Studying for the Algebra 2 Prelim

Gauss's Lemma.   A polynomial $f(x)\in \mathbb Z[x]$ is said to be primitive if there is no prime which divides all of the coefficients of $f$. Then a polynomial $f\in \mathbb Z[x]$ is irreducible in $\mathbb Z[x]$ if and only if it is primitive and irreducible in $\mathbb Q[x]$.

Gauss's Lemma.   Let $R$ be a UFD with $K$ its field of fractions and let $f(x) \in R[x]$ be a polynomial. Then $f$ is irreducible in $R[x]$ if and only if it is irreducible in $K[x]$ and the ideal generated by all its coefficients is $(1)$ (i.e. the polynomial is primitive).

Answer: The change of variables is $x \mapsto y - \frac{a}{3}.$

The Galois group of $f$ is a subgroup of $A_n$ if and only if the discriminant $D$ is a square in $K$.

Fix a splitting field $L$ for $f$. If $\alpha$ and $\beta$ are distinct roots of $f$ then $K(\alpha) \cong K(\beta)$ via a map sending $\alpha$ to $\beta$. This lifts to an automorphism of $L$ (by the isomorphism extension theorem) which fixes $K$ and sends $\alpha$ to $\beta$.

They are $S_4$, $A_4$, $D_4$ (dihedral group of order $8$), $K_4$ (the Klein $4$ group) and $C_4$ (the cyclic group of order$4$).

They are $S_5$, $A_5$, $F_{20}$ (the dihedral group of order $20$),

cycle type	1	2	(2,2)	3	(2,3)	4	5
$\mathbb Z/4\mathbb Z$	1						4
$D_5$	1		5				4
$F_{20}$	1		5			10	4
$A_5$	1		15	20			24
$S_5$	1	10	15	20	20	30	24

See the following document for a better table: cycle types

They are $S_3$ and $A_3$. The latter happens if and only if $D$ is a square over the base field.

There is a unique decomposition of $\sigma$ into disjoint cycles of lengths $n_1 \leq n_2 \leq ... \leq n_r$ (including $1$ cycles). The cycle type of $\sigma$ is the typle $(n_1,...,n_r)$.

If $p$ doesn't divide $D$, then there is a cycle $\sigma$ in the Galois group of $f$ whose cycle type is $(n_1,...,n_r)$ where $n_i = \deg f_i$.

Solution.   Every field extension of $\mathbb Q$ is separable, so simply find a field extension of $\mathbb Q$ which isn't Galois. $\mathbb Q(\sqrt[3]{2})/\mathbb Q$ works.  

Solution.   Every splitting field is normal, so one should look for the splitting field of an inseparable irreducible polynomial. Taking $\mathbb F_p(\sqrt[p]{t})/\mathbb F_p(t)$ works. It's the splitting field of $x^p - t$ and hence normal, but $x^p - t$ factors as $(x - \sqrt[p]{t})^p$ and is thus inseparable.  

Solution.   The primitive element theorem says that all finite separable field extensions are simple; hence we cannot consider finite field extensions of characteristic 0 field nor of finite fields.

Set $F = \mathbb F_p(x,y)$ and $K = \mathbb F_p(x^{1/p},y^{1/p})$. Then $[K:F] = p^2$. Now take an arbitrary rational function $f\in K$. Raising $f$ to the $p$th power leaves the coefficients of $f$ fixed while swapping $x^{1/p}$ for $x$ and $y^{1/p}$ for $y$, hence $f^p \in F$ and so $[F(f):F] \leq p$. This means $K/F$ is not simple.

