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Jan
29
comment If $\alpha$ is an algebraic element and $L$ a field, does the polynomial ring $L[\alpha]$ is also a field?
You mean the ring of polynomial expressions in $\alpha$. I don’t think it’s such a terrible solecism to refer to it the way you did, especially since your meaning was clear.
Jan
28
answered examples of unramified extensions of $\mathbb{Q}_p$
Jan
28
answered Extending $2$-adic valuation to a finite extension of $\mathbb Q$.
Jan
28
comment Isn't the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}(\sqrt3))$ just the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2))$?
The solidus (“$/$”) is just mathematical shorthand for “over”. When I saw that a student was thoroughly confused by the notation “$L/K$”, thinking it was like the quotient notation $G/H$ for $H$ a normal subgroup of $G$, I decided that when it was fields I was talking about,I would always write $L\supset K$ instead of $L/K$. (I notice that some people are now writing $L|K$ instead. That’s good too.)
Jan
27
answered Isn't the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}(\sqrt3))$ just the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2))$?
Jan
27
comment integer solutions for $\left( a+\sqrt{b} \right) ^ n=p+q\sqrt{b}$?
You’re merely finding the $n$-th power of a given element of $\Bbb Z[\sqrt b]$? I’d just use the Binomial expansion—should be perfectly straightforward.
Jan
27
comment Is there an “easier” way to find the factors of a polynomial than using Ruffini's method?
Doesn’t Ruffini just apply for finding roots (=linear factors)? You can use ordinary long division of polynomials just as well. The real questions present themselves when there are no roots but the polynomial factors very well, like $x^4+4=(x^2-2x+2)(x^2+2x+2)$.
Jan
27
answered Is this inheritance valid for algebraic extensions?
Jan
27
comment Is $\mathbb{Z}[\sqrt{-7}]$ a UFD or not?
And of course that theorem is hardly elementary…
Jan
27
comment Is $\mathbb{Z}[\sqrt{-7}]$ a UFD or not?
You just have to believe the Minkowski bound, which seems to tell me that I don’t need to check any ideals at all for principalness.
Jan
27
comment Is $\mathbb{Z}[\sqrt{-7}]$ a UFD or not?
On the other hand, the integers of $\Bbb Q(\sqrt{-7}\,)$ do form a UFD.
Jan
27
comment Why the limit of a quotient is not the quotient of a limit?
The limit of a quotient is the quotient of the limits, if the downstairs limit is nonzero. And if the limit of the denominator is indeed zero, the quotient of the limits is not defined.
Jan
26
comment Why does an argument similiar to 0.999…=1 show 999…=-1?
Often, when the radix is not (a power of) a prime, they’re called “$g$-adic numbers”, and OP may have success in searching for that term. Additionally, I might point out that $\sum_0^\infty10^n$ is convergent to $-1$ not only $10$-adically, but also $2$-adically and $5$-adically.
Jan
26
comment field properties from prime subfield
It is true that as an additive group (only), $\Bbb F_{p^n}\cong(\Bbb Z/p\Bbb Z)^n$. Not multiplicatively, and I don’t think that that was OP’s intent.
Jan
26
answered Eccentricity of a general ellipse
Jan
26
comment Path needed for solving these linear equations in Zn (my example Z105)
The congruence $49x\equiv21\pmod{105}$ is equivalent to $7x\equiv3\pmod{15}$, not what you wrote. It boils down to $x\equiv9\pmod{15}$. There are seven classes modulo $105$ that satisfy this, and among them is $84$.
Jan
26
comment Is the tensor product of 2 finite extension of $\Bbb Q$ isomorphic to a direct sum of fields?
And, @Thalanza, $f$ is to be irreducible over the base $k$, but may not be over $K_2$. I think that the example $K_1=\Bbb Q(\sqrt[3]5)$ and $K_2=\Bbb Q(\omega\sqrt[3]5)$, where $\omega$ is a primitive cube root of unity, should be eminently computable, and you can see what happens. In fact, maybe clearer to take $K_2=K_1$ here.
Jan
26
comment Is the tensor product of 2 finite extension of $\Bbb Q$ isomorphic to a direct sum of fields?
I think, @QiaochuYuan, that something is wrong, since $K_1=\Bbb F_p[x]/(x^p-a)$ is not a field. Namely, every $a\in\Bbb F_p$ has a $p$-th root, itself in this case. I think you needed a nonperfect field for a base rather than $\Bbb F_p$.
Jan
24
answered Completion of an extension $K|\mathbb{Q}$ w.r.t. a non archimedean abs. value, isomorphic to $K\cdot \mathbb{Q}_p$
Jan
24
comment Completion of an extension $K|\mathbb{Q}$ w.r.t. a non archimedean abs. value, isomorphic to $K\cdot \mathbb{Q}_p$
Actually, this is the very nub of the problem. I’ll explain in an answer when I get time.