Let $F$ be a field. Let $F[x]$ be the polynomial ring and $f(x)$ and $g(x)$ be 2 irreducible polynomials of same degree in $F[x]$. When do we have the 2 fields $$F[x]/&lt;f(x)&gt; \cong F[x]/&lt;g(x)&gt;$$   Well, I can think of $F[x]/&lt;f(x)&gt;$ as the field $F(\alpha)$ where $\alpha$ is the root of $f(x)$ and check if any algebraic combination of $\alpha$ with elements of $F$ is a root of $g(x)$. If it is then both the fields are the same/isomorphic. But is there any general condition on the 2 polynomials such that the isomorphism holds?
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!
56407
Let $f(x)$ a irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be a polynomial in $F[x]$. Prove that every irreducible factor of the composition $f(g(x))$ has a degree which is divisible by $n$.  I don't know even how to begin. I really need help. Thanks
52950
I have seen many textbooks state this result without proof.   $``$ If $E$ is the splitting field for the polynomial $f=x^p-1 \in \mathbb{Q}[X]$ where $p$ is prime, then the Galois group $Gal(E:Q)\simeq \mathbb{F}_{p}^{\times} \simeq\mathbb{Z}_{p-1} . "$  I would like to know how to prove this.  The things that I know from reading are that $E=Q(\zeta)$, where $\zeta$ is a root of the irreducible polynomial $x^{p-1} + x^{p-2} \dots + 1$. So $E$ a cyclotomic extension of Q of degree $p-1$ and that the roots of $x^p - 1$ are the $p^{th}$ roots of unity. Also that the Galois group is comprised of the automorphisms of $E$ that fix $\mathbb{Q}$ and permute the roots of $f$.  As a second question, if I am asked to prove that the Galois group $Gal(E/Q)$ is cyclic, when $E$ is the splitting field for some specific $f$ and $p$ is small, say $5$ or $7$, is there an easier way to prove this than first proving the more general statement above and then drawing the conclusion from that?   Any help will be much appreciated! Thanks.
1406
Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in   $K[x]$ or has a root in $K$.   My strategy was to split this up into a case for each characteristic.   The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.  The characteristic $0$ case  and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.   Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.
36429
Let $F$ be a field and $F[x]$ a polynomial ring. Let $p(x)$ be an irreducible polynomial.  Show that $\gcd(p(x),q(x)) = 1\Longrightarrow \exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x) = 1$.  I know the proof for Bezout's identity for , but this proof uses the notion of absolute value, which cannot be applied to a polynomial ring.   I am thinking about a way to prove the statement: Use the Euclidean algorithm to compute $\gcd(p(x),q(x))$ and show that the result is $1$, and somehow go backwards to obtain $r(x)p(x)+s(x)q(x) = 1$, but it is a bit messy, and I'm not yet able to see it completely.  Please offer some help. If you have some other methods, that would be great, too.
63636
I have a question, I think it concerns with field theory.     Why the polynomial $$x^{p^n}-x+1$$ is irreducible in ${\mathbb{F}_p}$ only when $n=1$ or $n=p=2$?   Thanks in advance. It bothers me for several days.
89883
If $\alpha$ is an algebraic element and $L \subset K$ are both field, does the polynomial ring $L[\alpha]$ is also a field?  I am trying to prove that the ring of fraction $L(\alpha)$ is equal to $L[\alpha]$. To do this, I use an exercise done in class : Let $K$ a field, $\alpha \in K$ and $L$ a subfield of $K$. Then $L(\alpha)$ is the smallest subfield of $K$ containing $L$ and $\alpha$.  Is anyone could help me?
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.
99129
Let $F$ be a field of characteristic $p$, and let $\alpha \in F$. Let $f \in \mathbb{Z}_p[x]$ be such that $f(\alpha) = 0$. Apparently, we have $f(\alpha^p) = 0$. This was mentioned but not proved in my lectures. Can someone explain why please?  Thanks.
104485
Consider the polynomial $p(x)=x^8-x$ in $F_3$:      (a) find the splitting field of $p$ over $F_3$ and factorize $p$ over $F_3$      (b) factorize p over $F_{81}$   (a) If the roots of $p$ are all distinct, we need a field extension of $F_3$ with at least $8$ elements, the smallest one is $F_9$, but does it contain all the roots of $p$ ?  $p(x)=x(x^7-1)=x(x+2)(x^6+x^5+x^4+x^3+x^2+x+1)$ since $1$ is a root of $p$ and $x^7-1$ is divided by $x-1\equiv x+2$ (using Ruffini's rule).  The sextic polynomial has no roots on $F_3$, so the possibile factors are three of deg 2, one of deg 2 and one of deg 4, two of deg 3. How to show if it is irreducible or not? On $F_2$ is easy to show that it is reducible since there is only one quadratic irreducibile polynomial, and two cubic irreducible polynomials.  It is correct to say that it is irreducible in this way: the multiplicative order of $3$ modulo $7$ is $6$, so there is at least one cyclotomic coset of cardinality $6$ that correspondes to that sixtic polynomial, and so it is irreducible, otherwise the multiplicative order should be less than $6$.  If the sextic polynomial is irreducible then the splitting field of $p$ over $F_3$ would be $F_{3^6}$.  (b) how to proceed ?
18171
Let $F$ be a field. Why does $F[x]$ have infinitely many irreducible elements?   For the case F has characteristic 0  Then x-a is irreducible for all a $\in F$ since x satisfies no non-trivial relations in F.  Obviously this argument fails for a finite field since there are only finitely many a to choose from.  So how may I construct irreducible polynomials in a finite field?  I figure it must involve higher powers of x, maybe $x^n-a$ ?
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
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.
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!
86347
I'm hoping someone can give me a nudge in the right direction...     Let $F$ be a finite field, and let $f(x)$ be a nonconstant polynomial whose derivative is the zero polynomial.  Prove that $f$ cannot be irreducible over $F$.   I've got that every root of $f$ is a multiple root and that for $F=\mathbb{F}_{p^r}$, the exponent of every term of $f$ is a multiple of $p$.
122193
So we've already had a question on . My question is - can we use a different finite field other than the one given by $x^8 + x^4 + x^3 + x + 1$ in $GF(2^8)$. In other words, would any irreducible polynomial over this field do the trick or are the special considerations and properties of that particular reducing polynomial?  If a more general discussion on appropriate fields for cryptographic operations, such as those used in ECC, is necessary, then I'd be happy to hear it. What I'm trying to get to is an understanding of whether there are any important properties in terms of the "randomness" of the operation - for example, are there weak fields to operate over? For example, $x^8 + x^3 + x + 1$ ought to also be irreducible - so would it be suitable? I ask partly because of the  which lead me to think not all fields might hold equal strength. Right? Wrong? Inconsequential?
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 $E$.