Let $\mathcal{I}$ be a proper ideal of $\mathbb{F}[x]$ where $\mathbb{F}$ is a field. Show that there is a polynomial $P_0(x)\in \mathbb{F}[x]$ with degree larger than $0$ with the property that $$\mathcal{I}=\{P_0(x)\cdot q(x):q(x)\in \mathbb{F}[x]\}.$$ (In other words, prove there is a polynomial $P_0$ with the property that $\mathcal{I}$ consists precisely of the multiples of $P_0$.)  Previously, I have proved     Let $\mathbb{F}$ denote the field described the set $$\mathbb{F}=\{r+si: r,s\in \mathbb{Q}\}$$   (Here $i$ is the infamous element of $\mathbb{C}$ with $i^2=-1$.) Consider   the function $\lambda: \mathbb{Q}[x] \to \mathbb{F}$ defined by   $$\lambda(p)=p(i) \qquad \text{ for all } p\in \mathbb{Q}[x]$$           1) Show that $\lambda$ is a ring     homomorphism.      2) Show that the kernel of $\lambda$ is the set of     all polynomials $p\in \mathbb{Q}[x]$ that are divisible by $x^2+1$.          Proof of (i): For $\lambda$ to be a ring homomorphism, we need to show that  > a) $\lambda(f+g)=\lambda(f)+\lambda(g)$      b) $\lambda(fg)=\lambda(f)\lambda(g)$      c) $\lambda(1)=1$         \textbf{\textit{Proof of a:}} Let $f,g \in \mathbb{Q}[x]$. Then we   need to show that $\lambda(f+g)=\lambda(f)+\lambda(g)$. So    $$\lambda(f+g)=(f+g)(i)=f(i)+g(i)=\lambda(f)+\lambda(g).$$      \textbf{\textit{Proof of b:}} Let $f,g \in \mathbb{Q}[x]$. Then we   need to show that $\lambda(fg)=\lambda(f)\lambda(g)$. So    $$\lambda(fg)=(fg)(i)=f(i)g(i)=\lambda(f)\lambda(g).$$      \textbf{\textit{Proof of c:}} Let $f=1 \in \mathbb{Q}[x]$. Then we   need to show that $\lambda(1)=1$. So $$\lambda(1)=1(i)=1.$$      Hence $\lambda$ is a ring homomorphism.      Proof of (ii):  We need to show   that the kernel of $\lambda$ is the set of all polynomials $p\in &gt; \mathbb{Q}[x]$ that are divisible by $x^2+1$.      Suppose $x^2+1|p(x)$. Then $p(x)=(x^2+1)g(x)$ for some $g(x)\in &gt; \mathbb{Q}[x]$. Using part (i), we can show that $p(x) \in &gt; ker(\lambda)$ (i.e. that $p(x)=0$):   $$\lambda(p(x))=(x^2+1)g(x)=(i^2+1)g(i)=0\cdot g(i)=0$$ Hence we have   show that $p(x)\in ker(\lambda)$.      Suppose $p(x) \in ker(\lambda)$. Using the Euclidean division   over $\mathbb{Q}[x]$ to divide $p(x)$ by $x^2+1$. We obtain:       \begin{equation*} \begin{aligned} 0 &amp; =\lambda(p(x)) \\ &amp; &gt; =\lambda(q(x)(x^2+1)+r(x))  \\ &amp; =\lambda(q(x))\lambda(x^2+1)+\lambda(r(x)) \\ &amp; =p(i)(i^2+1)+r(i)  \\ &amp; =r(i) \\ \end{aligned} \end{equation*} Hence we have show that the   remainder is $0$, which implies that $p(x)=(x^2+1)g(x)$.       Hence, the kernel of $\lambda$ is the set of all polynomials   $p\in \mathbb{Q}[x]$ that are divisible by $x^2+1$.   I know both proofs are related to one another. So what is needed to change in the previous proof for the new proof?
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.
