Do you know of a neat proof of the algebraic independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$? IMPORTANT: $f$ and $g$ are algebraic independent if for any nonzero polynomial $P \in \mathbb{C}[x, y]$, the function $P(f(t),g(t))$ is not the zero function.
121842
Friends,  Do you know of a neat proof of the linear independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$?
82335
It's pretty well known that $\text{trdeg}(\mathbb{C}/\mathbb{Q})=\mathfrak{c}=|\mathbb{C}|$.  As a subset of $\mathbb{C}$, of course the degree cannot be any greater than $\mathfrak{c}$. I'm trying to understand the justification why it cannot be any smaller. The explanation in my book says that if $\mathbb{C}$ has an at most countable (i.e. finite or countable) transcendence basis $z_1,z_2,\dots$ over $\mathbb{Q}$, then $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$. Since a polynomial over $\mathbb{Q}$ can be identified as a finite sequence of rationals, it follows that $|\mathbb{C}|=|\mathbb{Q}|$, a contradiction.   I don't see why the polynomial part comes in? I'm know things like a countable unions/products of countable sets is countable, but could someone please explain in more detail this part about the polynomial approach? Since $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$, does that just mean that any complex number can be written as a polynomial in the $z_i$ with coefficients in $\mathbb{Q}$? For example, $$ \alpha=q_1z_1^3z_4z_6^5+q_2z_{11}+q_3z^{12}_{19}+\cdots+q_nz_6z_8z^4_{51}? $$  Is the point just that the set of all such polynomials are countable?  Thanks,
77265
There proof given in my textbook is as follows:  Let T be a diagonalisable linear operator and let $c_1, ..., c_k$ be the distinct eigenvalues of T. Then it is easy to see that the minimal polynomial for T is the polynomial  $$ p = (x-c_1)...(x-c_k)$$  If $\alpha$ is an eigenvector, then one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into 0. Therefore  $$ (T-c_1 I)...(T-c_kI)\alpha = 0$$  for every eigenvector $\alpha$. There is a basis for the underlying space which consists of eigenvectors of T and hence  $$ p(T) = (T -c_1I)...(T-c_kI) = 0$$  I cannot understand this proof at all. I understand that one of the operators $T-c_1I,...,T-c_kI$ sends $\alpha$ into 0, but I can't see why we we multiply them together, it still works. And I also don't understand why there is a basis for the underlying space. Can someone kindly explain to me please. Thanks
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!
55527
If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.   This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.  I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!
17420
Let $F$ be a field and $x$ transcendental over $F$. Let $y=f(x)/g(x)$ be a nonconstant rational function with relatively prime polynomials $f,g\in F[x]$. Prove $[F(x):F(y)]=\max(\deg f,\deg g)$.  My attempt: By replacing $y$ with $\frac{1}{y}$ if necessary we assume $\deg g\geq\deg f$. By the Euclid algorithm we assume $\deg g&gt;\deg f$. Then $\deg g=\max(\deg f,\deg g)$, let $n=\deg g$. Let $R(t)$ be this polynomial $$R(t)=yg(t)-f(t)=\frac{f(x)}{g(x)}g(t)-f(t)$$ Then $\deg R=\deg g=n$ because $n=\deg g&gt;\deg f$. The coefficients of $R(t)$ lie in $F(y)$ and hence $R(t)\in F(y)[t]$. Clearly $R(x)=0$.  Now I only have to prove $R(t)$ is actually the minimal polynomial of $x$ over $F(y)$. I believe the next step is to obtain a contradiction that $R(t)$ being reducible can lead to $f,g$ being not coprime, but I can't figure out how to do that.  P.S. A hint is preferred over a full answer.
22490
I was just wondering whether anyone knows an answer to the following:  Suppose that ${\mathbb F}$ is a subfield of a field ${\mathbb G}$ and that $v_1,\ldots ,v_k$ are linearly independent vectors in ${\mathbb F}^n$ (over $\mathbb F$). Is it necessarily true that $v_1,\ldots ,v_k$ are also linearly independent when considered as vectors in ${\mathbb G}^n$ (over $\mathbb G$})?  Any help would be much appreciated. Thanks!
87181
Could you help me to show that the functions $\sin(x),\sin(2x),...,\sin(mx)\in V$ are linearly independent, where $V$ is the space of real functions?  Thanks.
