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May
8
comment Algebraic Number Theory - Rings of Integers
Yes to both questions, but something extra needs to be said for the second. The set you mention there is a $\Bbb Z$-basis of $\Bbb Z[\alpha]$ all right, but often one says that $\{\alpha\}$ is a generating set, because it generates $\Bbb Z[\alpha]$ as a ring.
May
6
comment Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$
Right you are, I’m in complete agreement. The moral of the story is that a good conjecture may need to be modified to take in phenomena more perfectly.
May
5
comment The ideal $(p)$ always factors in the ring of integers of $\mathbb Q(\sqrt{2},\sqrt{3})$
If a prime remains prime in a Galois extension, the Galois group must be cyclic. Look at the Frobenius.
May
5
comment Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$
I think your conjecture is false, see my comment below. But I gave you an up-arrow, because I think the conjecture is pleasing.
May
5
comment Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$
But I’m a little worried about units. Even for $i\in\Bbb Z[i]$, you have an integer that’s (naturally) relatively prime to its conjugate, but is not a perfect square. This seems to me to apply to the integer $i(1+2i)^2$ just as well.
May
5
comment Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$
@AlonsodelArte, how about $3\pm4i$ in $\Bbb Z[i]$?
May
5
comment Let $F/k$ be a Galois extension. Show there exists an $\alpha \in F$ such that $\{ \sigma(\alpha) | \sigma \in G (F/k)\}$ is a basis for $F$ over $k$.
This is known as the Normal Basis Theorem.
May
5
comment Solving $x^x \equiv x \pmod{17}$.
@RossMillikan, right on: I’d have liked to know the original formulation of the question. Maybe only the edited version supports my answer.
May
5
revised Solving $x^x \equiv x \pmod{17}$.
recover from a foolish arithmetic error
May
5
answered Solving $x^x \equiv x \pmod{17}$.
Apr
30
comment Is $3$ a prime element of $\mathbb{Z[\eta]}$?
In the fewest words, perfect.
Apr
30
comment Is $3$ a prime element of $\mathbb{Z[\eta]}$?
I like very much your having gotten to the nub of the matter in the first paragraph: that the splitting of $3$ is the same as the factorization of the cyclotomic polynomial over $\Bbb F_3$.
Apr
30
answered Is $3$ a prime element of $\mathbb{Z[\eta]}$?
Apr
30
comment By induction, show that $\sqrt{p}\notin\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\cdots,\sqrt{p_k})$
I’m sorry, I still have to disagree. OP asked about the field generated by the square roots, which of course is much larger than their $\Bbb Q$-linear span. To show that $\sqrt5\notin\Bbb Q(\sqrt2,\sqrt3,\sqrt7)$, it’s not enough to show that $\sqrt5$ is not of the form $a+b\sqrt2+c\sqrt3+d\sqrt7$.
Apr
29
comment By induction, show that $\sqrt{p}\notin\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\cdots,\sqrt{p_k})$
@user26857, I don’t agree. For instance $\sqrt{p_1p_2}$ is not in the span of the square roots of the primes.
Apr
28
comment Finding roots in finite fields.
But @KCd, there is a magical method in this case.
Apr
28
answered Finding roots in finite fields.
Apr
28
comment By induction, show that $\sqrt{p}\notin\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\cdots,\sqrt{p_k})$
For any answer to be helpful to you, we have to know how much you know already and the context in which you got to the question. Is this from a course? Homework?
Apr
28
comment By induction, show that $\sqrt{p}\notin\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\cdots,\sqrt{p_k})$
But OP is asking for a a much stronger conclusion than mere linear independence of the generators.
Apr
27
comment Techniques to prove that two field extensions are distinct
Yes, your objection is definitely well-taken. Would you like to e-mail me, and we can continue this discussion that way? Find my e-mail on my page, and the link is on my profile.