77272
I am trying to find the automorphisms of the polynomial ring $\mathbb{Z}[x]$.  So far, I have shown that if $\varphi$ is an automorphism, then $\varphi(n)=n$ for $n \in \mathbb{Z}$, and $\varphi(x)$ is a polynomial of degree one, say $mx+n$. If we consider what gets sent to $x$ by $\varphi$, we see that $m$ must be a unit of $\mathbb{Z}$, so $m=\pm 1$. So I think the automorphisms fix $\mathbb{Z}$ and send the indeterminate variable $x \mapsto x+n$ or $x \mapsto -x+n$ for $n \in \mathbb{Z}$. Is this correct?
15412
I was having some trouble finding an explicit isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$.  $\textbf{What I have noticed is}:$   7 is a prime element in $\mathbb{Z}[i]$ so $(7)$ is a maximal ideal in $\mathbb{Z}[i]$ and $\mathbb{Z}[i]/(7)$ is  a field.  7 is also a prime element in $\mathbb{Z}[\sqrt{-2}]$   $\textbf{What I have been trying to do is this}$   Find a surjective homomorphism between $\mathbb{Z}[i]/(7) \rightarrow \mathbb{Z}[\sqrt{-2}]/(7)$. Since $\mathbb{Z}[i]/(7)$ was a field, the kernel of this homomorphism will be either the whole ring or just $0$. In the latter case it would be a isomorphism.    I am having trouble finding this surjective homomorphism:  I have noticed that $\bar{{i}}^{2}=-1$ so the image of $\bar{{i}}$ must be sent to something whose square is $-1$. Any help would be appreciated. I may be missing some obvious insight.
118718
I have been trying to prove the following theorem:      Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is irreducible modulo $p$.    Idea: I tried to establish a homomorphism $\phi: \mathbb{Z}[x] \rightarrow \mathbb{F}$. Since $\mathbb{F}$ is a field it has characteristic $p$ and so integer prime p are mapped to 0 in $\mathbb{F}$. Hence $p\in \ker \phi$. Next we consider $\phi': \mathbb{Z}[x] \rightarrow \mathbb{Z_p}[x]$ and pick an arbitrary maximal ideal $M\in \mathbb{Z}[x]$. So, $\phi'(M)$ is maximal as long as $p \in M$ by correspondence. But now I am stuck at this stage and do not know how to proceed. I guess we might have to use primitivity given in problem but dont know how.
69298
On page 151 of Paolo Aluffi's Algebra: Chapter 0, an important property of the polynomial ring $\mathbb{Z}[x_1, \cdots, x_n]$ is introduced, namely that it's initial in the category of set functions from a fixed $n$-element set $A$ to commutative rings. (Or, to put it another way, a set function $f: A \to R$ with $R$ an arbitrary commutative ring factors uniquely through $\mathbb{Z}[x_1, \cdots, x_n]$.)  After establishing this property, two examples are given; the first, a special case when $n=1$, proves the existence of a unique ring homomorphism $\mathbb{Z}[x] \to S$ sending $x$ to an element $s \in S$ and extending the unique homomorphism from $\mathbb{Z} \to S$. So far so good.  The author then goes on to give a second example:     Let $a : R \to S$ be a fixed ring homomorphism, and   let $s \in S$ be an element commuting with $a(r)$ for all $r \in R$. Then there is a unique   ring homomorphism $\bar{a} : R[x] \to S$ extending $a$ and sending $x$ to $s$.   It is an exercise to provide the details of this construction. It's not hard to show what this map must be and prove that it's unique by "naive" methods, but, given its placement and the pedagogical bent of the text in general, I assume that there's a slicker way to define this map based solely on categorial considerations.   So: does this "extension homomorphism" satisfy a universal property?  (I've thought vaguely about this question for a little while but didn't hit upon anything; hints or full explanations would be much appreciated.)
