Let $F$ be an algebraically closed (hence infinite) field and $n$ a positive integer. For a subset $S$ of $F^n$, let us use the following notation:  $J(S):=\{f \in F[x_1,...,x_n] : f(a_1,...,a_n)=0$ for all $(a_1,...,a_n) \in S \}$  There is a theorem in Hungerford's Algebra, Section VIII.7, which asserts  $J(F^n)=\emptyset$,  with proof omitted. Is this an typo? I think, at least, the zero polynomial $0$ belongs to $J(F^n)$. On the other hand, how do I have to prove the inclusion  $J(F^n) \subset 0$?  It seems obvious, but I cannot write down rigorously.
123064
I am working on a problem from Ideas, Varieties, and Algorithms:     Show that if $f \in \mathbb{C} \left[x_1, \dots, x_n \right]$ vanishes at every point of $\mathbb{Z}^n$, then $f$ is the zero polynomial.   My attempt so far:  This proof is by induction on $n$. Let $f \in \mathbb{C}\left[x\right]$ vanish at every point in $\mathbb{Z}$. Then, it must be identically the zero polynomial, as a polynomial of degree $d$ has at most $d$ roots. Hence, $f$ vanishes at every point at $\mathbb{C}$. Now, assume that $g \in \mathbb{C} \left[x_1, \dots, x_{n-1} \right]$ vanishes at every point in $\mathbb{Z}$. We may collect terms and rewrite $f$ as  $$ f(x_1, \dots, x_n) = \sum_{i=0}^m x^i_n g(x_1, \dots, x_{n-1}), $$ for some functions $g_1,\dots,g_n\in \Bbb C[x_1,\dots,x_{n-1}]$.  How do I proceed from here? I was trying to adapt an earlier proof with no success.
48871
I would like to show that if $R$ is a field, then $R(x)$ is a proper subset of $R((x))$, where $R(x)$ is the ring of rational functions, and $R((x))$ is the ring of formal Laurent series.   If $f \in R(x)$, then $f(x) = f_1(x)f_2^{-1}(x)$, where $f_1(x), f_2(x) \in R[x]$. So I wrote this as $$f(x) = \frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}\;,$$ and I would like to show that I can write $f$ in the form $\sum_{k=r}^{\infty}c_kx^k$. However, I am unsure how to manipulate $f$ in order to show this. What I was thinking was to find some formal power series expansion for $f_2^{-1}(x)$, multiply out the summation with $f_1(x)$, then rearrange the coefficients and terms to obtain the desired form. However, I can't seem to derive a formula for the inverse of a polynomial in general that I could use for this. How can I go about manipulating $f_2^{-1}(x)$ to show this? Any suggestions?  Thanks!
121137
Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$   Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?  I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$ I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$. The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity. Any ideas ?
14317
I'm trying to solve this question:  Let $$M=\{f\in C([0,1])\ | \ f(1)=0\}.  $$  Is $M$ closed in $(C[0,1]),d_1)$?  At first, I assumed that $M$ was closed, so I tried to prove that if there is a sequence $f_1,f_2,f_3,...\in M,$and $\ f_n\rightarrow f\,$ with respect to $d_1$, then $f \in M$.    However, I got stuck and checked out the answers, and the answer I was given was just a counterexample of a sequence $(f_n)_n$ where $f_n \in M,\ f_n\rightarrow f$ wrt $d_1$, and $f\notin M$.    I understand that it's a lot easier to give a counterexample, but unfortunately I suck at analysis and cannot tell whether a set is closed or not upon inspection or come up with a counterexample right away.  Are there other ways of proof, like proof by contradiction, to show that this set is not closed?
123135
Let $n \in \mathbb{N}$ and $p$ be a prime number. Let $f(x),g(x) \in \mathbb{Z}_{p^n}[x]$ such that, in $\mathbb{Z}_{p^n}[x]$, $f(x)g(x)=0$. Prove that $a_ib_j=0$ for all $i,j$, where $a_i$ and $b_j$ are the coefficients of $f(x)$ and $g(x)$.  I have the set of equations (for degree $r$ and degree $s$): $$a_rb_s=0, a_rb_{s-1}+a_{r-1}b_s=0,...,a_0b_0=0.$$ From these, how do I conclude $a_ib_j=0$ for all $i,j$? Also, I found that for $n=1$, the polynomial ring is an integral domain, so the conclusion holds. I had an idea to use induction on $n$, but do not know how to proceed.
16690
Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence.
