I'm trying to prove unique factorization for the set of polynomials in x, with integer coefficients.   I have tried proving that the analogous Euclidean algorithm works for such polynomials, using “degree” instead of “norm" if that makes sense, but couldn't quite get it. If anyone could help me out that would be great! Thanks :)
74797
I'm trying to prove that $A$ a UFD implies that $A[T]$ is a UFD.   The only thing I am sure I could try to use is Gauss's lemma.  Also, how can we deduce that the polynomial rings $\mathbb{Z}[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_n]$ are UFDs?
115182
Suppose that a polynomial $p(x,y)$ defined on $\mathbb{R}^2$ is identically zero on some open ball (in the Euclidean topology). How does one go about proving that this must be the zero polynomial?
33610
Let $P(x)$ be a nonconstant polynomial whose coefficients are positive integers. If $P(n)$ divides $P(P(n) - 2015 )$ for every natural number n, then prove that $$P (-2015) = 0$$  I've observed that the proof depends on  $ P(n)$ divides $ P(P(n)-2015)-P(-2015)\,$ but I cannot see how to prove that. Could someone please elaborate on how to proceed.
113920
Excuse me for the confusing title. I was asked to find $gcd(x^2+3x+2,x^2+x)$  What i did is i factorized both polynomials $x^2+x=(x+1)x$  $x^2+3x+2=(x+1)(x+2)$  So i expected the gcd to be $x+1$   But using the euclidean algorithm i found out the gcd to be $2x+2$. Why is factorizing wrong? Is it because $K[X]$ is not factorial ? Would the euclidean algorithm also work if the polynomials are in $\Bbb Z[X]$ ???
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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!!
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The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)  So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.  Is the following a valid argument for there only being at most 1 solution per degree? Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.  Finally, is there any elegant way to prove that no odd solutions exist?
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I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:     In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.   My understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:  Proceeding with Euclid's algorithm:  $$ x^3 + 2x + 1  =(x^2 + 1)(x) + (x + 1) \\\\   x^2 + 1 = (x + 1)(2 + x) + 2$$  We stop here because 2 is invertible in $\mathbb{Z}_3[x]$. We rewrite it using a congruence:  $$(x+1)(2+x) \equiv 2 \mod{(x^2+1)}$$  I don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?  Wikipedia has a page on this we a , but it's still not clear in my mind.  Note that  has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.  Thanks.
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Suppose $p$ and $q$ primes and $p$ is odd. Then, are there nice and clever ways to factorize polynomials of the form $$f(x)=1+x+\cdots +x^{p-1}$$ in the ring $\mathbb{F}_q[x]$ ? What about the case when $q=2$ ?   I know that there are factorization algorithms but they are too general. I want to know if there are clever ways to do this for these special type of polynomials. Like for example, in $\mathbb{F}_2$ one might add terms $x^r+x^r$ and rearrange them so that $f(x)$ gets factorized.  In case there are no good methods to factorize, are there nice ways to check whether $f(x)$ is irreducible in $\mathbb{F}_q[x]$ ?
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Suggest an algorithm for factorizing an integer such that each of its factor is .  Thanks,
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I am struggling to use the Euclidean algorithm for polynomials.  Given something like $$GCD(x^5+1, x^3+1)$$  I can easily use it as follows:   $$x^5+1 = x^2(x^3+1) -x^2 +1 \\ x^3+1 = -x(-x^2+1) + x +1 \\ -x^2+1 = (x+1)(-x+1)$$  GCD = x+1  But for something like $$GCD(2x^2+6x+3, 2x+1)$$   I cannot figure out how to do it using the same method. I start like this:  $$2x^2+6x+3 = x(2x+1) + 5x+3\\ 2x+1 = \frac{2}{5}(5x+3) - \frac{1}{5}\\$$  and I am not sure how to continue. Any help would be appreciated, sorry for my poor attempt at using LaTeX.
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If R was a unique factorization domain, can we deduce that for a nonzero element d in R, d has a finite number of divisors?   I need this in solving this question  "If R is a unique factorization domain then there are only finite number of distinct principal ideals that contain the ideal (d)."  My idea to solve it: if a divides d then (d) is contained in (a), so if we have finite divisors then we have finite principal ideals. Am I wrong?
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.
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I know that you can't factor a real polynomial into $\Pi_{i=1}^N(x-a_i)$ in general. But is it possible to factor every real finite polynomial into this form:    $(\Pi_{i=1}^N a_ix^2 + b_ix + c_i) (\Pi_{i=1}^Mx-d_i)$  where the second term is possibly empty?  This is at least true for polynomials with odd powers because they must have at least one real root. (My reasoning was false, edit see below) So I wonder if it is also true for even powers?    Edit: Berci has pointed out a lapse in my reasoning. For an odd power, once you factor out the first root, it is conceivable that you may hit an even-degree polynomial that is irreducible.
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The question that I am trying to answer is :  Suppose is $p(x)$ is a polynomial with integer coefficients. Show that if $p(a) = 1$ for some integer a then $p(x)$ has at most two integer roots.  I have no idea how to get started. Any help would be awesome!
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Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than $5$.   This is what I tried:  It's evident that $x,x+1$ are irreducible. Then, use these to find all reducible polynomials of degree 2. There ones that can't be made are irreducible. Then use these to make polynomials of degree 3, the ones that can't be made are irreducible. Repeat until degree 5.  Doing this way takes way too long and I just gave up during when I was about to reach degree 4 polynomials.  My question is: is there any easier way to find these polynomials?  P.S.: this is not exactly homework, but a question which I came across while studying for an exam.
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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 $0$.
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There is a question I have based on the fact: If you take a quadratic polynomial with integer coefficients, take the set $\{1,2,3,4,5,6,7,8\}$, make a partition $A=\{1,4,6,7\}$, $B=\{2,3,5,8\}$, and then evaluate the polynomial with the elements of each set you got an equality.  If now with a cubic polynomial and the set $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16\}$, how to pick a suitable partition that accomplish the same fact? I mean is there a general formula for bigger degree?