Let $F$ be a finite extension of the finite field $K$ and $\alpha \in F$. The mapping $L: \beta \in F \mapsto \alpha \beta \in F$ is a linear transformation of $F$, considered as a vector space over $K$. Prove that the characteristic polynomial $g(x)$ of alpha over $K$ is equal to the characteristic polynomial of the linear transformation $T$; that is $g(x) = det(xI - L)$, where $I$ is the identity transform.
123202
This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:  Let $K$ be a finite extension of $F$ of degree $n$. Let $\alpha$ be an element of $K$. Prove that the minimal polynomial for $\alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_{\alpha}$. In this problem, $T_{\alpha}$ is an $F$-linear transformation of $K$ that arises from $\alpha$ acting by left multiplication on $K$.  I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!
72517
I have problems to prove the following statement.     Let $K$ be a field and let $K(u,v)$ be an algebraic extension of $K$. If $u$ is separable over $K$ then $K(u,v)$ is a simple extension.   (My incomplete) proof: We can suppose that $char K=p$, because in characteristic $0$ all finite  extensions are simple. Let's divide the problem in $3$ cases (the third is problematic):  1) $v$ is separable. In this case $K(u,v)$ is a separable extension of $K$ and so it is simple.    2) $v$ is purely inseparable. We prove that $K(u,v)=K(u+v)$; by definition exists $m\in\mathbb N$ such that $v^{p^m}\in K$, so $(u+v)^{p^m}=u^{p^m}+v^{p^m}\in K(u^{p^m})$. So $K(u^{p^m})\subseteq K(u+v)$, but $u$ is separable on $K$ and follows that $K(u^{p^m})=K(u)$. Moreover $v=(u+v)-u\in K(u+v)$ and we can conclude that $K(u+v)=K(u,v)$.  3) $v$ is neither separable nor purely inseparable. ???    I would appreciate any hint about the point 3), but also a more organic proof without any subdivision.  Thanks in advance.
98032
I am trying to prove that if $L/K$ is an algebraic extension and if $\alpha \in L$, then    $\alpha$ is separable over $K$ if $\mathrm{char}(K)=0$. This is clear because $K$ is perfect which in turn implies that $L/K$ is seperable .  Now if $\mathrm{char}(K)=p$ is prime, then the statement is: $\alpha$ is separable if and only if $K(\alpha) =K(\alpha^p)$.  This problem is from a past examination and looks very interesting, but somehow I am not able to connect what's happening with the extension of $K$ with $\alpha$ and $\alpha^p$ .   I need some help here.    This somehow makes me think that the minimal polynomial of $\alpha$ and $\alpha^p$ have different roots and $\alpha$ is given by some root of minimal polynomial of $\alpha^p$ .   Note that I am using the definition of separability as follows :      An extension $L/K$ is called separable if for every $\alpha \in L$ the minimal polynomial of $\alpha$ has distinct roots in $L$ .    Thanks for helping.
72712
Let $F$ be a finite field. If $f : F \to F$, given by $ f(x) = x^3$ is a ring homomorphism, then    $ F = \mathbb{Z} / \mathbb{3Z}$ $ F = \mathbb{Z}/ \mathbb{3Z}$ or  $ F = \mathbb{Z}/  \mathbb{2Z}$ $ F = \mathbb{Z}/  \mathbb{2Z}$ or characteristic of $F$ is 3. Characteristic of $F$ is $3$   For ring homomorphism, $f(x+y)= f(x)+ f(y)$ and $f(xy)= f(x) f(y)$ for $x,y \in F$. So, Charcteristic of $F$ must be 3 so that $ 3x^2y + 3xy^2= 0$, so $4$ option should be true but how to discard option 1?
112741
If $V_1$ and $V_2$ are vector spaces over the field of rational numbers and $f \colon V_1 \to V_2$ is a map such that $f(x+y)=f(x)+f(y)$ for all $x, y \in V_1$, show that $f$ is a linear transformation.   It is easy to show that $f(0)=0$ and $f(-x)=-x$. Here scalars are from rationals. If scalar is an integer then clearly $f(\alpha x)=\alpha f(x)$. Hence $f$ is a linear transformation.  Now I have to show that when $\alpha$ is not an integer then $f(\alpha x)=\alpha f(x).$
87223
Assume that $F$ is a field and $\operatorname{char}(F)=p$. Let $a$ be an element in $F$ without $p$th root, then the polynomial $$x^{p^n}-a$$ is irreducible and inseparable over $F$ for all $n$.    I have proved the inseparable part by considering the derivative of the polynomial, but I'm having trouble with the irreducible part. Any help?
