If $f$ and $g$ are irreducible polynomials in $K[X,Y]$ that are not associates, show that the zero set $Z(f,g)$ is either empty or finite.   Here is what I been told to do: If $(f,g)≠K[X,Y]$, show $(f,g)$ contains a nonzero polynomial in $K[X]$ and similarly a nonzero polynomial in $K[Y]$.   and what i should do before that is let $R=K[X]$ and $F=K(X)$, and apply Gauss’s Lemma to show $f$ and $g$ are relatively prime in $F[Y]$.  Can any one help please, thanks a lot.
106515
Let $k$ be any field, and let $f,g\in k[x,y]$ be two irreducible polynomials such that $g$ is not divisible by $f$. Prove that $V(f,g)\subseteq A_k^2$ is finite.
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.
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.
110296
The hint is: Let F has as elements $a_1,...,a_n$. Note that if $$f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)$$, then $f_i(a_j)=0$ for $i\neq j$ and the value $f_i(a_j)$ can be controlled by the choice of $c\in F$. Use this to show that every function on F is a polynomial function.  Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!
91368
Let $R$ be a sub-ring of a ring $S$. Let $f,g$ be non-zero polynomials in $R[x]$ and assume that the leading coefficient of $f$ is a unit in $R$. If $f$ divides $g$ in $S[x]$, show that $f$ divides $g$ in $R[x]$.   Thoughts: If $f$ divides $g$ in $S[x]$, then  $g(x) = f(x)q(x)$ for some $q(x)$ in $S[x]$. From here I'm not sure what the question is asking. Do I have to show that $q(x)$ is an element of $R[x]$ ? Insights appreciated.
3442
I know that $\mathbb{Q}[y][x]=\mathbb{Q}[y,x]$; thus, we can see this polynomial as a polynomial with variable $x$ and with coefficients in $\mathbb{Q}[y]$. Someone explains to me that we could use Gauss's Lemma to reduce to proving irreducibility over the $\mathbb{Q}(x)[y]$ where $\mathbb{Q}(x)$ is the fraction field and after apply Eisenstein for the prime $y+1$. I am blocked for a while on this problem.   Could anyone solve this problem?
15942
Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$.   How to prove? I really have no idea... Thank you a lot.
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.
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.
94023
I need your help with this: Let $f$, $p$, and $g$ be a polynomials in $F[x]$, assume that $f$, $g$, and $p$ are nononzero.  If $\gcd(f,p)=1$ and $fg$ is divisible by $p$ with no remainder, I need to prove that $g$ is  divisible by $p$ with no remainder  Thanks.
48811
Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.
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
55256
A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?  It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.
112281
Let $K=\mathbb{Z}_p(t)$. How to prove $f(x)=x^p-t$ is irreducible in $K[x]$?
20388
Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be irreducible over $D$ if, whenever $f(x)$ is expressed as a product $f(x)= g(x)h(x)$, with $g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$.  This is the definition of the irreducible polynomial in Gallian's book. What I don't understand is how this definition translate to the fact that the polynomial cannot be factored into a product of polynomials of lower degree when $D$ happens to be a field.  There is also this examples that the polynomial $f(x)=2 x^2+ 4$ is irreducible over $\mathbb R$, but reducible over $\mathbb{C}$.  To me it seems that it is also irreducible over $\mathbb{C}$ since in its factorization $2(x^2+2)$, $2$ is unit over $\mathbb{C}$.
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.
92660
Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.   Here $[F:K]$ denotes the dimension of $F$ over $K$ as a vector space.  I already know it is true for $K=F$, so the problem here is when $f$ is irreducible in $K[X]$.