Let $E$ be a field and $F \subseteq E$ a subfield of $E$ and let $p(x), q(x) \in  F[x]$ two polynomials with coefficients in $F$. Show that if $p(x)$ divides $q(x)$ on $E$ $(= \exists h(x) \in E[x]$ t.q. $q(x) = p(x) · h(x)$), then $p(x)$ divides $q(x)$ on $F$ $(= \exists h(x) \in F[x]$ t.q. $q(x) = p(x)·h(x)$). I tried to do the following : by an euclidian division on $F[x]$ $$q(x) =p(x) ·b(x) +r(x) $$ and by the given supposition : $$q(x) = p(x)·h(x)$$ Since, $p(x) ·b(x) +r(x) \in E[x]$, $$p(x)·h(x) = p(x) ·b(x) +r(x)$$ $$\iff p(x) [b(x)-h(x)]+r(x)=0$$ Then I don't know how to show $r(x) = 0$ to finish the proof. I thought of saying that since the remainder and the quotient are unique then $r(x)=0$ and $b(x)=h(x)$ but I feel there is something I am doing wrong... Maybe, I could say that $r(x) = p(x)[h(x)-b(x)]$. If $r\ne0$, then $h-b\ne0$, and so $\deg(p(h-b))\ge \deg(p) &gt; \deg(r)$. Thus, $r(x)=0$.
70708
I would like to verify the validity of the following line of thought:  Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).  Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.  Is there any other simpler argument than that?  Added:  Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge &lt;l$. Then $x^p-a=(m_b(x))^q (x-b)^r$. Can i conclude that $(x-b)^r \in K[x]$ and so $r=0$ by the minimality of $l$?  Thanks.
22758
Let $F=\mathbb{F}_p(X,Y)$ be the field of rational functions in variables $X,Y$ over the finite field of $p$ elements. Let $K=\mathbb{F}_p(X^p,Y^p)$ be a subfield. Note that for any $f\in F$, $f^p\in K$. Deduce from this that $F/K$ is not a simple extension.
31274
This is from Pinter, A Book of Abstract Algebra, p.265.   Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.  This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.
101390
I want to solve:  Show that, in $\mathbb Q[x]$ (rationals), the polynomial $(x-1)^2$ divides a $zx^{z+1} – (z+1)x^{z}+1$, where $z \geq 1$ is any integer  I am really lost with this one, don't know how to aproach it
14967
I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks          Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.      Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.  This is where I got stuck. Hope some one can help me solve this. Thanks.
32234
In the integers, it follows almost immediately from the division theorem and the fact that $a | x,y \implies a | ux + vy$ for any $u, v \in \mathbb{Z}$ that the least common multiple of $a$ and $b$ divides any other common multiple.  In contrast, proving $e|a,b \implies e|gcd(a,b)$ seems to be more difficult. In Elementary Number Theory by Jones &amp; Jones, they do not try to prove this fact until establishing Bezout's identity.  has a proof without Bezout's identity, but it is convoluted to my eyes.  I tried my hand at it, and what I got seems no cleaner:  Proposition: If $e | a,b$, then $e | gcd(a,b)$.  Proof: Let $d = gcd(a,b)$. Then if $e \nmid d$, by the division theorem there's some $q$ and $c$ such that $d = qe + c$ with $0 &lt; c &lt; r$.  We have $a = k_1 d$ and $b = k_2 d$, so by substituting we obtain $a = k_1 (qe + c)$ and $b = k_2 (qe + c)$. Since $e$ divides both $a$ and $b$, it must divide both $k_1 c$ and $k_2 c$ as well. This implies that both $k_1 c$ and $k_2 c$ are common multiples of $c$ and $r$.  Now let $l = lcm(r, c)$. $l$ divides both $k_1 c$ and $k_2 c$. Since $l = \phi c$ for some $\phi$, we have $\phi | k_1, k_2$, so $d \phi | a, b$.  But we must have $\phi &gt; 1$ otherwise $l = c$, implying $r | c$, which could not be the case since $c &lt; r$. So $d \phi$ is a common divisor greater than $d$, which is a contradiction. $\Box$  Question: Is there a cleaner proof I'm missing, or is this seemingly elementary proposition just not very easy to prove without using Bezout's identity?