 

Solution:   First recall that we can take quotients by the generators of an ideal in any order, so and since $K[x,y]/g(y) \cong (K[y]/g(y))[x]$, we have that Because $g$ is irreducible and $K[y]$ is a PID, $K[y]/g(y)$ is a field. Together with the above isomorphism, this implies $K[x,y]/(g(y),f(x))$ is a field precisely when $f$ is irreducible over $K[y]/g(y)$. We then easily have the following equivalences:

Solution to (1):   The $n$th cyclotomic extension $\mathbb Q(\zeta_n)/\mathbb Q$ has Galois group $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \left(\mathbb Z/n\mathbb Z\right)^\times$, so $\mathbb Q(\zeta)/\mathbb Q$ has Galois group $G \cong (\mathbb Z/9\mathbb Z)^\times$. There are $6$ integers between $0$ and $8$ coprime to $9$, hence $G$ has order $6$. The only groups of order 6 are $S_3$ and $\mathbb Z/6\mathbb Z$, and since $S_3$ isn't Abelian, $G\cong \mathbb Z/6\mathbb Z$. The subgroup

is then also Abelian and hence normal, so by the Galois correspondence, its fixed field $K \subset \mathbb Q(\zeta)$ is also a Galois extension of $\mathbb Q$ and $\operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q) \cong G/H \cong \mathbb Z/3\mathbb Z$. Hence $K$ is a subextension of $\mathbb Q(\zeta)/\mathbb Q$ which is Galois and order $3$.

Solution to (2):   We solve this problem in two parts: we find a primitive element for $K$ and then we find its minimal polynomial. Whatever the primitive element of $K$ is, it must be a polynomial in $\zeta$. Hence it makes sense to start looking for polynomials in $\zeta$ which are fixed by $H$, the subgroup of $\operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)$ which fixes $K$. The group $H$ consists of the identity and an involution which sends $\zeta \mapsto \zeta^{-1} = \zeta^{8}$. An obvious candidate primitive element for $K$, therefore, is $\zeta + \zeta^8$, as this is fixed exactly when $\zeta$ is fixed or sent to $\zeta^8$. Our hypothesis is therefore

To find the minimal polynomial of $\zeta + \zeta^8$, we write down powers of $\zeta + \zeta^8$ up to degree 3. Note that because $H$ fixes $\zeta + \zeta^8$ we automatically know $\zeta + \zeta^8 \in K$, hence the order of $\zeta + \zeta^8$ over $\mathbb Q$ is either $1$ or $3$. Since there are automorphisms of $\mathbb Q(\zeta)$ which don't fix $\zeta + \zeta^8$ (for instance $\zeta \mapsto \zeta^2$) we know it must be degree 3.

The 9th cyclotomic polynomial is $x^6 + x^3 + 1$, hence $\zeta^3 + \zeta^6 = -1$. This implies has $\zeta + \zeta^8$ as a root. It must be the minimal polynomial because, as we noted above, $\zeta + \zeta^8$ is order $3$ over $\mathbb Q$.

Solution to (3):   A polynomial has nonzero discriminant if and only if it is separable; so $g$ is separable. The statement that $g(x) \equiv f(x) \mod p$ means we should keep $K$ from the previous parts in mind as we solve this. This insight, together with the fact that one of the given hypotheses involves the number 9, indicates that we should consider the cyclotomic extension $\mathbb F_p(\zeta)$.

Assume now that $p \equiv -1 \mod 9$. Squaring both sides gives us $p^2 \equiv 1 \mod 9$, which implies that $9 ~|~ p^2 - 1$. Using the general fact that $d~|~n \iff x^d - 1~|~ x^n -1$, we have the following chain of divisions:

This implies that $\mathbb F_p(\zeta)$ is a subfield of $\mathbb F_{p^2}$, and hence the minimal polynomial of $\zeta$ over $\mathbb F_p$ is at most degree 2, meaning that Since $g(x) \equiv f(x) \mod p$, the splitting field of $g$ is precisely $\mathbb F_p(\zeta + \zeta^8)$. If $\operatorname{Gal}(\mathbb F_p(\zeta)/\mathbb F_p) \cong 1$ then we're done, so assume it's $\mathbb Z/2\mathbb Z$. The only nontrivial automorphism of this group is $\zeta \mapsto \zeta^{-1}$ which fixes $\zeta + \zeta^8$, hence $\mathbb F_p(\zeta + \zeta^8) = \mathbb F_p$.

