Consider the irreducible polynomial $g = X^4 + X + 1$ over $F_2$ and let $E$ be the extension of $F_2 =$ {0, 1} with a root $α$ of $g.$  Could $E$ have a subfield of order 8  When working through my work Ive come across this question which has stumped me and I cant get my head aound if this is true of not so any help will be appreciated.
123119
Consider the irreducible polynomial $g = X^4 + X + 1$ over $F_2$ and let $E$ be the extension of $F_2 =$ {0, 1} with a root $α$ of $g.$  Could $X^3 + X + 1$ have a root in $E?$  When working through this sheet Ive come across this question in which im not sure how to answer so any help will be appreciated
57204
Suppose $E/F$ is an algebraic extension, where every polynomial over $F$ has a root in $E$. It's not clear to me why $E$ is actually algebraically closed.  I attempted the following, but I don't think it's correct:   I let $f$ be an irreducible polynomial in $E[X]$. I let $\alpha$ be a root in some extension, so $f=m_{\alpha,E}$. Since $\alpha$ is algebraic over $E$, it is also algebraic over $F$, let $m_{\alpha,F}$ be it's minimal polynomial. I now let $K$ be a splitting field of $m_{\alpha,F}$, which is a finite extension since each root has finite degree over $F$.  If $m_{\alpha,F}$ is separable, then $K/F$ is also separable, so as a finite, separable extension, we can write $K=F(\beta)$ for some primitive element $\beta$. By assumption, $m_{\alpha,F}$ has a root in $E$, call it $r$. Then we can embed $F(\beta)$ into $r$ by mapping $\beta$ to $r$. It follows that $m_{\alpha,F}$ splits in $E$. Since $f\mid m_{\alpha,F}$, we must also have the $f$ is split in $E$.  But what happens if $m_{\alpha,F}$ is not separable? In such case, $F$ must have characteristic $p$. I know we can express $m_{\alpha,F}=g(X^{p^k})$ for some irreducible, separable polynomial $g(X)\in F[X]$. But I'm not sure what follows after that.  NB: I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$.
91474
I want to prove that $f = X^4 + \overline{2}$ is irreducible in $\mathbb{F}_{125}[X]$.  I know that $\mathbb{F}_{125}$ is the splitting field of $X^{125} - X$ over $\mathbb{F}_5$, and that this is a field extension of $\mathbb{F}_5$ of degree 3, and that $\mathbb{F}_{125} = \mathbb{F}_5[X]/(g)$ where $g = X^3 + X - \overline{1}$.  If we put $\alpha = X + (g)$ then $\{\overline{1}, \alpha, \alpha^2\}$ is a basis for $\mathbb{F}_{125}$ considered as a vector space over $\mathbb{F}_5$. Hence every element $x \in \mathbb{F}_{125}$ can be written $x = a_0 + a_1\alpha + a_2\alpha^2$, with $a_i \in \mathbb{F}_5$.  How do I proceed?
51163
I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it &lt;.&lt;  Let's consider a polynomial $f(x)= x^4+ax^2+b \in \mathbb{Z} $. Let $ \pm\alpha , \pm\beta$ denote the roots of $f(x)$. Note that $f$ is irreducible iff $\alpha^2, \alpha\pm\beta \notin \mathbb{Q}$. So let's suppose that $f(x)$ is irreducible. Let's denote $G$ the galois group of $f$ (the Galois group of the splitting field $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$)  $i)$ $G\cong V$ the Klein 4-group iff $\alpha\beta \in \mathbb{Q}$  $ii)$ $G\cong \mathbb{Z}_4$ iff $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)$  $iii)$ $G\cong D_8$ the dihedral group of order 8, iff $\alpha\beta \notin \mathbb{Q}(\alpha^2)$  I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that $u\in K/F$ a galois extension, the other roots of the minimal polynomial of $m_u(x)\in F[x]$ are precisely the elements $ \sigma(u)$ , where $\sigma \in Gal(K/F)$. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on $\mathbb{Q}(\alpha,\beta)$, so I have to know the values of both. But I have to respect the algebraic relations involving $\alpha , \beta$ I proved that in $i),ii)$ $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) $ So I have to know the values on $\alpha$.   My first question is: If in both cases $i),ii)$, I have an extension of degree 4 (so $|G|=4$) I have to map $\alpha$ to determine the automorphism, but I have also to map $\alpha$ to all of it's four roots. Why the galois group in this case are different?  My second question is about $iii)$ I have no idea How to attack it.
