Where I've got to so far:   Let $m(X)$ ε I be a non-zero polynomial of least degree.  For any $a(X)$ ε $\mathbb{F}$$[X]$, $a(X)m(X)$ ε I as I is a principal ideal of $\mathbb{F}$$[X]$ thus $&lt;m(x)&gt;$ $⊆$ I.  Let us now consider $b(X)$ ε I. By the division theorem for polynomials, $b(X) =q(x)m(x) + r(X)$ where $q(X), r(X)$ ε $\mathbb{F}$$[X]$ and $r(X) = 0$ or $deg(r(X)) &lt; deg(m(X))$.  Observe that $r(X) = b(X) - q(X)m(X)$ ε I.  If $r(X) = 0$ then we are done: We have shown that $b(X) = q(X)m(X)$ which means $I ⊆ &lt;m(X)&gt;$ hence I $=$ $&lt;m(X)&gt;$.  If $deg(r(X)) &lt; deg(m(X))$ then observe that $m(X)$ was chosen to have the least degree in I thus $deg(r(X)) = 0$ which means $r(X)$ is a constant polynomial.  So where I am so far is $r(X)$ = $b(X) - q(X)m(X)$ where $r(X)$ is a constant polynomial. How do deduce $b(X) = q(X)m(X)$ in this situation?
122248
Let $F$ be a ﬁeld and $R = F[X]$. Suppose $I$ is an ideal of $R$. Show that $I = (p(X))$ for some $p(X)$ in $F[X]$. (Hint: consider a polynomial $p(X)$ of least degree in $I$.)  I'm trying to do this question but can't seem to start. if anyone could give me a hint that would be appreciated.  My instinct was to take a monic polynomial (which exists since we are in a field) of degree $1$ and divide given the first two parts of the question were about polynomial division.  Thanks.
24054
Trying to solve a problem, I got stuck on the following question.     If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $$\mathbb{R}[X]/(P)\cong\mathbb C?$$   I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than . It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.  Here's what I have tried.   First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have  $$P(x)=x^2+bx+c$$  for some real numbers $b,c$ such that $b^2&lt;4c$.   I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$  I have   $$ [x^2+bx+c-1]\mapsto -1, $$  So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $$(kx+l)^2-(x^2+bx+c-1)\in (P),$$  that is $$ (kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c) $$  for some real number $a$, and further  $$ (k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0. $$  This gives   $$ \begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases} $$  The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?
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)$?
100951
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
77699
Let $A$ be a commutative ring. I am trying to show that if $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor then there exists $a$ in $A-\{0\}$ such that $af=0$ in $A[x_1,x_2,\ldots, x_r]$.    What I have tried so far as following. I am using induction on $r$ (not fixing the ring). The base case is for $r=1$ which I did assuming the minimal degree of $g$ for which $fg=0$ holds. So my induction assumption is for $r \geq 2$ whenever $f$ is in $A[x_1,x_2,\ldots x_r]$, $n&lt;r$ is a zero divisor there is $a$ in $A-\{0\}$ such that $af=0$. So for the final step I take $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor. Now since $A[x_1,x_2,\ldots, x_r]=A[x_1][x_2,x_3,\ldots, x_r]$ by induction hypothesis there is $g$ in $A[x_1]-\{0\}$ such that $fg=0$.      I cannot proceed further. Am I correct so far? Please help me. Thank you.
38717
Let $E$ be an extension field of $F$. If $a \in E$ has a minimal polynomial of odd degree over $F$, show that $F(a)=F(a^2)$.  let $n$ be the degree of the minimal polynomial $p(x)$ of $a$ over $F$ and $k$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$.   Since $a^2 \in F(a)$, We have $F(a^2) \subset F(a)$, then $k\le n$  In order to prove the converse:  $q(a^2)=b_0+b_1a^2+b_2(a^2)^2\ldots+b_k(a^2)^k=0$   implies   $q(a)=b_0+b_1a^2+b_2a^4\ldots+b_ka^{2k}=0$ Then   $p(x)|q(x)$, because $p(x)$ is the minimal polynomial of $a$ over $F$.  If I prove that $n|2k$ we done, since $k$ is odd, we have $n|k$ and $n\le k$ and finally $n=k$.  So I almost finished the question I only need to know how to prove that $n|2k$ It should be only a detail, but I can't see, someone can help me please?  Thanks
64532
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.  I have to prove that:     $f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.   I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.
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.
68891
Let $F$ be a perfect field, i.e. every irreducible polynomial over $F$ has distinct roots in the algebraic closure of $F$. Suppose that $K$ is an algebraic extension of $F$ with the property that every non-constant $p(X) \in F[X]$ has a root in $K$.   I want to show that $K$ is algebraically closed, i.e. that $K = \overline{F}$, the algebraic closure of $F$.  If $K$ were a normal extension of $F$, this would follow immediately: Let $\alpha \in \overline{F}$ and $m(X)\in F[X]$ its minimal polynomial. By our assumption, $m(X)$ has at least one zero in $K$. Since $K$ is a normal extension of $F$, it thus contains all other roots of $m(X)$ including $\alpha$.  I don't have an idea as to how to show that $K$ is a normal extension.
10631
Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant.