Solution:   Let $K = \mathbb F_p\left(\frac{f(x)}{g(x)}\right)$ and consider the polynomial $h(t) = \frac{f(x)}{g(x)}g(t) - f(t)$ in $\mathbb K[t]$. It has $x$ as a root. This implies the minimal polynomial of $x$ over $K$, call it $a(t)$, divides $h$. Because $\deg a(t) \leq \max\{f(t), g(t)\} < p$, $a'(t) \neq 0$. This implies that $a(t)$ and $a'(t)$ have no common factor in $K[t]$ (trivially, by the irreducibility of $a(t)$), and hence $(a, a') = 1$. Thus $a(t)$ is separable, so $\mathbb F_p(x)$ is a separable extension of $K$.

Solution:  

Solution:   Let $L$ be the Galois closure of $K$ over $\mathbb Q$. By the primitive element theorem we know that $K \cong \mathbb Q(\theta)$ for some $\theta$ algebraic over $K$ with $f(x) \in \mathbb Q[x]$ its minimal polynomial. Then $\deg f = n$ and $L$ is the splitting field of $f$ ($K/\mathbb Q$ is separable by virtue of the fact that $\operatorname{char}\mathbb Q = 0$, hence we need only adjoin the remaining roots of $f$ to ensure the extension is Galois). This implies $[L:\mathbb Q] \leq n!$.

Every subfield of $K$ must be contain $1$, and hence also contains $\mathbb Q$. The number of subfields of $K$ is therefore bounded by the number of subfields of $L$ containing $\mathbb Q$, which in turn is equal to the number of subgroups of $\operatorname{Gal}(L/\mathbb Q)$ by the Galois correspondence. The number of subgroups is certainly bounded by the number of subsets, which is $2^{|\operatorname{Gal}(L/\mathbb Q)|}$. Hence the number of subfields of $K$ is at most $2^{n!}$.

Suppose now that $K = \mathbb Q(\alpha, \beta)$. For each $0\leq m\leq 2^{n!}$, $\mathbb Q(\alpha, \beta)$ is a subfield of $K$. There are $2^{n!}+1$ possible choices for $m$, and hence by the pigeonhole principle and our above bound, there must exist distinct integers $a$ and $b$ between $0$ and $2^{n!}$ such that $\mathbb Q(\alpha + a\beta) = \mathbb Q(\alpha + b\beta)$. This then implies that

is contained in $\mathbb Q(\alpha + a\beta)$, and thus so is $\alpha$. We conclude that $\alpha + a\beta$ is a primitive element of $K$.

Solution to (a):   Note first that $[\mathbb Q(\alpha):\mathbb Q] = 4$ since $f(x)$ is irreducible by Eisenstein's criterion. The field $\mathbb Q(\alpha)\subseteq \mathbb R$ clearly does not contain $\beta$, hence the factorization of $f$ into irreducibles over $\mathbb Q(\alpha)$ is $(x - \alpha)(x + \alpha)(x^2 - \beta^2)$. Thus $\beta^2 \in \mathbb Q(\alpha)$ (implying that $\beta$ is purely imaginary) and $\beta$ has minimal polynomial $x^2 - \beta^2$ over $\mathbb Q(\beta)$. This means

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Solution to (b):   The Galois action on the roots of $f$ is transitive since $f$ is irreducible^[1]. This means $\operatorname{Gal}(E/\mathbb Q)$ is isomorphic to one of the five transitive subgroups of $S_4$: $S_4$ itself, $A_4$, the dihedral group of order 8 $D_4$, the Klein 4 group $K_4$ or $C_4$. These last two groups are of order 4 while the first two are order $24$ and $12$ respectively. Since $|\operatorname{Gal}(E/\mathbb Q)| = [E:\mathbb Q] = 8$, we conclude $\operatorname{Gal}(E/\mathbb Q)\cong D_4$, the dihedral group of order 8.