4074
Let $R$ be a Dedekind domain, $\mathfrak{m}$ be a maximal ideal in $R[x]$ is of the form $\mathfrak{m} = (\mathfrak{p},f(x))$ where $\mathfrak{p}$ is a maximal ideal in $R$, and $f$ is a polynomial in $R[x]$ which is irreducible in $(R/\mathfrak{p})[x]$. Let $g(x) \in \mathfrak{m}$.  I am know that exist a ideal $\mathfrak{a}$ in $R$ such that $(\mathfrak{p},\mathfrak{a}) = 1 $, i.e $\mathfrak{p} + \mathfrak{a} = R$ and $\mathfrak{p}\mathfrak{a} = \mathfrak{p}\cap \mathfrak{a} = (p)$. (chapter 16, corollary 19, page 768, Abstract Algebra, Dummit Foote...)  Easily, since $g(x)\in \mathfrak{m}$, we have $ g(x) = l(x) + f(x).k(x)$ where $l(x) \in \mathfrak{p}[x]$. The questions is: Can we make a "strong" form as to show that there exists an ideal $\mathfrak{a}$ in $R$ such that $g$ can be show as the form $$ag(x) = p h(x) + a f(x) k(x)$$ where $p\in \mathfrak{p}$ such that $(p) = \mathfrak{p} \mathfrak{a}$, $(\mathfrak{p},\mathfrak{a}) = 1$ and $a\in \mathfrak{a} - \mathfrak{a}\mathfrak{p}$.
116701
From Berkeley Problems in Mathematics, Spring 1999, .  Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.   (BELOW IS NONSENSE, SHOULD BE IGNORED)  Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.   Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here.
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?
4774
From a bank of past master's exams I am going through:     Let $F$ be a finite field. Show that any function from $F$ to $F$ is a polynomial function.   I know that finite fields are fields of $p$ elements for $p$ prime [EDIT: It's actually $p^n$ for $p$ prime; see comment below]. Since I have $p$ choices for the $p$ elements to map to, then I have $p^p$ distinct functions. I think every function can be written in the form $f(x) = a_{p-1}x^{p-1} + \dots + a_0x^0$. For then given the values $f(0), f(1), \ldots f(p-1)$, I can solve for the coefficents by the linear system of equations  $$ a_0 + \sum_{i=1}^{p-1} n^i a_i = f(n).$$  This then gives me a $p-1 \times p-1$ square matrix over the field $\mathbb{F}_p$:  $$\left( \begin{array}{ccccc} 1&amp; 0 &amp; 0 &amp; \ldots &amp; 0 \\ 1&amp; 1 &amp; 1 &amp; \dots &amp; 1 \\ 1&amp; 2 &amp; 4 &amp; \dots &amp; 2^{p-1} \\ \vdots &amp; \vdots &amp; \vdots &amp; &amp; \vdots \\ 1 &amp; p-1 &amp; (p-1)^2 &amp; \dots &amp; (p-1)^{p-1} \end{array} \right) \left( \begin{array}{c} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{p-1} \end{array}\right)  = \left( \begin{array}{c} f(0) \\ f(1) \\ f(2) \\ \vdots \\ f(p-1) \end{array} \right)$$  If I can show this matrix is invertible, then I can always find the $a_i$. But I am a bit stumped on how to show this (partially because I don't think I've ever done linear algebra in a vector space over a finite field). It does not seem easy to show linear independence, or nonzero determinant, or full row rank.    Alternatively (I just thought of this), can I show this is true by arguing that the map between the two sets (the set of polynomials of degree $p-1$; and the set of functions $F \to F$) is injective, and that it must be a bijection because the sets have the same cardinality $p^p$?
15429
Can someone please give me a hint how to prove that the quotient ring $\mathbb{F}_p[x]/{\langle f\rangle}$, where $f$ is a irreducible polynomial of degree $k$ and $p$ is a prime (and $\langle f\rangle$ the ideal generated by $f$), has cardinality $p^k$ and how to determine the dimension of $\mathbb{F}_p[x]/{\langle f\rangle}$ as a vector space ?
86430
Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.  Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.  I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this.   I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$.   Help/suggestions are appreciated.
