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Fields as algebraic objects. For vector and tensor fields, use eg. [dg.differential-geometry]. For physical fields, use eg. [mp.mathematical-physics] or [quantum-field-theory].
15
votes
Accepted
Factorization of an irreducible polynomial in the field extension it defines
Let us show that for the partition $2+1+1=4$, there is no such $f$.
If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group …
11
votes
Accepted
Surjectivity of bilinear forms.
The answer to Question B is no, as I'll show below.
Let $U=V=\mathbf{Q}^3$ and $W=\mathbf{Q}^4$. Define
$$\beta((u_1,u_2,u_3),(v_1,v_2,v_3))=(u_1 v_1,u_2 v_2,u_3 v_3, (u_1+u_2+u_3)(v_1+v_2+v_3)).$$
…
4
votes
A question on function fields (extending my previous question)
The complete set of solutions consists of
$$(1,0), (0,1),$$
$$(a,x), (x,a), (1/a,-x/a), (-x/a,1/a), (1/x,-a/x), (-a/x,1/x),$$
$$(b,y), (y,b), (1/b,-y/b), (-y/b,1/b), (1/y,-b/y), (-b/y,1/y).$$
Let $C$ …
24
votes
Accepted
When f(x)-a and f(x)-b yield the same field extension
The answer is that over a number field $k$, an equivalence class can be finite, and in fact it is usually so for $f$ of moderately large degree. Consider $f(x):=x^7+x$ over $\mathbf{Q}$, for example. …