46247
I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.  My question is as follows:  Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$  I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.  What I thought of:  Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?  Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
95299
A complex number $z$ is said to be algebraic if there are integers $a_0, ..., a_n$, not all zero, such that  $a_0z^n+a_1z^{n-1}+...+a_{n-1}z+a_n=0$. Prove that the set of all algebraic numbers is countable.  The Hint is: For every positive integer $N$ there are only finitely many equations with $n+|a_0|+|a_1|+...+|a_n|=N$.  Here is a proof I have. The problem though, is that I did not use the hint provided in the text, so maybe this proof is invalid or that there is an alternate (simpler) proof? Please help me out here. Thanks in advance.  Proof:  The set of integers is countable, we have this following theorem:  Let $A$ be a countable set, and let $B_n$ be the set of all n-tuples $(a_1,...,a_n)$, where $a_k \in A, k=1,...,n,$ and the elements $a_1,...,a_n$ need not be distinct. Then $B_n$ is countable.  So by this theorem, the set of all $(k+1)$-tuples $(a_0,a_1,...,a_k)$ with $a_0 \neq 0$ is also countable.   Let this set be represented by $\mathbb Z ^k$. For each a $a \in \mathbb Z ^k$ consider the polynomial $a_0z^k+a_1z^{k-1}+...+a_k=0$.  From the fundamental theorem of algebra, we know that there are exactly $k$ complex roots for this polynomial.  We now have a series of nested sets that encompass every possible root for every possible polynomial with integer coefficients. More specifically, we have a countable number of $\mathbb Z^k s$, each containing a countable number of $(k + 1)$-tuples, each of which corresponds with $k$ roots of a $k$-degree polynomial. So our set of complex roots (call it $R$) is a countable union of countable unions of finite sets. This only tells us that $R$ is at most countable: it is either countable or finite.  To show that $R$ is not finite, consider the roots for $2$-tuples in $\mathbb Z^1$. Each $2$-tuple of the form $(-1, n)$ corresponds with the polynomial $-z + n = 0$ whose solution is $z = n$. There is clearly a unique solution for each $n \in \mathbb Z$, so $R$ is an infinite set. Because $R$ is also at most countable, this proves that $R$ is countable.
61251
Let $n\in\mathbf{N}$ be fixed, and $f$ entire and $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$ for every $w\in\mathbf{C}$. Then $f$ is a polynomial of degree at most $n$.    I try to prove this statement, and I think one can prove it as follows: consider $f(1/z)$. $0$ cannot be an essential singularity of $f(1/z)$, for the big Picard theorem would imply that on any neighborhood of $0$ $f(1/z)$ takes on all possible complex values (with at most one exception) infinitely often, but this is contrary to $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$. Then $f(1/z)$ has a pole of order $k$ say, and since $f$ is holomorphic it is a polynomial of degree $k$ (for its principal part vanishes). By the fundamental theorem of algebra, $k\leq n$.      Is it possible here to avoid using Picard's theorem?
98034
I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:  Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.  Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.  I don't understand the following points:   Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist? Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?   Thanks in advance for any help!
109459
Define the content of a polynomial (over an arbitrary commutative ring $A$) to be the ideal generated by its coefficients, denoted $c(f)$. I want to show that   $$c(fg) = c(f)c(g).$$      (I was told this is true.)    What I was able to show was that $c(fg) \subseteq c(f)c(g)$ (this is obvious), and that their radicals are the same. My reasoning for the latter was as follows: let $f = a_0 + \dotsb + a_nx^n,$ $g = b_0 + \cdots + a_mx^m,$ and consider the matrix  $$ \begin{pmatrix}  a_0 b_0 &amp; \cdots &amp; a_0b_m\\ \vdots &amp; \ddots &amp; \vdots\\ a_nb_0 &amp; \cdots &amp; a_nb_m \end{pmatrix} .$$  Then $c(f)c(g)$ is generated by the entries of this matrix, while $c(fg)$ is generated by the sums along the diagonals. Let $\mathfrak p$ be a prime ideal of $A$ not containing $c(f)$ or $c(g)$, and let $i,\,j$ be minimal such that $a_i,\, b_j \notin \mathfrak p$.  Then all the terms in the generator of $c(fg)$ corresponding to the coefficient of $x^{i+j}$ are in $\mathfrak p$ except for $a_ib_j$, so that $c(fg) \not\subset\mathfrak p$. It follows that $c(fg) \subseteq \mathfrak p \Longleftrightarrow c(f)c(g) \subseteq\mathfrak p$, and hence $\sqrt{c(fg)} = \sqrt{c(f)c(g)}$.     Can I go all the way and show that in fact $c(fg) = c(f) c(g)$?
75469
Let $k$ be an algebraically closed field.  Let $I$ be the map that takes algebraic sets in $k^n$ to the ideal generated by them: $I : \{$ algebraic sets $\} \to \{$ ideals of $k[x_1,\dots, x_n] \}$, $I(X) = \{ f \in k[x_1, \dots, x_n] : f(X) = \{0\}\}$.  Similarly let $Z$ be the map that takes ideals in $k[x_1,\dots, x_n]$ to algebraic sets.  Then $I(Z(I)) = \sqrt{I}$.  How does that follow from Hilbert's Nullstellensatz?  This version of Nullstellensatz: If $k$ is an algebraically closed field then, the maximal ideals of $k[x_1,\dots, x_n]$ are of the form $(x_1 - a_1, \dots, x_n - a_n)$ where $a = (a_1, \dots, a_n)$ is a point in $k^n$.
