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let $a$ a complex number , and $f$ be an irreducible polynomial with integer coefficients such that : $ f(a)=0$

  • 1) Show that the set : $\{ g(a) \mid g \in \mathbf{Z}[X]\}$ is a ring isomorphic to $\mathbf{Z}^{n}$ respect to their group structure where $n=\deg f$

  • 2) show that every non null Ideal of the previous ring is isomorphic to $\mathbf{Z}^{n}$ respect to their group sturcture

Help me to solve these questions with some hints please :)

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Is $f$ in $\mathbf Q[X]$? What does $\mathbf Q(a)/\mathbf Q \in \mathbf Z[X]$ mean? –  Dylan Moreland Jul 23 '11 at 20:11
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@mathfan: "Show that $Q(a)/Q\in Z[X]$ is a ring isomorphic to..." What?! First, what is $Q$ and what is $Q(a)$? Second, if $Q(a)/Q$ is an element of $Z[X]$, then it's a polynomial, not a ring. –  Arturo Magidin Jul 23 '11 at 20:12
    
@ arturo Magidin : sorry it is a latex problem : the set { Q(a)/Q \in Z[X]} is the set of images of a by all polynomials in Z[X] –  mathfan Jul 23 '11 at 20:21
    
@mathfan: No, it's a concept problem: if something is an element of a polynomial ring, then chances are that it is a polynomial, not a ring itself. What set? What is $Q$? (To get { and } to show up, you need to use \{ and \}, but even with them, $\{Q(a)/Q \in Z[x]\}$ is still confused and confusing). –  Arturo Magidin Jul 23 '11 at 20:22
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You also want $f$ to be monic, I think (What is the rank of $\mathbf Z[\frac 12]$?). –  Dylan Moreland Jul 23 '11 at 20:40

2 Answers 2

What you are trying to construct is the quotient ring $\mathbb{Z}[x]/(f(x))$, which is the ring of polynomials, modulo the equivalence where two polynomials which differ by a multiple of $f(x)$ are viewed the same. The ring structure is inherited by the ring structure of $\mathbb{Z}[x]$. Let $R$ be the ring you defined in your question. If you have worked with quotient rings before, then a good first step is to verify that the map $\mathbb{Z}[x]/(f(x))\to R\subset\mathbb{C}$ defined by $g(x)\mapsto g(a)$ is an isomorphism (or, by the first isomorphism theorem, that the kernel of the map $\mathbb{Z}[x]\to \mathbb{C}$ given by evaluation at $a$ is exactly the ideal generated by $f(x)$.

The second step is to try to show that $\mathbb{Z}[x]/(f(x))\cong \mathbb{Z}^n$ as abelian groups, where $n$ is the degree of $f(x)$. For this to hold, you MUST have that that $f(x)$ is monic. In this case, $1,x,x^2,\ldots, x^{n-1}$ will be a basis for the ring viewed as an abelian group.

To understand the ideals, it is helpful if you know that subgroups of free abelian groups are free (this follows from the structure theorem for finitely generated abelian groups, but an independent proof is given here, which does not require finite generation). Then, it suffices to show that if $g(x)$ is a polynomial not in $(f(x))$, then $g(x), xg(x), x^2(g(x), \ldots, x^{n-1}g(x)$ are all $\mathbb{Z}$-linearly independent in the ring. This requires the fact that $\mathbb{Z}[x]$ is a unique factorization domain.

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Aaron ,in the source of this problem they don't mention that f is monic , so i got stuck on the first question , can you give a counter example with f a non monic polynomial? , thank you very much –  mathfan Jul 24 '11 at 0:01
    
If you take $f(x)=2x-1$, your ring is $Z[1/2]$, which contains all the dyadic rationals. This is not even finitely generated as an abelian group: if you take any finite generating set, you can only get the denominators to be as large as the largest denominator of a generator. The problem is that you can't divide a polynomial by an arbitrary polynomial when you're working over $\mathbb{Z}$ and still get a remainder of smaller degree (with integer coefficients). However, the division algorithm works fine if you divide by a monic polynomial. This is the key to step 2. –  Aaron Jul 24 '11 at 0:28
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@Aaron What "requires $\mathbb Z$ is a UFD", and why? –  Bill Dubuque Jul 24 '11 at 14:55
    
@Bill That was a typo. I meant $\mathbb{Z}[x]$. Essentially, we need that the element $f(x)\in \mathbb{Z}[x]$ is not only irreducible but prime. That way, if some non-zero linear multiple of $h(x)g(x)$ with $\deg(h)<n$ were in $(f(x))$, we could find a lower degree minimal polynomial for $a$, and so the multiples of $x^kg(x)$ of $g(x)$ above must be linearly independent. –  Aaron Jul 24 '11 at 20:19

This is pretty confused.

First: it's not enough to just give a set and ask about "ring structure" or "group structure"; you need to say what the operations are. Presumably, you want the operations to be inherited form $\mathbb{C}$ (where the values of $g(a)$ lie).

Second: the problem is incorrect as stated because you have placed no condition son $f$. Take $f(x) =x-\pi$, a polynomial that is irreducible in $\mathbb{C}[x]$ (you never said what kind of polynomial it was, and since you open by considering complex numbers, the obvious interpretation would be that $f$ is a polynomial with complex coefficients). Then $a=\pi$ satisfies the desired condition, but the set $\{g(a)\mid g(x)\in\mathbb{Z}[x]\}$ is isomorphic to $\mathbb{Z}[x]$, and its additive group is free abelian of countably infinite rank, and not isomorphic $\mathbb{Z}^1$, as the statement requires.

So presumably, you want $f(x)$ to be a polynomial with integer or rational coefficients. (Which?) But even more is required, as Pete's comment shows.

So my first hint (nay, strong suggestion), which is very applicable to all math problems is: make sure you write down all the hypothesis!

Now: part (1) has two parts: you want to show that the set is a ring (under the addition and product of complex numbers); and you want to show that as an abelian group it is free of rank $n=\deg f$.

For the first part of part 1, you need to show that the set is nonempty, and that if $g(a)$ and $h(a)$ are in the set, then so is $g(a)-h(a)$; i.e., that you can find a polynomial $k(x)\in\mathbb{Z}[x]$ such that $k(a)$ evaluates to the same thing as $g(a)-h(a)$. This shows it's a subgroup of $\mathbb{C}$. Then you need to show that it is closed under multiplication: if $g(a),h(a)$ are in the set, then so is $g(a)h(a)$; that is, there is a polynomial $k(x)\in\mathbb{Z}[x]$ such that $k$, evaluated at $a$, equals the product of the values of $g$ and $h$ when evaluated at $a$. This part should be easy.

For the second part, try showing that the group is freely generated by $1$, $a$, $a^2,\ldots,a^{n-1}$; you'll need to figure out exactly what the correct conditions on $f$ are supposed to be, otherwise this is going to be very hard (if not false; see above).

For part 2, suppose that you have a nonzero element in the ideal, and think about what happens when you multiply it by $1$, $a$, $a^2$, $a^3,\ldots,a^{n-1}$.

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