42079
Let $\phi(x) = x^p$ be the Frobenius automorphism on $\mathbb F_{p^n}$. We can view $\mathbb F_{p^n}$ as an $n$-dimensional vector space over $\mathbb F_p$. In this case, $\phi$ is a linear transformation. What are its characteristic and minimal polynomials? What are its Jordan and rational canonical forms? I know that $\phi$ satisfies $x^n - 1$, so its minimal and characteristic polynomials must divide this one. But, I can't seem to make much progress beyond this. Any suggestions?
47249
Let $f:F_n\to F_m$ be an epimorphism ($n\geq m$). Then it is true that there is a basis $X=X_1\sqcup X_2$ in $F_n$ such that $f$ maps $\langle X_1\rangle$ isomorphically onto $F_m$, and maps $X_2$ to identity. One can probably deduce this statement from the Grushko-Neumann theorem, but somehow I cannot get an elegant proof of it.   I can see that by the Grushko-Neumann we can assume that $m=1$, and hence our epimorphism factors through the abelianization. And then an ugly linear algebra appears... Any suggestions how to make the proof slick?
55318
Let $T$ be a linear transformation of an $n$-dimensional vector space $V$ over a field $k$. It's pretty easy to define the minimum polynomial of $T$ and make sure its degree is between $1$ and $n^2$, inclusive.     Observe $ I = \{ p(x) \in k[x] : p(T) =0\}$ is an ideal in $k[x]$. Indeed, $I$ is the kernel of the evaluation homomorphism $\mathrm{eval}_T: k[x] \to \mathrm{End}(V)$.  Notice also that:         $\mathrm{eval}_T$ is unital homomorphism, so $I$ is a proper ideal.    The $n^2 + 1$ transformations $I, T,T^2,T^3,\ldots, T^{n^2}$ must be linearly dependent, since $\mathrm{dim}(\mathrm{End}(V)) = n^2$, so there exist scalars $a_0,\ldots,a_{n^2}$, not all zero, such that $a_0I + a_1 T + \ldots + a_{n^2}T^{n^2} = 0$, whence the nonzero polynomial $p(x) = a_0 + a_1 x + \ldots + a_{n^2}x^{n^2}$ belongs to $I$.           Since $k[x]$ is a p.i.d., we may define the minimum polynomial $m(x)$ of $T$ to be the monic generator of the ideal $I$. By the preceding two observations, we have $1 \leq \mathrm{deg}(m(x)) \leq n^2$.    Now, of course, we know that the degree of $m(x)$ actually satisfies $1 \leq \mathrm{deg}(m(x)) \leq n$. One way to see this is to use the Cayley-Hamilton theorem which shows that the characteristic polynomial $c(x)=\det(xI - T)$, whose degree is $n$, annihilates $T$, whence $m(x)$ divides $c(x)$.     Question: Is there another way to see that $T$ is annihilated by a polynomial of degree $\leq n$ which does not require use of the characteristic polynomial?
106058
I have a homework question that plays off my las question about the characteristic function $\chi_A(x)$.  It is:  Prove that there is a function $f$ that gives a one-to-one correspondence between $\mathcal{P}(X)$ and $2^X$.  Here the power set $\mathcal{P}(X)=\{A|A\subseteq{X}\}$ and $2^X$ is the set of all functions that map elements into the set $\{0,1\}$ Now i know that $|\mathcal{P}(X)|=2^{|X|}$, but the hint in the example mentions this:  "Now let us define a function $f$ on $\mathcal{P}(X)$ into $2^X$ by taking as the image a subset $A$ of $X$ the characteristic function of $A$," so then this mapping looks like this  $$f:\mathcal{P}(X)\rightarrow{2^X}$$ $$f(A)=\chi_A{x}$$  From here I must show injectivity and surjectivity.  So $$f(A)=f(B)\Rightarrow{A=B}$$  So by the definition of the characteristic function we can show that injectivity holds since if $$x\in{A}, \chi_A(x)=1=\chi_B(x) \text{ if } x\in{B}$$ and $$x\in{X-A}, \chi_A(x)=0=\chi_B(x) \text{ if } x\in{X-B}$$ But how do we create a surjectivity argument?
117739
Let $A$ be an $n\times n$ diagonal matrix with characteristic polynomial $(x-c_1)^{d_1}\cdots(x-c_k)^{d_k}$, where $c_1,\ldots,c_k$ are distinct. Let $V$ be the space of $n\times n$ matrices $B$ such that $AB=BA$. Prove that the dimension of $V$ is $d_1^2+\cdots+d_k^2$.
48957
Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$. My work $\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$ Q1) Is it right ? Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$. Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ? The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$. Q4) How to conclude that $G_f$ act transitively on the roots ? I don't have particular problem for the converse.