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Solution to (c):   Recall that $E = \mathbb Q(\alpha,\beta)$. Since $E/\mathbb Q$ is a finite Galois extension, the proof of the primitive element theorem demonstrates that there is some integer $n$ such that $\alpha + n\beta$ is primitive in $E$^[2]. We therefore need only identify an element of the form $\alpha + n\beta$ which isn't fixed by any element of $\operatorname{Gal}(E/\mathbb Q)$.

Since $E = \mathbb Q(\alpha, \beta)$, an automorphism $\varphi\in \operatorname{Gal}(E/\mathbb Q)$ is entirely determined by the image of $\alpha$ and $\beta$. Since $\varphi(-\alpha)$ is determined by $\varphi(\alpha)$, if we choose $\varphi(\alpha)$ first there are only two choices remaining for $\varphi(\beta)$. Keeping this in mind, we can easily see that choosing $n = 3$ does the trick by examining a few cases.

Note first that $\varphi$ fixes $\alpha + 3\beta$ precisely when

This shows that $\alpha + 3\beta$ isn't fixed by any element of $\operatorname{Gal}(E/\mathbb Q)$, and hence $\mathbb Q(\alpha + 3\beta) = E$.

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Solution to (d):   As far as I am aware, the best way to proceed here is in the obvious way: compute the subgroup lattice of $G = \operatorname{Gal}(E/\mathbb Q)$ and determine which subgroups are normal. I haven't finished typing this answer up yet.

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[1]	This follows from the fact that if $\alpha$ and $\beta$ are any two roots of an irreducible polynomial over a field $K$, then $K(\alpha)\cong K(\beta)$ via the map $\alpha \mapsto \beta$.

[2]

The proof of the primitive element theorem for infinite fields can be reduced to the case of $K(\alpha,\beta)/K$, in which case, one can show that there exist distinct elements $x,y\in K$ such that $K(\alpha + x\beta) = K(\alpha + y\beta)$. It readily follows that $\alpha + x\beta$ is a primitive generator for $K(\alpha,\beta)$.

Solution.   Let $K = \mathbb F_2(t)$ for convenience. If $f$ were not irreducible then it would factor as the product of either two quadratics or a linear and a cubic. I don't see a better way to approach this other than by examining these two cases separately; generalized Eisenstein's criterion fails under all tricks I tried. Note that $a = -a$ for all $a\in K$ since $\operatorname{char} K = 2$.

Case 1: Suppose $f(x) = (x + a)(x^3 + bx^2 + cx + d)$ for some $a,b,c,d \in K$. Then

Comparing coefficients, we see that

• $b + a = 0\implies b = a$,

• $c + ab = 0 \implies c = a^2$,

• $d + ac = 0 \implies d = a^3$ and

• $ad = t \implies t = a^4$.

However, $t$ is not a fourth root in $K$, so this is impossible.

Case 2: To arrive at a contradiction this time, we'll need to resort to using the valuation $\deg$ on $\mathbb F_2(t)$. Suppose then that that $f(x) = (x^2 + ax + b)(x^2 + cx + d)$ for some $a,b,c,d\in K$, so

Comparing coefficients gives us the following:

• $c + a = 0\implies c = a$.

• $d + ac + b = 0 \implies d = a^2 + b$.

• $ad + bc = 1 \implies a(a^2 + b) + ba = a^3 + 2ba = a^3 = 1$. One can show that the only cube root of $1$ in $\mathbb F_2(t)$ is $1$ itself, since $x^3 - 1 = (x+1)(x^2 + x + 1)$ and $x^2 + x + 1$ is irreducible over $\mathbb F_2(t)$.

• $bd = t\implies b(a^2 + b) = b(1 + b) = 1.$ Comparing degrees we get $\deg(b) + \deg(b+1) = \deg(1)$, so $2\deg(b) = \deg(t) = 1$ and hence $\deg(b) = \frac12$. This is impossible. Hence $f(x)$ is irreducible over $\mathbb F_2(t)$.

