I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.  Now I've managed to prove this using the primary decomposition theorem and then restricting $T$ to $ker({f_i}^{m_i})$ and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of $f_i$ gives us the result.  However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general?  We know that $m$ divides $\chi$  and so certainly $\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g$ but then how do we show that $g$ must have only $f_i$ as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on $k$, i.e. if $k$ is algebraically closed then it is easy because the polynomials split completely into linear factors.  Help is much appreciated, Thanks
121038
Let $V$ be a vector space over a field $F$ (not necessarily algebraically closed), and $T:V\rightarrow W$ be a $F$-linear map. If $c(T)$ and $m(T)$ denote the characteristic and minimal polynomial respectively, then I want to show that $c(T)$ and $m(T)$ have the same set of irreducible factors.  I know that the claim is true if $F$ is algebraically closed, because then the two polynomials have the same roots. The typical reasoning one gives when $F$ is not algebraically closed is: Take an algebraic closure of $F$, say $K$, and look at $m(T)$ and $c(T)$ in $K$ to conclude the proof. But as far as I know, the proof that any root of $c(T)$ is a root of $m(T)$ requires the fact that $V$ is a $K$-vector space. But we only know that $V$ is an $F$-vector space. I can't understand how to proceed after this.
77261
I'm working on the following problem:     Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then           the minimal polynomial of $A$ can only be of degree $2$.     the minimal polynomial of $A$ can only be of degree $3$.     either $A = I$ or $ A = -I$.     there are uncountably many such $A$.        By taking $A=I $ we can eliminate options (1) &amp; (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take   $$A = \begin{bmatrix}      1 &amp; 0 &amp; 0 \\      0 &amp; -1 &amp; 0 \\      0 &amp; 0 &amp; -1 \end{bmatrix}$$  Then $A^8=I $ but $A$ is neither $I$ nor $-I$. So option (3) is eliminated. Now I don't know how to proceed about with option  (4). Can someone help me please? Thanks in advance.
119646
Let $\Phi_n(x) \in \mathbb{Z}[x]$ denote the $n$-th cyclotomic polynomial, and let $\mathbb{F}_q$ be the finite field with $p^k = q$ elements ($p$ prime). Let $\Phi'_n(x)$ be the reduction of $\Phi_n(x)$ mod $p$ (i.e., $\Phi'_n(x)$ is the image of $\Phi_n(x)$ in $\mathbb{F}_q[x]$). Is it true that $\Phi'_n(x)$ is the $n$-th cyclotomic polynomial in $\mathbb{F}_q$; that is, are the roots of $\Phi_n'(x)$ precisely the primitive $n$-th roots of unity in $\mathbb{F}_q$? If so, I want to prove this is the case, but I'm not sure how I'd go about doing this.
106784
Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.  To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.  But there is a hint to this question that uses different approach: Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.  My questions regarding the hints: 1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible? 2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?  Thank you for any explanations!    Edit: Thanks for all the answers.  I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is: Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.  Is it inevitable to use Second Gauss Lemma? Is there any other way around?
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 $g(x)$.
77259
Let $A$ and B be $n\times n$ matrices over a field $F$. Then prove or disprove  $(i)\, AB$ and $BA$ have same characteristic values.  $(ii)\, AB$ and $BA $ have same characteristic polynomial.  $(iii)\, AB$ and $BA $ have same minimal polynomial.  Proof of $(i)$   Let $c$ be an eigen value of AB and v be corresponding eigen vector. Then    AB(v)=cv,    B(ABv)=B(cv),    BA(Bv)=c(Bv),   this implies $Bv$ is an eigenvector of $BA$ corr. to eigen value $c$.  But this statement holds if $Bv$ is nonzero.  What if $Bv=0$?  $(ii)$ If $A$ is invertible then ($A$ inverse)$(AB)A= BA$, i.e. $AB $ and $BA$ are similar and hence have same charateristic polynomial.   Similar proof when $B$ is invertible.  But what if both $A$ and $B$ are not invertible?  I have no idea about $(iii)$.  Please help me to complete these proofs.
102650
Let $T$ be a linear operator on a finite dimensional space $V$, and let $p=p_{1}^{r_{1}} \cdots p_{k}^{r_{k}} $ be the minimal polynomial for $T$, and let $V= W_{1} \oplus\cdots\oplus W_{k}$ be the primary decomposition for $T$, i.e., $W_{j}$ is the null space of $p_{j}(T)^{r_{j}}$. Let $W$ be any subspace of $V$ which is invariant under $T$. Prove that  $W= (W \cap W_{1})\oplus (W \cap W_{2})\oplus \cdots \oplus (W \cap W_{k})$.