109312
Let $p$ be a prime. Prove that $\mathbb{F}_p[X]/(X^2+X+1)$ is a field iff $p \equiv 2\bmod3$.    So:   If $p \equiv 2 $ mod $ 3$, I have to show that every element of $\mathbb{F}_p[X]/(X^2+X+1)$ has an inverse.  If $\mathbb{F}_p[X]/(X^2+X+1)$ is a field I have to show $p \equiv 2 $ mod $ 3$.
29348
I don't see how to solve the following problem:     Let $R$ be a commutative and unitary ring. If there exists a monic polynomial $f(x) \in R[x]$ so that $R[x]/(f(x))$ is a field, show that $R$ is a field. What about if $f(x)$ isn't monic?   I don't have a clue about how proving it. I see that if $R$ is a field and $f(x)$ a monic and irreducible polynomial in $R[X]$ then $\frac{R[x]}{(f(x))}$ is a field, but that is not the question!  Thank you in advance for your help
6121
Suppose $K$ is a finite field, $K = \mathbb F_{p^s}$. If we take an irreducible polynomial $f$ of degree $d$ over $K$, then the splitting field $L$ of $f$ is $K(\alpha)$ where $f$ is the minimal polynomial of $\alpha$. But then $L = \mathbb F_{p^{sd}} $. Since $\mathbb F_{p^{sd}}$ is unique, we see that this is the splitting field of every irreducible polynomial of degree $d$ over $K$.   Take $K = \mathbb F_2$ and let $P(X) = X^3 + X + 1$, $Q(X) = X^3 + X^2 + 1$. Let $L$ be the splitting field of $P$ and $L&#39;$ be the splitting field of $Q$. The above tells us that $L$ and $L&#39;$ are isomorphic. I would like to construct an explicit isomorphism between $L$ and $L&#39;$.   I know that $L \cong \mathbb F_2[X] /(X^3 +X + 1)$ and $L&#39; \cong \mathbb F_2[X] / (X^3 + X^2 + 1)$. Intuitively, I want to find an isomorphism $\phi : \mathbb F_2[X] \to \mathbb F_2[X]$ such that $\phi((X^3 + X + 1)) = (X^3 + X^2 + 1)$. A little playing around gives me $\phi(X) = X+1$. It now feels like I'm falling at the last hurdle: how do I finish the construction of an isomorphism between $L$ and $L&#39;$? I don't think $\phi$ makes sense as a map from $L$ to $L&#39;$, yet it seems the map I want.
73433
Suppose $z_1, .....z_{m+1}$ are distinct elements of $F$ and $w_1,....,w_{m+1} \in F$. Prove that there exists a unique polynomial $p \in P_m(F)$ such that $p(z_j) = w_j$ for j=1,...m+1.  Any ideas on where to start/go with this one. I'm having quite some trouble with it.
91629
Suppose $K/F$ is an extension field, and $E_1$ and $E_2$ are two intermediate subfields of finite degree. Let $E$ be the subfields of $K$ generated by $E_1$ and $E_2$. I'm trying to prove that $$ [E:F]\leq[E_1:F][E_2:F].$$  Since $E_1$ and $E_2$ are finite extensions, I suppose they have bases $\{a_1,\dots,a_n\}$ and $\{b_1,\dots,b_m\}$, respectively. If $E_1=F$ or $E_2=F$, then the inequality is actually equality, so I suppose both are proper extension fields. I think $E=F(a_1,\dots,a_n,b_1,\dots,b_m)$. Since $\{a_1,\dots,a_n,b_1,\dots,b_m\}$ is a spanning set for $E$ over $F$,  $$[E:F]\leq n+m\leq nm=[E_1:F][E_2:F]$$ since $m,n&gt;1$.  Is this sufficient? I'm weirded out since the problem did not ask to show $[E:F]\leq [E_1:F]+[E_2:F]$ which I feel will generally be a better upper bound.
