My teacher says-     The degree of the zero polynomial is undefined.   My book says-     The degree of the zero polynomial is defined to be zero.    says-     The degree of the zero polynomial is $-\infty$.   I am totally confused and want to know which one is true or are all true?
122556
I wonder why the degree of the zero polynomial is $-\infty$ ?  I heard that, it is $-\infty$ to make the formula $\deg(fg)=\deg(f)+\deg(g)$ hold when one of these polynomials is zero. However, if that was the only reason we could have said that it is $\infty$ instead.
51577
I don't think I fully understand.  Let's say there is a root $x_0 \in K=\mathbb{F}_p(t)$, where $p$ is a prime number.  Then $x_0 = \frac{P(t)}{Q(t)}$ for some polynomials $P,Q \in \mathbb{F}_p[t]$. We can assume $\gcd(P,Q)=1$  and $x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0$, so coefficients of $P(t)^p, tQ(t)^p$ must be identical, which contradicts $\gcd(P,Q)=1$,   hence such $x_0$ does not exist and the polynomial has no root in $K$.   am I about right?    anyway, I would appreciate an explanation about $\mathbb{F}_p(t)$, what is $t$? what is the meaning of a variable which does not belong to any specific "world"? I cannot use $t$ as if it was a member of $\mathbb{F}_p$ and hence cannot assume $t^{p-1} = 1 \pmod p$...
27127
Let $f$ be a complex function with the property that $f^{(n)} = 0$.  Does this imply that $f$ is a polynomial?  If so, why?  Upon thinking about this problem myself, I can easily observe that every $n$ degree polynomial satisfies $f^{(n+1)} = 0$.  Still, this doesn't show that some non-polynomial function couldn't also have this property.
48147
Why is the zero set in $\mathbb{C}\times\mathbb{C}$ of a polynomial $f(x,y)$ in two complex variables always non-discrete (no zero of $f$ is isolated)?
59179
If we consider a multivariate polynomial of $n$ variables with degree $N$, how do we show that the maximal number of monomials is $$\binom{N+n}{n} = \frac{(N+n)!}{n!\, N!}\quad ?$$  The hint our teacher gave was that the number of monomials is the number of independent coefficients. So how many independent coefficients does a multivariate polynomial have (at most)?  Thank you.
109759
It's a really basic question,in these days, I've been thinking why a polynomial $p(x)\in F[x]$ ($F$ a field) with degree $n$ can have at most n roots. It seems easy to prove, but I've been trying to prove this since yesterday, maybe I forgot some important details necessary to prove, I don't know. I'm solving some questions about Field Theory and I noticed that almost every question I should use this theorem, I need help.  Thanks
2916
Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$.    Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.
122716
$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients.  If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.  I tried by assuming a fifth degree polynomial but got stuck after that.  The question was asked by my friend.
4265
For what values of $n$, where $n$ is an integer, the polynomial $x^3+nx+1$ is reducible over $\Bbb Z$. My attempt:  When $n= 0,-2 $, the given polynomial is reducible over $\Bbb Z$ as $x=-1$ and $x=1$ are zeros of the polynomial. But I couldn't find whether there exists any integer $n$ for which the polynomial $x^3+nx+1$ is reducible over $\Bbb Z$. How can we proceed from here? Is the polynomial irreducible over $\Bbb Z$ if $n$ is not in $\{0,-2\}$?
44175
Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.   Is it possible to prove this without using induction on degree? If so, how can I do this?
84041
Let $F$ be a field and let $a \neq 0$ be a zero of the polynomial $a_0 + a_1x + . . . +a_nx^n$ in $F[x]$.  I want to show that $\frac{1}{a}$ is a zero of the polynomial $a_n + a_{n-1}x + . . . + a_0x^n$  How can I do this?
48676
Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$  Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.   Is this property (if true) a famous one?  Thank you for your help.
61839
Let $f$ be an entire function such that $|Re(f(z))| \leq |z|^n, \forall z$   where $n$ is a positive integer.  Could anyone advise me how to show $f$ is polynomial of degree at most $n \ ?$  Hints will suffice, thank you very much.
121582
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
37931
How can I prove that $\sqrt{2}+\sqrt[3]{3}$ is algebraic?  I have been trying various things, and have failed as of now. I know that the degree of the polynomial satisfying this equation has to be $\leq 6$. But that is all I know. Clearly, the simple idea of building the polynomial from the root itself, like in the case of $\sqrt{2}+\sqrt{3}$ does not work here.   Thanks in advance!
2364
I shall assume all rings to be commutative in this question. The impatient can scroll down to the "blockquote" to read the actual question.  Whenever we have a polynomial over a ring, it defines a function from the ring to itself by evaluation. It's reasonable to ask when two different polynomials define the same function.  From the factor theorem it follows that an $n^\text{th}$ degree polynomial over an integral domain has at most $n$ roots. Then it's easy to show this:  Theorem. Let $R$ be an infinite integral domain and let $f \in R[X]$ such that $f(a)=0$ for all $a \in R$, then $f = 0$. Proof. $f$ has infinitely many roots, so it must be the zero polynomial. $\quad\square$  For finite rings a kind of opposite situation occurs:  Theorem. For any finite ring $R$ there are polynomials over $R$ that are different but agree on all elements. Proof. There are only finitely many functions from $R$ to itself, but $R[X]$ is infinite. $\quad\square$  If we make further assumptions it's of course possible to prove more, as Pete L. Clark wrote in this post: []  Then there is the question of infinite rings that are not integral domains. It's relatively easy to come up with examples of a ring $R$ with positive characteristic and a nonzero polynomial that evaluates to the zero function, e.g.: $$ R := \bigoplus_{n=1}^\infty \mathbb{Z}/6\mathbb{Z} \quad\text{and}\quad f(X) := X^3-X.$$  The Question:     This leaves open the case alluded to in this post's title: Is there a commutative ring of characteristic $0$ (hence infinite) such that a nonzero polynomial evaluates to the zero function?
90268
The question is, how many times does $f(C_r)$ go around $0$ for $C_r = \{x \in \mathbb{C}: |x|=r\}$ and a polynomial with complex coefficients $f$.    The answer is clear to me when $f(x) = ax^n + b$ but I don't see how exactly middle terms affect the result. How to see this without a graph? Can this be found for any $r$?    Many thanks