99750
Let $\mathcal F$ be a field. Suppose that there is a set $P \subset \mathcal F$ which satisfies the following properties:   For each $x \in \mathcal F$, exactly one of the following statements holds: $x \in P$, $-x \in P$, $x =0$.  For $x,y \in P$, $xy \in P$ and $x+y \in P$.    If such a $P$ exists, then $\mathcal F$ is an ordered field.   Define $x \le y \Leftrightarrow y -x \in P \vee x = y$.   Exercise: Prove that the field of complex numbers $\mathbb C$ cannot be given the structure of an ordered field.   My Work So Far: (Edit 1 note: This section and the Question is at the beginning, simply leaving this up for reference as to where I started)  Let $i$ be such that $i \in P, i \ne 0 \Rightarrow i &gt; 0$. But $i^2 = -1 \notin P$.   My Question: I am not sure how much I need to redefine, and how I go about rigorously making this patchwork argument airtight. I am aware that I have not addressed how I assumed that $-1 \notin P$, but I'm not sure how to distinguish between $1$ and $i$ in this proof.     Edit #1    1st Step: Showing that $-1 \notin P$, observe that $(-1)(-1) = 1$ therefore if $-1 \in P$, both $x, -x \in P$, a contradiction.   2nd Step: To show $i \notin P$, we have that if $i \in P \Rightarrow i^2 \in P$, but $i^2 = -1 \notin P$, so $i \notin P$.   3rd Step: To show $-i \notin P$, we have $(-i)(-i) = i^2 \notin P$, so $-i$ cannot be in $P$.   Conclusion: Since $i \ne 0$, and $i, -i \notin P$, there is no set $P \subset \mathbb C$ that satisfies the above properties, thus $\mathbb C$ is not ordered.   Thank you André Nicolas and Eric Stucky for your help!
6414
If I have a field $K$ and an extension $L$ of $K$ such that all (non-constant) polynomials in $K[X]$ have a root in $L$, is the set of algebraic elements  of $L$ over $K$ (the sub-field of all the elements of $L$ which are roots of a polynomial in $K[X]$) algebraically closed ?  Do you have a counterexample ?
123230
I am looking at   and trying understand the proof for the theorem that states:  Given a nonzero ideal I in $P(F)$, there is a monic polynomial $p(t)$ such that   $$I=\left\{q(t)p(t) | q(t) \in P(F) \right\}$$  Proof goes like  Since $I$ contains at least one nonzero element, we can let $p(t)$ be a nonzero element of $I$ of minimum degree. We can show that every element of $I$ is a multiple of $p(t)$. Let $f(t) \in I$. Then there exist $q(t),r(t) \in P(F)$ such that  $$f(t) = q(t)p(t) + r(t)$$ and deg $r(t)$ &lt; deg $p(t)$. Since $p(t) \in I$, so is $q(t)p(t)$, and therefore so is $$r(t) = f(t) -q(t)p(t)$$.  By the choice of $p(t),r(t) = 0$, so $f(t)$ is a multiple of $p(t)$.  What I am having trouble understanding is why $r(t)$ must equal $0$. If someone can explain why it is, I'd appreciate it.
98026
Let $A$ be a commutative ring and $S$ a multiplicative subset of $A$ generated by $s\in A$ which is not a zero-divisor. Consider the polynomial ring $A[x]$.      Given a polynomial $f\in A[x]$, suppose that $f(1/s)=0$ for $s\in S$. Does it follow by some application of the first isomorphism theorem that $f(x)=(sx-1)g(x)$, with $g(x)\in A[x]$?   On the one hand I think this can be proven by constructing an isomorphism from $A[x]$ modulo the relation $x=1/s$ (really the replacement of $x$ with $1/s$). This is evidently an isomorphism from $A[x]/(x=1/s)$ into $S^{-1}A$. It is tempting to conclude (though I'm not certain that this should work), that there must be some surjective homomorphism $\phi: A[X]\to S^{-1}A$ such that $\ker(\phi)=(sx-1)$. [Since generally $(A/B\cong A/C) \nRightarrow (B\cong C)$, it is not sufficient to conclude that, say, the kernel of $\psi: f\to f(1/s)$ is $(sx-1)$.]  I am especially interested in answers as to whether, say, $\psi$ has kernel $(sx-1)$. But for anyone who can follow my own attempt at a proof, I'm also interested in whether my argument breaks down or can be made to work rigorously.
13252
Let's start with a definition:     Definition: Let $F$ be a field, $I$ a nonzero ideal in $F[x]$, and $g(x)$ an element of $F[x]$. Then, $I= \langle g(x)\rangle$ if and only if $g(x)$ is a nonzero polynomial of minimum degree in $I$.   Now it gets completely foggy what is meant with "minimum degree".I constantly come across the "minimal polynomial". This seems to be linked, but it seems strange that there is a division here.   Could someone give a pointer to what the second line actually means. Maybe some example or diagram (if possible) would help a lot here.