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.
16179
Is there a division algorithm for polynomials in R[x], where R is a commutative ring with unity?    All the algebra books I read mention division algorithm for polynomials in F[x], where F is a field. Since the leading coefficient of a non zero polynomial in R[x] is not necessarily invertible, I find it difficult to proceed the same way we do for polynomials in F[x].  Can I get some help?
54881
I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $  So far I have that:  $\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$  $\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$  $|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$   So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$  Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$.   So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$  Given that we consider the same polynomial each time, relabel to consider:   $f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$  We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$  $\alpha^2 \in K \Leftrightarrow a = 0_K$  I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$.   I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$  Any help would be greatly appreciated.
89703
Question is :     Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]   What I have done so far is :  Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :  $p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$  i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.  so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$  I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(  Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$   i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$   $p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$ $p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$  $p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$    Once I prove that those are the only divisors then I would consider :    $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$ $p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$ $p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$ $p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$ $p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$ $p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$   So, no divisor of $2$ is a root..  Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$  suppose I have something like :  $$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$  But then, $bd=2\Rightarrow \text { b,d  are  a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$  But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$  So, only possibilities are   $x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$ $x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$   But these are not possible...  This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...  So, please help me to clear this..  Thank you...   P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain.
74901
I'm trying to show that $f(x)=1+x^p+x^{2p}+\dots +x^{p(p-1)}$ is irreducible over $\mathbb{Q}[X]$. I'm well aware that cyclotomic polynomials are irreducible, however the (many) proofs of this statement are quite involved. Since this exercise appears in a textbook that only covers basic algebra, there should be a more direct approach to this exercise.  I tried all criterions that I know to solve this (Eisenstein, linear substitutions, reducing to $\mathbb{Z}_p[X],\dots$). Only one remains. We know that $f(x)=\frac{1-x^{p^2}}{1-x^p}$, or equivalently $1-x^{p^2}=(1-x^p)f(x)$. We know that the roots of $1-x^{p^2}$ are given by $$e^{\frac{2\pi i k}{p^2}}, \mbox{ for $0\leq k\leq p^2-1$ }. $$ Similarly, the roots of $1-x^{p}$ are given by $$e^{\frac{2\pi i kp}{p^2}}, \mbox{ for $0\leq k \leq p-1$}.$$  Hence the roots of $f(x)$ are given by $e^{\frac{2\pi i k}{p^2}}$ where $p\not|  \;k$ and $0\leq k\leq p^2-1$. One way to show that $f(x)$ is irreducible over $\mathbb{Q}$ is to show that any product of linear factors of $f(x)$ is not a polynomial over $\mathbb{Q}$. I tried to show this, but it is too difficult.  Does anyone know a elementary way of solving this exercise?
118198
I want to prove that $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})=\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ for $p_{1},p_{2}$ primes. I know this was proved before for the generalized case of $p_{1},....,p_{n}$ primes here:    And every proof of this kind of equality involves Galois theory but I still don't know Galois theory and I'm supposed to prove this with only basic  Field theory(field extensions, irreducible polynomials, algebraic extensions, etc).  By definition $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt p_{1}+\sqrt p_{2}$, also $\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ is the smallest field containing $\sqrt p_{1}$ and $\sqrt p_{2}$. Or  $$\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}})=\{a+b\sqrt{p_{1}}+c\sqrt{p_{2}}+d\sqrt{p_{1}p_{2}} \mid a,b,c,d\in\mathbb{Q}\}$$   $$\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}) = \lbrace a+b(\sqrt{p_{1}}+\sqrt{p_{2}}) \mid a,b \in \mathbb{Q} \rbrace  $$  Also I know that $[\mathbb{Q}(\sqrt p_{1},\sqrt p_{2}):\mathbb{Q}]=4$ from this:    Still don't know how to proceed proving this two extensions are the same.