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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.
Jan
24
comment Completion of an extension $K|\mathbb{Q}$ w.r.t. a non archimedean abs. value, isomorphic to $K\cdot \mathbb{Q}_p$
Exactly how do you define $K\cdot\Bbb Q_p$? I presume you’re speaking of a compositum of two fields, and to define this, you’ll need them both to dwell inside some bigger field.
Jan
23
comment Finite extensions of $\mathbb F_p(t)$
Right. I wonder whether the quickest way to exhibit two different fields, in this situation, might be to show that their automorphism groups are different. Over $k=\Bbb F_2$, for instance, $k(t)$ has exactly $6$ automs, but the fraction field of $k[x,y]/(y^2+y+x^3 +x)$ has many more, since it has five $k$-rational points, and you have the translations by these, thus a subgroup of order $5$.
Jan
23
answered Are the $p^n$-adic numbers isomorphic to the $p$-adic numbers?
Jan
22
comment Inseparable extensions
There is a classification of the intermediate fields when you have an extension $K\supset k$ and $k\supset K^p$, maybe restricted to the case of finite degree. You look at certain derivations, I think $k$-linear derivations of $K$ into $K$. To tell the truth, I’ve forgotten it. But the extensions that are “deeper” in some sense can get very complicated.
Jan
22
answered Inseparable extensions
Jan
22
comment Distributive Law and how it works
My usual response to questions of this form, i.e. “Why isn’t it such-and-such a way?” is to say that In mathematics, nothing is true. Except when there’s a proof. There are proofs for the proper distributive rule, in various contexts, and of course there’s all the experimental evidence.