65879
Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $\textrm{Hom}_K(L,M)$ denote the set of all $K$ - algebra homomorphisms from $L$ to $M$. Since $L/K$ is separable we know that the number of elements in $\textrm{Hom}_K(L,M)$ is equal to $[L:K]$. Now I want to prove that we have an isomorphism of $M$ - algebras   $$\varphi : M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$$  Proof that they are isomorphic as $K$ - modules : Write $L = K(a)$ for some $ a\in L$ (we can do this via the primitive element theorem). Then  $$\begin{eqnarray*} M \otimes_K L &amp;=&amp; M \otimes_K K[a] \\ &amp;\cong&amp; M\otimes_K K[x]/(f) \hspace{3mm} \text{where $f$ is the minimal polynomial of $a$ over $K$} \\ &amp;\cong&amp; M[x]/(f)\\ &amp;\cong&amp; M[x]/(f_1\ldots f_{[L:K]}) \hspace{3mm} \text{where the $f_i$ are the distinct}\\ &amp;&amp; \hspace{1.5in} \text{irreducible factors of $f$ since $M/L$ is Galois} \\ &amp;\cong&amp; M^{[L:K]}\end{eqnarray*}$$  where  the last step was using the Chinese remainder theorem. For the third last step, we consider the ses   $$ 0 \to (f) \to K[x] \to K[x]/(f) \to 0$$  and tensor with the exact functor $-\otimes_K M$ to get $$0 \to (f) \otimes_K M \to K[x] \otimes_K M \to K[x]/(f) \otimes_K M \to 0$$  and so $$\begin{eqnarray*} M \otimes_K K[x]/(f) &amp;\cong&amp; K[x]/(f) \otimes_{K} M \\ &amp;\cong&amp; \frac{K[x] \otimes_{K} M}{f \otimes_K M}\\ &amp; \cong&amp; M[x]/(f) \end{eqnarray*}$$ where $(f)$ is now viewed as an ideal of $M[x]$.          My question is: The tensor product $M \otimes_K L$ is a left $M$ - module, but why is it also an $M$ - algebra? Also why are the isomorphisms above isomorphisms of $M$ - algebras and not just $K$ - modules?
102540
Let $M$ be a finite extension of $E$ and let $E$ be a finite extension of $F$. Then $M$ is a finite extension of $F$ and $[M:F] = [M:E][E:F]$.   Is there an easy explanation and/or proof for this theorem? My instructor gave an incomplete proof in class but got stuck and I'm kind of confused of what's going on.
22188
Let $K=F(x)$ be the rational function field over a field $F$ of characteristic $0$, let $L_1=F(x^2)$, and $L_2=F(x^2+x)$. How to show that $L_1\cap L_2 = F$?
118984
If I understand correctly, Arturo Magidin says in  that the following is true.     If $E\subset F$ is a finite field extension, then $|\operatorname{Aut}(F/E)|\leq (F:E).$   I understand why this is true when the extension is simple, but the general case eludes me. (I also understand that my particular extension in that question is simple, so this is irrelevant to the problem, but I would like to understand this.)  Let $E\subset F$ be a finite extension. Any finite extension is finitely generated, so let $$F=E(a_1,\ldots,a_n).$$ Let $$E=F_0\subset F_1\subset\ldots\subset F_n=F,$$ with $F_i=F_{i-1}(a_i)$ for $i=1,\ldots,n.$ Let $f_i$ be the minimal polynomial of $a_i$ over $F_{i-1}$, and let $g_i$ be the minimal polynomial of $a_i$ over $E$. Then $$(F:E)=(F_n:F_{n-1})\cdots (F_1:F_0)=\deg(f_n)\cdots\deg(f_1).$$  I think I can't use $f_i$ to bound the order of $\operatorname{Aut}(F/E).$ Surely, $$|\operatorname{Aut}(F/E)|\leq \deg (g_n)\cdots\deg (g_1),$$ because the generators must be mapped to roots of their minimal polynomials over $E$ (taking non-minimal polnomials here would only make things worse), but this doesn't work because the inequality we have is $$\deg (f_i)\leq\deg (g_i),$$ not the other way around.
79826
My class notes has as theorem (without proof): "Let $K/F$ be finite field extension, with $K=F(\alpha_1,\ldots,\alpha_n)$ and $\alpha_k$ is separable for all $k$. Then $K/F$ is separable".  My question is: in either finite field, or field characteristic $0$, will the $\alpha_k$ always be separable? (by "$\alpha_k$ separable" I mean that its minimal polynomial is separable). Also, could somebody give me an example for which an $\alpha_k$ is not separable?