Notice that $f$ is separable since $f'(x) = 1$, and thus its splitting field – call it K – is indeed a Galois extension.

 

Proof.   Our strategy will first be to find a degree 5 Galois extension $K/\mathbb Q$. By the Fundamental Theorem of Galois theory $K$ is necessarily the splitting field of some polynomial over $\mathbb Q$, and because $K$ is necessarily simply by the primitive element theorem, it is in particular the splitting field of some degree 5 polynomial over $\mathbb Q$. Thus, we first find Galois extension $\mathbb Q(\alpha)/\mathbb Q$ of degree 5 and then identify the minimal polynomial of $\alpha$.

Recall that every finite Abelian group is the subgroup of some cyclotomic extension of $\mathbb Q$ and let $\zeta_n$ be a primitive $n$th root of unity. It is a standard result that $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q) \cong \left(\mathbb Z/n\mathbb Z\right)^\times$. Furthermore, if $G$ is an Abelian group whose order is divisible by a prime $p$, then $G$ contains a copy of $\mathbb Z/p\mathbb Z$. This means we need only find $n$ such that $\left(\mathbb Z/n\mathbb Z\right)\times$ has order divisible by $5$. Taking $n = 11$ works: in this case

Consider the element $\zeta_{11} + \zeta_{11}^{-1}$. The only nontrivial automorphism $\sigma\in\operatorname{Gal}(\mathbb Q(\zeta_{11})/\mathbb Q)$ which fixes it is the one defined $\sigma:\zeta_{11}\mapsto \zeta_{11}^{-1}$. Hence $\mathbb Q(\zeta_{11}+\zeta_{11}^{-1})$ is the fixed field of $\left\langle \sigma \right\rangle \cong \mathbb Z/2\mathbb Z$, and thus by degree arguments, Thus $\mathbb Q(\zeta_{11} + \zeta_{11}^{-1})$ is a good candidate field. All that remains is to identify the minimal polynomial of $\zeta_{11}+\zeta_{11}^{-1}$; set $\zeta = \zeta_{11}$ and $\mathbb Q(\zeta + \zeta^{-1}) = K$ for convenience. We **know** the minimal polynomial of $\zeta + \zeta^{-1}$ is of degree 5, so to compute it, we first find the powers of powers of $\zeta+\zeta^{-1}$ up to 5. Note that $\zeta^{-1} = \zeta^{10}$, so using the binomial theorem, we get Recall that $\zeta$ is a root of the $11$th cyclotomic polynomial, so Set $\alpha = \zeta + \zeta^{-1}$ and notice that and since the sum of this is precisely $\Phi(\zeta) - 1$, the polynomial has $\alpha$ as a root and is degree $5$. Since $K/\mathbb Q$ is Galois and hence normal, and because $K$ contains one of the roots of $f$ (namely $\alpha$) it must contain them all; $f$ splits over $K$. We know that $f$ must be irreducible over $\mathbb Q$ as otherwise $[K:\mathbb Q]$ would be strictly smaller than $5$; hence no roots of $f$ are in $K$. This implies, again by degree considerations, that $K$ is the splitting field of $f$.

 

Proof.   The TLDR; is that these fields are not isomorphic because $\mathbb Q(\sqrt{3})$. Does not contain a number which squares to $2$.

Suppose we had an isomorphism $\varphi:\mathbb Q(\sqrt{2}) \to \mathbb Q(\sqrt{3})$. Then there is some $a,b\in \mathbb Q$ such that

and hence since $\varphi$ must "fix" the rationals (it is also a $\mathbb Q$-algebra homomorphism). This implies The second equation implies that either $a = 0$ or $b = 0$. If $a = 0$, then $b = \sqrt{2}/\sqrt{3}$, which can't happen since $b$ is rational. If $b = 0$, then $a = \sqrt{2}$ which again can't happen since $a$ was assumed to be rational. We conclude that $\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$ are not isomorphic.

 

Proof.   A bit of high school algebra gives us and thus