Tag Info

Hot answers tagged

10

In fact, they can be isomorphic. For example $\mathbb{R}(X^2) \subset \mathbb{R}(X)$ is a field extension of degree $2$ but the two fields are isomorphic (as fields) by $X^2 \mapsto X$. This isomorphism (as fields) does however not induced an isomorphism of $F$-vectorspaces between $F$ and $K$, which is why $[K:F]\neq [F:F]$ is not a problem even when ...


6

In general we would guess that $\sqrt a+\sqrt b+\sqrt c$ has eight conjugates, obtainable by toggling signs individually for the surds. However, in this special case we see that $\sqrt a\sqrt b\sqrt c=30$, which cannot change its sign. Hence once we picked the sign of two of the surds, the sign of the third is determined.


4

$$\sqrt{6}+\sqrt{10}+\sqrt{15}=\sqrt{2\cdot 3}+\sqrt{2\cdot 5}+\sqrt{3\cdot 5} = \frac{1}{2}\left(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2-(2+3+5)\right)$$ where $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an algebraic number of degree $8$ over $\mathbb{Q}$, having conjugates $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$, whose minimal polynomial is an even function. It follows ...


4

Here's an approach using the fact that the discriminant is a symmetric polynomials in the roots. First, consider the ring $R=\mathbb{Z}[T_1, \ldots, T_n]$, and look at the "universal" monic polynomial $P(X) = \prod_{i=1}^n (X-T_i) \in R[X]$. Its discriminant is $D=\prod_{1\leq i<j\leq n}(T_i-T_j)^2$. This is a symmetric polynomial in the $T_i$, so it ...


4

The argument given in this answer on MO shows that if $A$ is any set equipped with a collection of finitary operations which satisfy some equational axioms such that $2\leq|A|\leq\aleph_0$, then there is an infinite-dimensional contractible CW-complex $X$ which can be equipped with corresponding finitary operations which are continuous and which satisfy the ...


4

What makes you think that $$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$ is true? The powers of $\zeta_{12}$ repeat modulo 12. In fact since the $12$th cyclotomic polynomial is $\Phi_{12}=x^4-x^2+1$, we have that $$\begin{align*} \zeta_{12}^5&=\zeta_{12}^5+0\\\\ ...


2

The polynomial $x^{16}-x\in\mathbb{F}_2[x]$ has roots precisely equal to the elements of $\mathbb{F}_{16}$, and the subfields of $\mathbb{F}_{16}$ are $\mathbb{F}_{16}$, $\mathbb{F}_{4}$, and $\mathbb{F}_{2}$, which have degrees $$[\mathbb{F}_{16}:\mathbb{F}_2]=4\qquad [\mathbb{F}_{4}:\mathbb{F}_2]=2\qquad [\mathbb{F}_{2}:\mathbb{F}_2]=1$$ Therefore ...


2

Right before that he says Let us keep the hypotheses of prop. 9. If $\mathfrak{P}$ is a non-zero prime ideal of $B$... Proposition 9 says Proposition 9. If $A$ is Dedekind then $B$ is Dedekind. and in a Dedekind domain, every non-zero prime ideal is maximal. Thus $\mathfrak{P}$ and $\mathfrak{p}=\mathfrak{P}\cap A$ are indeed maximal, and thus ...


2

The conjugates of an algebraic number are (by definition) the roots of its minimal polynomial. The number of (distinct) roots of an irreducible polynomial over the rationals is equal to its degree, that is four. Thus once you know the minimal polynomial "it is clear." There is some wiggling room as one might or might not count the number itself among its ...


1

The only place that you use the fact that $F$ is algebraic over $\mathbb{Q}$ is in the primitive element theorem. If for some reason your irreducible polynomial will have only one parameter (instead of three), then your solution will still work. Suppose that your polynomial is $f(x)=x^3+Ax^2+Bx+C$. First consider the polynomial ...


1

Yes, $\mathbb{C}_p\cong\mathbb{C}_q\cong\mathbb{C}$. However, the isomorphism is only an isomorphism of fields. The natural topology on the three fields are different, so they are nonisomorphic as topological fields. However, the isomorphism is impossible to fully specify using a finite number of symbols/words. Yes, for $x\in\mathbb{C}_p$, it does make ...


1

No. A counter-example is given by the algebraic closure $\overline{\mathbb{F}}_p$ of a finite field $\mathbb{F}_p$ of characteristic $p>2$. Indeed, since any automorphism $\sigma$ of $\overline{\mathbb{F}}_p$ fixes all elements of the prime field $\mathbb{F}_p$, then this $\sigma$ is an element of the absolute Galois group of $\mathbb{F}_p$, which is ...



Only top voted, non community-wiki answers of a minimum length are eligible