Problem: Let $f$ be an irreducible polynomial in $F[x]$. Prove that $f$ has no multiple root in any field extension of $F$ unless the derivative $f'(x)$ is the zero polynomial. My work so far: I'm assuming that $f'(x)=0$ and I want to should that $f$ has a multiple root. I know that if $f$ has a multiple root $\alpha$, then $\alpha$ is a common root of $f$ and $f'$. In general, for $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0, f'(x)=na_nx^n{n-1}+\cdots +a_1$ and $deg{f}\geq def{f'}$. If $f$ has a multiple root, then there is a factor $(x-\alpha)^k$ s.t. $k\leq n$ and $(x-\alpha)|f$ and $(x-\alpha)|f'$. I know I want to do something s.t. I can say $deg{f}\leq deg{f'}$ to get their degrees to be equal. In the textbook I'm using (Algebra by Artin, pg 458), it says &quot;Since it is irreducible, $f$ will have a nonconstant factor in common with another polynomial $g$ only if $f|g$.&quot; I'm struggling to wrap my mind around how I can see $f|g$. In general, $a|b$ and $a|c$ doesn't imply $b|c$. Thanks in advance!
37895
Let $E/F $ be a field extension and $f,g$ $\in$ $F[x]$ (the polynomial ring with coefficients in $F$ ). Let's denote with $\gcd_F(f,g)$ the greatest common divisor of $f$ and $g$ in $F[x]$. Is it true that $\gcd_F(f,g)$ $=$ $\gcd_E(f,g)$ ?
66456
Let $f(x)=x^3+x+1$ and set $E=z_2[x]/(x^3+x+1)$ prove that $f(x)$ splits in E that is find 3 distinct roots of $f(x)$ in $E$    Questions found $[x^2+1]$ $[x^2+x+1]$ on my sloppy work. There are more than 3 roots if my notes are correct. If that is the case do we just use the irreducibles??  Another question I have is that what is [x]?? Is it a variable, some sort of constant. I know the book uses idertimenatn and has special properties.
4760
I want to argue that argue that $\pi(\alpha)=\alpha^3+3\alpha+3$ is an irreducible polynomial over the finite field with 5 elements $\mathbb{F}_5$. My approach was just to check that $\pi$ has no roots in $\mathbb{F}_5$. Is this right?
105331
Let $F \in K[x_1, \dotsc, x_n]$, $n\ge 2$, with $K$ an algebraically closed field and $\deg F \ge 1$. Prove that $F$ has infinitely many zeroes.   I would consider $F$ as a polynomial in the single variable $x_1$, so that from the algebraic closure of $K, \exists c \in K$ s.t. $F(c) = 0$ considered as a polynomial of the only variable $x_1$. But then I can vary the remaining variables as I like in $K$, thus obtaining infinite zeroes. However this argument doesn't convince me: can I vary the variables $x_2, \dotsc, x_n$ and keep fixed $x_1$ still obtaining a zero of $F$?
123169
Using the affine linearized polynomial $f(x)=x^p-x-b$ is irreducible over $\mathbb{F}_q$ if and only if $\operatorname{Tr}_{\mathbb{F}_q/\mathbb{F}_p} \neq 0$. But here $\mathbb{F}_3$ is not a field extension of $\mathbb{F}_5$.  Also we have that: Let $a \in \mathbb{F}_q$ and let $p$ be the characteristic of $\mathbb{F}_q$. Then the trinomial $x^P - x - a$ is irreducible in $\mathbb{F}_q[x]$ if and only if it has no root in $\mathbb{F}_q$.
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.
32765
This is an assignment. There are two related (I think) problems. Please solve one of them and I will try to solve the other.   Let $\alpha, \beta $ be algebraic over $k$ whose irreducible polynomials are $f(x), g(x)$. Suppose $\gcd(\deg(f),\deg(g))=1$. Show that $g(x)$ is irreducible in $k(\alpha)[X]$. $f(x) \in k[x]$ is irreducible with degree $n$ and let $[K:k]=m$, where $(n,m)=1$. Show that $f$ is irreducible in $K[x]$.
84054
Two part question:  1) There are three irreducible cubic polynomials $\mathbb F_2$  Aproach: Bruteforce, there are not as much cubic polynomials, just 8 of them:  $x^3$  $x^3+1$  $x^3+x$  $x^3+x+1$  $x^3+x^2$  $x^3+x^2+1$  $x^3+x^2+x$  $x^3+x^2+x+1$  Only three of them are irreducible(??): $x^3+1$,$x^3+x+1$ and $x^3+x^2+1$  Reality check - do i miss point of irreducible polynomial here, and should consider adding $x^3+x^2+x+1$ to irreducible list? It is reducible, bit different way then other polynomials: $x^3+x^2+x+1=(x+1)^3$  2) Only irreducible polynomial of degree 2 over $\mathbb F_2$ is $x^2+x+1$. Does it help, with conjuction of part 1, to find all irreducible polynomials of degree 5?
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!
60717
Let $f(t_1, ... , t_n)$ be a polynomial over a field $F$. Suppose there exist infinite subsets $X_1, ... , X_n$ of $F$ such that $f(x_1, ... , x_n) = 0$ for all $(x_1, ... , x_n) ∈ X_1× \cdots ×X_n$. Prove that $f$ is the zero polynomial.   Not sure how to start on this one!
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.
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.
73163
Prove that each irreducible factor of $f(x)=x^{2^n}+x+1$ in $\mathbb Z_2[x]$ has degree $k$, where $k\mid 2n$.   Edit. I know I should somehow relate the question to an extension of $\mathbb Z_2$ of degree $2n$, say $GF(2^{2n})$. By this way I will be able to correspond each irreducible factor of $f(x)$ to a subfield of $GF(2^{2n})$ that obviously has degree $k$, where $k\mid 2n$.
19236
Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)  This question is worth $7$ marks out of a possible $75$, this is my solution..  $f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.  Would this be a sufficient answer? am I missing something?
105931
Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case?
44151
I'd like some help making this argument complete and rigorous (if it's correct - if not, help with that would be nice).  Here $k$ is a field.     Let $A_1,\ldots,A_n \subseteq k$ be infinite subsets. Then any polynomial in $k[x_1,\ldots,x_n]$ that vanishes on $A_1\times\cdots\times A_n\subseteq k^n$ must be $0$ (as a polynomial).   This is what I have ...  For the case $n=1$, a non-constant polynomial can only have as many roots as its degree, and in particular, it must have a finite number of roots. The only polynomial in one variable that has an infinite number of roots is $0$, so if a polynomial in $k[x_1,\ldots,x_n]$ vanishes on an infinite subset then it must be $0$.  For the inductive step, suppose the proposition is true for less than $n$ subsets and variables. Let $p\in k[x_1,\ldots,x_n]$ vanish on $A_1\times\cdots\times A_n$. Fix $x_n$ as some $a\in A_n$, and we have a polynomial in $n-1$ variables that vanishes on the set $A_1\times\cdots\times A_{n-1}$, so by the inductive hypothesis it must be identically $0$.  (Now it gets sketchy). Since this is true for any of the infinite values in $A_n$, and , $p$ must be $0$.
111893
Let $f(x)=x^n+5x^{n-1}+3$ where $n\geq1$ is an integer. Prove that $f(x)$ can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree $\geq1$.   If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that $f(x)$ is irreducible over real numbers. But since this condition is given,therefore, we can't exclude the case when $f(x)$ is reducible over real numbers but not with polynomials of integer coefficients. Anyone with idea how to proceed? Thanks in advance!!