85504
Let $n\in\mathbb{N}$ and $A\in M_n(\mathbb{F})$. Assume that the minimal polynomial of the square matrix $A$ is defined as the minimal polynomial of the linear operator $x\mapsto Ax$ for all $x\in\mathbb{F}^n$, on $\mathbb{F}^n$. Now, assume that the minimal polynomial of $A$ splits over $\mathbb{F}$. Show that $A$ and $A^T$ are similar.  Two matrices $A$ and $B$ similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$. Finding such an $S$ for $B=A^T$ will solve the problem. Since the minimal polynomial of $A$ splits over $\mathbb{F}$, let us denote it by $$m_A(t) = (t-\lambda_1)^{m_1}...(t-\lambda_k)^{m_k}$$ for some $k\in\mathbb{N}$. We know that $m_A(A)$ corresponds to the zero linear operator, i.e. $m_A(A)x=0$, $\forall x\in \mathbb{F}^n$. I quickly noticed that if matrices $P_1,P_2,...P_q$ commute, then $(P_1P_2...P_k)^T = P_1P_2...P_k$. It may be useful to replace each $P_i$ with $(A-\lambda_i)^{m_i}$? I'm not sure how to proceed from here. I'd appreciate any hints!
45109
I'm currently trying to solve this problem.      Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.   I solved the ideal part.   Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$.   Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$.   Therefore $f(P)$ is an ideal.   The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime.   I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.  Any help would be appreciated.  Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.
94279
Let $F$ be a field, and $p(x)$ an irreducible polynomial in $F[X]$. Define $K=F[X]/\langle p(x)\rangle$. For every $a\in F$, denote $\bar{a}=\left&lt;p(x)\right&gt;+a$. How do I show that $\bar{x}\in K$ is a root of $p(x)$?.
88414
$F$ is an extension field of a field $K$. Let $F$  be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that $(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$ $(ii)$  The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$. $(iii)$  The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$    So, here is my attempt for (i). Let  $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$.  Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well.  Thanks.  ADDED:  Attempt at (iii).   but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$ hints...  The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$. Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...
55731
I've been trying to show that   $$ K[X,Y]/(Y-X^2)\cong K[X] $$  where $K$ is a field, $K[X]$ and $K[X,Y]$ are the obvious polynomial rings over the indeterminates $X$ and $Y$ and $(Y-X^2)$ is the ideal generated by the polynomial $Y-X^2$.  Though I'm sure there's a fairly easy way to find an explicit isomorphism between the two rings, the following argument jumped out at me:   If we substitute in a value for $X$ - $x$, say, then the ideal $(Y-x^2)$ is a maximal ideal of $K[Y]$.  So the quotient $K[Y]/(Y-x^2)$ is a field; in fact, the homomorphism $K[Y]\to K:P(Y)\mapsto P(x^2)$ and is clearly surjective, so the quotient is isomorphic to $K$.    I'd like to be able to deduce from this that $K[X,Y]/(Y-X^2)\cong F[X]$, but I can't see a nice way to do it.  I know that the 'substitution' maps $P(X,Y)\mapsto P(x,Y)$ are homomorphisms, but I can't see a nice way of pulling all these homomorphisms back to the polynomial ring in two variables.    Or maybe I'm completely wrong and there is no way to turn this into a valid argument.  Can anyone help me?
89566
We know the following fact from field theory. Let $F$ be a field and $p(X)$ an irreducible polynomial in $F[X]$. Then we can find a field extension $L$ of $F$ such that $p(X)$ has a root in $L$. Proof: Clearly the ideal $\langle p(X)\rangle$ generated by $p(X)$ is maximal in $F[X]$. So $F[X]/\langle p(X)\rangle$ is a field. Let us denote it by $L$. So $L$ is a field extension of $F$ since we have $F \hookrightarrow F[X] \xrightarrow{\rm \pi} L$ in a natural way. Note that $F[X] \leq L[X]$ and $\pi(X)$ is a root of $p(X)$ in $L[X]$.  Is it true that $\pi(X)$ is the unique root of $p(X)$ in $L$ ? What is the relation between $\pi(X)$ and other roots of $p(X)$ if $deg(p) &gt; 1$ ?  Thanks!