53797
I want to show that if $K$ is a field then $K[x]$ is principal  Here is what I did:  I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:  $q(x) = p(x)a(x) + r(x)$ with $deg(r) &lt; deg(p)$  Now if $r= 0$ we obviously have $p \in (q)$   But I don't know how to show that it is principal if $r \neq 0$  Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that
104444
Question is to check :  which of the following is sufficient condition for a polynomial   $f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ to have a root in $[0,1]$.   $a_0 &lt;0$ and $a_0+a_1+a_2+\dots +a_n &gt;0$ $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$ $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$   First of all i tried by considering degree $1$ polynomial and then degree $2$ polynomial and then degree $3$ polnomial hoping to see some patern but could not make it out.  And then, I saw that $a_0= f(0)$ and $f(1)=a_0+a_1+a_2+\dots +a_n$.  So, if $f(0)&lt;0$ and $f(1)&gt;0$ it would be sufficient for $f$ to have root in $[0,1]$  In first case we have $a_0 &lt;0$ i.e., $f(0)&lt;0$ and $f(1)&gt;1&gt;0$.  So, first condition should be implying existence of a root in $[0,1]$  for second case, let $f(x)$ be a linear polynomial i.e., $f(x)=a_0+a_1x$  Now, $a_0+\frac{a_1}{2}=0$ implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}&lt; 1$ So, this might be possibly give existence in case of linear polynomials.  Now, $\frac{a_0}{1.2}+\frac{a_1}{2.3}=0$implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}&lt; 1$ So, this might be possibly give existence in case of linear polynomials.  So, for linear polynomials all the three conditions imply existence of a root in $[0,1]$.  But, i guess this can not be generalized for higher degree polynomial.  I think there should be some "neat idea" than checking for roots and all.  I am sure about first case but I have no idea how to consider the other two cases.   please provide some hints to proceed further.
105952
We have the ring $R$ and the polynomial $a=x^3+x+1$ in $R[x]$. I want to check the following statements:  If $R=\mathbb{R}$ then $a$ is irreducible in $\mathbb{R}[x]$. This statement is false, since according to Wolfram the polynomial has a real solution, right? But how can we calculate this root without Wolfram, by hand?  If $R=\mathbb{Z}_5$ then $a$ is irreducible in $R[x]$. The possible roots are $0,1,2,3,4$. We substitute these in $a$ and we get the following: \begin{align*}&amp;0^3+0+1=1\neq 0\pmod 5 \\ &amp;1^3+1+1=3\neq 0\pmod 5 \\ &amp;2^3+2+1=11\equiv 1\pmod 5\neq 0\pmod 5 \\ &amp;3^3+3+1=31\equiv 1\pmod 5\neq 0\pmod 5 \\ &amp;4^3+4+1=69\equiv 4\pmod 5\neq 0\pmod 5\end{align*} So since none of these elements is a root of $a$, the statement is correct, right?  If $R=\mathbb{C}$ then $a$ is irreducible in $R[x]$. This statement is wrong, because from the first statement we have that $a$ is reducible in $\mathbb{R}[x]$, and so it is also in $\mathbb{C}[x]$. Is this correct?  If $R=\mathbb{Q}$ then $a$ is not irreducible in $R[x]$. Since this is a cubic polynomial, it is reducible if and only if it has roots. By the rational root test, the only possible rational roots are $\pm 1$. Since neither of these is a root, it follows that $a$ is irreducible over $\mathbb{Q}$, right?  If $R=\mathbb{C}$ then $a$ has no root in $R$. This statement is wrong, since from statemenet 3 we have that $a$ is reducible in $\mathbb{C}[x]$ and so it has roots in $\mathbb{C}$, or not?
11021
I just need a check for my reasoning:  If I have a ring $R=\mathbb{F}_p[X]$ and elements $x,y\in R$ it is true $$(x+y)^p=x^p+y^p?$$ It is true since every other element of expansion has a multiple of $p$ as coefficient.  What about field $R=\mathbb{F}_p[X]/(f)$ where $f$ is polynomial of degree $n$. I think rule still holds for the same reason.   Also what can be said about $x^p$? Is there any general rule about that element? It seems to me that $x^p=x$ since $(\mathbb{F}_p,\cdot)$ has order $p-1$. Am I missing something?
111112
I've been asked to prove that if a polynomial $f\in \mathbb{Q}[x]$ has degree $n$, then the splitting field of $f$ has degree less than or equal to $n!$. That is, $[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb Q]\leq n!$, where $\alpha_i$ are the roots of $f$.   My approach to this question is to use the tower law, to arrive at the equation: $$[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}]=[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}(\alpha_1,...,\alpha_{n-1})][\mathbb{Q}(\alpha_1,...,\alpha_{n-1}):\mathbb{Q}(\alpha_1,...,\alpha_{n-2})]\cdots[\mathbb{Q}(\alpha_1):\mathbb{Q}]$$  Clearly $[\mathbb{Q}(\alpha_1):\mathbb{Q}]\leq n$, since the minimum polynomial of $\alpha_1$ divides $f$.  I claim that in general: $[\mathbb{Q}(\alpha_1,...,\alpha_i):\mathbb{Q}(\alpha_1,...,\alpha_{i-1})]\leq n-i, \forall i= 1,2,...,n$. My question is, whether this claim is indeed true, and if so how would I prove it. I've tried using induction but I can't seem to make it work.  Also, if anyone has a proof for the original question without using my claim, then can you please post that too.
40763
I am stuck with the following problem. Every polynomial of degree $d$ can be expressed as $$ p(x) = p_d \binom{x}{d}+ p_{d-1}\binom{x}{d-1} + \cdots + p_0 \binom{x}{0} $$     What is the representation of    $$ \binom{x}{i} \binom{x}{j}  $$   in such a basis?   I have no idea how to solve this...
77261
I'm working on the following problem:     Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then           the minimal polynomial of $A$ can only be of degree $2$.     the minimal polynomial of $A$ can only be of degree $3$.     either $A = I$ or $ A = -I$.     there are uncountably many such $A$.        By taking $A=I $ we can eliminate options (1) &amp; (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take   $$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.