99442
Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $\operatorname{rank} A=2$ and $A^3=A^2\neq 0$  Suppose $A$ is not diagonalizable.  Then which one is true   $A^2=A\neq 0$ $\exists$ a vector $v$ such that $Av\neq 0$ but $A^2v= 0$ the characteristic polynomial of $A$ is $x^4-x^3$   My try:  If $A^2=A$ then minimal polynomial of $A$ has two distinct roots and hence $A$ is diagonalisable hence false. Since $A^3=A^2\neq 0$ so the characteristic roots of $A$ are $0,0,1$ Since rank of $A$ is $2$ $1$ must appear twice so the characteristic polynomial of $A=x^4-(0+0+1+1)x^3+(0.0.1+0.1.1+1.1.0+1.0.1)x^2+(0.0+0.1+1.1+...)x+(0.0.1.1)$ hence 3 is false  2 I am not sure Please find the error in my solution and don't refer it to some previous solution
41753
Let $K$ be any field. Is it true that:   If $f_1,\dots, f_k \in K[x_1,\dots,x_n]$ are polynomials of degree $1$, then $I=(f_1,\dots, f_k)$ is all $K[x_1,\dots,x_n]$, or, if it is a proper ideal, then is it prime? In particular, is it true that $K[x_1,\dots,x_n]/I$ is isomorphic to a polynomial ring (in the case that $I$ is proper and hence prime by the statement above)?   If they are true, how can i prove these statements in an elementary but rigorous way?  P.S.: Birth of my question: I found on my book that "affine varieties given by linear equations are irreducibile", so I translated (I don't know if I have made it right) this geometric statement into the analogous algebraic version.
63235
Let $y=\frac{x^4+1}{x^2} \in F(x)$. Then $g(x)=0$ for the polynomial $g(s) = (s^4+1)-ys^2$. How to show that it is the minimal polynomial over the field $F(y)$?
67146
Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.   This  to a related question mentions the result is due to Capelli.  I can prove the result if $F$ has characteristic $p$ as follows.  Suppose $f$ is reducible: $f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m&lt;p$. Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have $$f(X)=X^p-\alpha^p=(X-\alpha)^p$$ so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$.  So $\alpha\in F$ because $m$ is invertible modulo $p$.  How would you show the result in other characteristics?
91066
Suppose $L: V \rightarrow V$ self-adjoint (on a finite-dimensional space $V$); $b$ is a real number. Show   The minimal polynomial has distinct roots; $\ker(L) = \ker(L^k)$ for $k\geq1$; $\operatorname{im}(L) = \operatorname{im}(L^k)$ for $k\ge 1$; $\ker(L-bI) = \ker((L-bI)^k)$ for $k\geq1$;     My thoughts:  1) I think I'm supposed to assume root is repeated and reach a contradiction but I'm not sure how to proceed mathematically with that.  2-4) pretty sure there's some huge theorem that I'm missing.
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?
55256
A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?  It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.
40010
Let $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$ denote the field of 2 elements   (a) Is $x^4+x^2+1$ irreducible in $\mathbb{F}_2[x]$? Find a complete factorization.  (b) How many irreducible polynomials of degree 4 are there in $\mathbb{F}_2[x]$?  For the first part of the problem, I just plug in $0$ and $1$ into $x^4+x^2+1$, and know that there is no root for that polynomial, so this polynomial is irreducible.   For the second part, I just list polynomials of degree 4 and having no zero in $\mathbb{F}_2[x]$, which are $x^4 + x+1, x^4 +x^2 +1, x^4 + x^3 +1$ , and $x^4 + x^3 +x^2 + x +1$.   But I am not sure about whether these approaches are right, so could you help me to figure it out? Thank you!
21358
I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question.  True or False:   let $A,B \in \mathbb{Q}^{n \times n }$  Suppose $xI-A$ and $xI-B$ are equivalent then $\det(xI-A) = \det(xI-B)$.     I wrote in my notes this is true because any two matrices are equivalent in $\mathbb{Q[x]}^{n \times n}$ if and only if they have the same invariant factors.  Since the characteristic polynomial is a product of the invariant factors it follows that $\det(xI-A) = \det(xI-B)$.        True or False:   let $A,B \in \mathbb{Q}^{n \times n }$. If $\det(xI-A) = \det(xI-B)$ (in $\mathbb{Q}[x]$) then $xI-A$ and $xI-B$ are equivalent  in $\mathbb{Q}[x]^{n \times n}$   I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example.
2836
I have started working through the textbook Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea and I am stuck on one part of an introductory question.  The question begins by getting one to show   (1) If $f(t)$ and $g(t)$ are polynomials of degree $\le n$ in $t$, then for $m$ large enough, the monomials  $$[f(t)]^e[g(t)]^f$$ with $e+f\le m$ form a linearly dependent set in $k[t]$.  The next part of the question asks one to deduce      (2) If $C\,:\,x=f(t),\,y=g(t)$ is any polynomial parametic curve in $k^2$, then $C$ is contained in $\textbf{V}(F)$ for some $F\in k[x,y]$.   I have demonstrated (1), but I am at a loss as to how to use that fact to prove (2). It doesn't seem like it is supposed to be a hard problem, but never the less I am having trouble wrapping my head around it. Additionally I assume that for (2) we are not interested in the trivial solution $F=0$.  [Added for completeness]  To do part (1), we recognize that if $\deg(f),\deg(g)\le n$, then $$\deg\left([f(t)]^e[g(t)]^f\right)\le ne+nf\le nm$$ Hence if we have more than $nm+1$ polynomials of the form $[f(t)]^e[g(t)]^f$ then we have a dependent set (since $d+1$ is the dimension of the vector space of polynomials of degree less than or equal to $d$). Now we simply count the number of polynomials in this form (this was an exercise before the problem) - it is $\binom{m+2}{2}=\frac{1}{2}(m+1)(m+2)$ from a standard application of . Ensuring $m$ large enough (specifically $m&gt;2n-3$) gives us the conclusion.