope to find some linear combination of powers of it that give you either $\sqrt 3$ or $i$ (or both). Check that $\alpha^2 = 2i\sqrt 3 + 2$ and $\alpha^3 = 8i$, so ${1\over 8}\alpha^3 = i$ and thus $i$ is in the LHS. > But then $\sqrt 3 = (\alpha^2-2)\cdot {1\over 2i}$ is a linear combination of stuff in the LHS, so $\sqrt 3$ is in the LHS. - For any $a, b\in \mathbf{Z}$ with $a,b>0$, prove that $\mathbf{Q}[\sqrt a, \sqrt b] = \mathbf{Q}[\sqrt a + \sqrt b]$. *Hint: you may assume $a\neq b$; prove subset containment both ways.* > $\subseteq$: this is clear, since some elements in the RHS are of the form $a_0\sqrt a + a_1 \sqrt b$, so take $a_0=1, a_1=1$. > $\supseteq$: harder! - Find a *power basis* for $K = \mathbf{Q}[\sqrt 3, \sqrt 7]$, which is defined to be an element $\alpha\in K$ such that the powers $\left\{\alpha^0, \alpha^1,\cdots, \alpha^n \right\}$ form a basis for $K$ as a vector space over $\mathbf{Q}$. *Hint: $K = \mathbf{Q}[\alpha]$ for $\alpha = \sqrt 3 + \sqrt 7$ by a previous problem. What is the minimal polynomial of $\alpha$? What is the dimension of $K$?* > Let $\alpha = \sqrt 3 + \sqrt 7$ as in the hint. Check that $\alpha^2 = 2\sqrt{21}+10$ and $\alpha^4 = 40\sqrt{21} + 184$, so $\alpha^4 = 20\alpha^2 - 16$ meaning it is a root of $f(x) = x^4 - 20x^2+16$. Since we know $K = \mathbf{Q}[\sqrt 3 + \sqrt 7]$ and $f$ is degree 4, we can take as a basis $B = \{\alpha^0=1, \alpha, \alpha^2, \alpha^3 \}$. This is linearly independent since $\alpha$ doesn't satisfy any degree 3 or lower polynomials, so $\sum_{i=0}^3 c_i \alpha^i = 0 \implies c_i = 0$ for all $i$, and is spanning since $K$ is 4-dimensional and we have 4 linearly independent vectors. - Find a basis for > General strategy for $\mathbf{Q}[\alpha]$: try to find the minimal polynomial $f$ for $\alpha$ by looking for a combination of $\alpha, \alpha^2,\alpha^3,\cdots$ that add up to zero, then in most cases $n = \deg f$ is the degree of the extension, i.e. the dimension of $\mathbf{Q}[\alpha]$ as a vector space over $\mathbf {Q}$. So go looking for $n$ basis vectors, usually of the form $\alpha^0, \alpha^1,\alpha^2,\cdots, \alpha^{n-1}$. Note that $1$ is always a basis element, for free! - $\mathbf Q[i\sqrt7]$ over $\mathbf Q$. > Let $x=i\sqrt 7$, check $x^2 = -7\implies x^2+7=0$ so we want 2 basis elements. Take $B = \{1, i\sqrt 7\}$, this is enough because the dimension is 2. It is linearly independent since $a_0 \cdot 1 + a_1\cdot i\sqrt 7=0$ with $a_0, a_1\neq 0 \implies {a_0\over -ia_1} = \sqrt 7 \implies i{a_0\over a_1} = \sqrt 7$, but the LHS is purely imaginary and the RHS is purely real, a contradiction. So $a_0 = a_1 = 0$. > This is a spanning set because it's a "big enough" linearly independent set, namely 2, the dimension. - $\mathbf Q[\sqrt{-3}]$ over $\mathbf Q$. > Let $x=\sqrt{-3}$ to get $x^2+3 = 0$, so $\dim = 2$. Take $\{1, \sqrt{-3}\}$. Note $\sqrt{-3} = i\sqrt 3$. > Carefuly! Don't mistakenly try to take a basis like $\{1, i, \sqrt 3\}$, this is too big (3, vs 2 dimensions). - $\mathbf Q [i\sqrt d]$ over $\mathbf Q$, where $d\in \mathbf{Z}$ is a *squarefree* positive integer (i.e. if $d = \prod_{i=1}^m p_i^{e_i}$ is the prime factorization of $d$, then every $e_i = 1$, so $d$ is not divisible by any perfect square). > Let $x=i\sqrt d$, so $x^2 = -d \implies x^2+d=0$ and $\dim =2$. Take $\{1, i\sqrt d\}$. - $\mathbf Q[\sqrt d + i]$ over $\mathbf Q [\sqrt d]$. > Whoops! This one might be harder than expected. Let $\alpha = \sqrt d + i$, and check that it has minimal polynomial $f(x) = x^4-4x^2+16$ over $\mathbf{Q}$, so $[\mathbf Q[\sqrt d + i]: \mathbf Q] = 4$. Since $[\mathbf Q[\sqrt d]: \mathbf Q] = 2$, this forces $[\mathbf Q[\sqrt d + i]: \mathbf Q[\sqrt d] ] = 2$ and you can take $B = \{1, \sqrt d + i\}$ (and this is enough because of the 2).