Rings and Fields

n ≡ 4 (mod 15) gives n ≡ 1 (mod 3) and n ≡ 4 (mod 5). With n ≡ 1 (mod 7), CRT gives n ≡ 1 (mod 21); n ≢ 1 (mod 5) kills the mod-35 claim.

89 is prime and $(\mathbb{Z}/89)\ast$ is cyclic of order 88. For a generator $g,{\sum }_k$ $k^{37}={\sum }_j$ $g^{37j}$ is a geometric sum equal to 0 because 88 ∤ 37 (so $g^{37}\ne 1)$.

By CRT, $\mathbb{Z}/105\mathbb{Z}\cong \mathbb{Z}/3\times \mathbb{Z}/5\times \mathbb{Z}/7;x^2=1$ has exactly 2 solutions in each odd prime field, giving $2^3=8$.

$(1)\mathbb{Z}[x]/(x)\cong \mathbb{Z}$, an integral domain, so (x) is prime. $(2)\mathbb{Z}$ is not a field, so (x) is not maximal. $(3)\mathbb{Z}[x]/(2,x)\cong$ 𝔽${}_2,a$ field, so (2, x) is maximal — note this also shows a prime ideal can be properly contained in another. (4) Fails: (2, x) itself is not principal, so $\mathbb{Z}[x]$ is not a PID, even though $\mathbb{Z}$ and 𝔽[x] both are.

Units are the residues coprime to $20:\phi (20)=20(1-1/2)(1-1/5)=8$, namely 1, 3, 7, 9, 11, 13, 17, 19.

$\mathbb{Z}[\surd -5]$ is not a UFD: $6=2\cdot 3=(1+\surd -5)(1-\surd -5)$.

(1) Gauss's theorem makes $\mathbb{Z}[x]a$ UFD, while the ideal (2, x) is not principal. $(2)6=2\cdot 3=(1+\surd -5)(1-\surd -5)$ gives two genuinely different factorisations into irreducibles. (3) In a Euclidean domain every nonzero ideal is generated by any element of least norm in it. (4) fails: (x, y) is not principal, by the same argument as in $\mathbb{Z}[x]$.

The derivative $3x^2+3>0$, so the cubic is strictly increasing: exactly one real root.

(1) is the Sophie Germain trap: $x^4+4=(x^2+2x+2)(x^2-2x+2).(2)$ Eisenstein at p = 2; alternatively, a cubic with no rational root is irreducible. (3) Degree 2 with no rational roots. (4) is the 5th cyclotomic polynomial ${\Phi }_5$: substitute x → x + 1 and apply Eisenstein at p = 5.

𝔽${}_8=$ 𝔽${}_{2^3}$. Its subfields are 𝔽${}_{2^d}$ for d dividing 3, namely 𝔽${}_2$ and 𝔽${}_8$ — so (1) fails, since 2 does not divide 3. (2) The multiplicative group of any finite field is cyclic, here of order 8 − 1 = 7. (3) 𝔽${}_8$ is a 3-dimensional vector space over 𝔽${}_2$, since |𝔽${}_8$| $=2^3.(4)\mathbb{Z}/8\mathbb{Z}$ is not a field — 2 is a zero divisor — so it cannot be isomorphic to 𝔽${}_8$.

$(1)[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ and $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$, so the tower multiplies to $4.(2)\alpha =\sqrt{2}+\sqrt{3}$ has ${\alpha }^3=11\sqrt{2}+9\sqrt{3}$, so $\sqrt{2}=({\alpha }^3-9\alpha )/2$ and $\sqrt{3}=(11\alpha -{\alpha }^3)/2$ lie in $\mathbb{Q}(\alpha )$ — the primitive element made explicit. $(3)\sqrt{6}=\sqrt{2}\cdot \sqrt{3.}(4)$ fails: K is the splitting field of $(x^2-2)(x^2-3)$, and splitting fields of polynomials are normal.

$x^3-2$ is irreducible over $\mathbb{Q}($Eisenstein at 2) and has $2^{1/3}$ as a root, so it is the minimal polynomial and the degree is its degree, 3. Note $\mathbb{Q}(2^{1/3})$ $\subset \mathbb{R}$ does not contain the complex cube roots — it is not the splitting field, which has degree 6.

$K=\mathbb{Q}(\sqrt[3]{2},\omega )$ has degree 6 over $\mathbb{Q}$ with non-abelian Galois group $S_3$; over $\mathbb{Q}(\omega )$ the degree is 3, so the group is $\mathbb{Z}/3$.