91399
let $f$ be a polynomial function such that $f(\mathbb{Q})\subset \mathbb{Q}$ and $f(\mathbb{R}-\mathbb{Q}) \subset \mathbb{R}- \mathbb{Q}$ . Prove that $f(x)=ax+b$ for some $a,b\in\mathbb{Q}$   My approach :  I started by defining $g(x)=f(x)-f(0)$ .   Now $g(x)$ satisfies the properties $g(\mathbb{Q})\subset \mathbb{Q}$ ,  $g(\mathbb{R}-\mathbb{Q})\subset\mathbb{R}-\mathbb{Q}$ and $g(0)=0$ . $g$ is a polynomial and $g(0)=0$ . $0$ is root of $g$ and hence $g(x)=x h(x)$ for some polynomial $h(x)$ . It will be enough if one can prove that $h(x)$ is constant . I'm stuck here .   Any form help will be highly appreciated.
8092
let $P \in \mathbb{R}[X]$ be a real polynomial divisible by a polynomial $Q \in \mathbb{R}[X]$ in $\mathbb{C}[X]$. How can I easily show that $P$ is also divisible by $Q$ in $\mathbb{R}[X]$?  A simple argument without using higher algebraic theorems is desirable. If I could use instruments of higher algebra, the exercise I have to do in whole would be done in two lines. But I'm not allowed to use. I think there would be an easy argument which I can't see yet because of my mental fogginess that I have sometimes.  Thank you beforehand.
93453
Hypothesis:   $f(x) = a_0 + a_1 x + \ldots + a_n x^n \in F[x]$ s.t. $f(x)$ irreducible over field $F$. $deg(f(x)) = n \ge 1$. Let $V = (f(x)) \subset F[x]$.   Goal: Show that $\mathcal{B} = \{1 + V, x + V, x^2 + V, \ldots , x^{n-1} + V\}$ forms a basis for $E = F[x]/V$  Attempt:   Linear Independence: If this set weren't linearly independent, then a non-trivial $b_i \in F$ would satisfy  $$ b_0 + b_1x + \ldots + b_{n-1} x^{n-1} + V = 0 + V $$   But this would then imply that $b_0 + b_1x + \ldots + b_{n-1} x^{n-1}  \in 0 + V$. But this would contradict $V = (f(x))$ since it is impossible that $f(x)\,\,|\,\,b_0 + b_1x + \ldots + b_{n-1} x^{n-1}$ of strictly lower degree.  Now how would I show that this set spans $F[x]/V$?
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.
118804
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
93148
Prove that: $\forall a,b\in\mathbb{C} \exists c,d \in \mathbb{C}: c+d = a \land cd = b$   We just learned about the characteristic/minimal polynomial and diagonalization but I am not sure if it has something to do with this question.  I'd be glad for guidance!
41353
If $v$ is algebraic over $K(u)$ for some $u\in F$, $F$ is an extension over $K$, and $v $ is transcendental over $K$, then $u$ is algebraic over $K(v)$.  I came across this problem in the book Algebra (Hungerford), I feel it shouldn't be too hard but I just can't work it out. Hope you can help me. Thank you!
93989
Let $K\subset \mathbb{C}$ be a subfield and $f\in K[t]$ an irreducible polynomial. Show that $f$ has no multiple roots in $\mathbb{C}$.   If I understand this question correctly, I must show that there is no $a \in \mathbb{C}$ such that $(t-a)^n|f$ in $F[t]$ with $n&gt;1$. So suppose $(t-a)^2|f$ and $f=(t-a)^2h$. Then we have $f'=2(t-a)h+(t-a)^2h' \Rightarrow (t-a)|f'$ so $\gcd(f,f'$) is not constant. Therefore $f$ is divisible by some square of non-constant polynomial in $F[t]$, which is a contradiction. Is my argument correct? Thank you.
120249
I need help proving this theorem:     Given the field extension: $\mathbf{K} \subseteq \mathbf{L}$, for $\alpha \in \mathbf{L}$ and $g(x) \in \mathbf{K}[x]$, $\alpha$'s minimal polynomial over $K$,   and $f(x) \in \mathbf{L}[x]$, $\alpha$'s minimal polynomial over $L$,   then the degree of $g$ is bigger than the degree of $f$ and $f(x)$ divides $g(x)$.