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
113379
I must find the irreducible factors of $f(x) = x^8 - x$ in $Z/2Z[x]$ and that's what I did:  $f(x) = x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$  but of course the correct solution is:  $f(x) = x(x-1)(x^3 + x^2 + 1)(x^3 + x + 1)$  I do not understand how I can reach the correct solution by logical reasoning (without going on attempts). The solution of the exercise says:  1) "The irreducible factors are all the irreducible polynomials of degree 1 and 3 alone" Why?  2) "You find exactly $6/3 = 2$ irreducible polynomials of degree 3". Does this happens because the 6th degree polynomial I found is not reducible (since the only ones have degree 1 or 3) and it has no zeros?
33881
Let $f(x) \in \mathbb Q[x]$ be irreducible of degree 4. Let $\alpha$ be a root of $f(x)$. Let $E = \mathbb Q(\alpha)$ and let $K$ be the splitting field of $f(x)$ over $\mathbb Q$. Prove that there is no field properly between $\mathbb Q$ and $E$ if and only if $G(K/\mathbb Q) \cong A_4$ or $S_4$.  At this point in the course we've just gone through the Fundamental Theorem of Galois Theory. I'm fairly lost, so any hints, sketches of a proof would help. Thanks!
41057
Find all primes $p&gt;2$ for which $x^2+x+1$ is irreducible in $\mathbb{F}_p[x]$   Attempt. Since $x^2+x+1$ is of degree 2, it is reducible iff it has a root in $\mathbb{F}_p$. It has a root in $\mathbb{F}_p$ iff $x^3-1=(x-1)(x^2+x+1)$ has a root other than 1. The latter has a root other than $1$ iff $\mathbb{F}_p$ has elements of order $3$, which happens iff $3\mid |\mathbb{F}_p^{\times}|=p-1$, which happens iff $p\equiv 1\bmod 3$. So $x^2+x+1$ is irreducible iff $p\not \equiv 1\bmod3 $.   I am getting a different answer than my professor, who has $p\equiv 1\bmod 3$. If I am incorrect, were am I going wrong?  Update. Now I am unsure about my own reasoning. What if $1$ is a root of $x^2+x+1$? I don't know why I am considering roots of $x^2+x+1$ other than $1$ when $1$ can be a root.
67525
Show that   $f_p (x^{p^{e-1}})$ is irreducible over $\mathbb{Q}$. I know it is obviously true for $ p=2$. Please suggest me any hints
73277
What algorithm can be used find an irreducible polynomial of degree $n$ over the field $GF(p)$ for prime $p$. The reason I ask is I want to make a program for finite field arithmetic, but creating a field $GF(p^k)$ requires finding a irreducible polynomial. (I'm mostly interested in $GF(2^k)$.)
28718
Let $A$ be a UFD, $K$ its field of fractions, and $f$ an element of $A[T]$ a monic polynomial.  I'm trying to prove that if $f$ has a root $\alpha \in K$, then in fact $\alpha \in A$.  I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.
90131
Would you please help me to prove this problem.  Let $G=\mathrm{Hol}(C_8)$. And let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $8$, $L$ a splitting field of $f(x)$ over $\mathbb{Q}$ and suppose that $\mathrm{Gal}(L/\mathbb{Q})\cong G$.     Show that there is a subgroup of $S_8$ isomorphic to $G$.   I am reviewing for my test and i find this good exercice    Thanks.
124213
Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?  ($\cong$ denotes an isomorphism that is the identity on restriction to $K$).  Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
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.
120366
this is a really easy question but I cant find an answer;  In need to see if $x^4+x^2+x+1$ is an irreducible polynomial over $\mathbb{Q}$
48704
This is problem 3.18 from Patrick Morandi's Field and Galois Theory.     Give an example of fields $k \subseteq K \subseteq L$ and $l \subseteq L$ for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has been asked twice, namely  and , though neither had any answers, and only some inconclusive comments. Normally I wouldn't ask the question again, but after struggling with it for weeks at this point, I emailed Morandi, and he gave me some hints.  He told me that he remembers finding an example, but that he isn't sure. He told me that it would need to be over characteristic $p$ and that the extensions (at least some of them I suppose) could not be separable. Moreover, he told me that $K = k(x)(a)$, where $a$ is a root of an irreducible polynomial $f(x,y) \in k(x,y)$ (I'm not 100% what the root of a two variable polynomial is). Finally, he said that $l$ is an algebraic extension of $k$, where $f$ becomes reducible over $l$. Still however, I can't come up with anything. I was hoping that with these extra hints, someone would be able to figure it out.  He did point out that there is a chance that there is an error in the question, but I haven't managed to prove that such an example cannot exist.
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.