(March 2009) Let \(F\) be a field, let \(R=F[x]\) be the ring of polynomials in one variable with coefficients in \(F\), and let \(f(x)\in R\) be a polynomial of degree \(2009\). Let \[ I=\{g(x)f(x)\mid g\in R\} \] be the set of all polynomials which are multiples of \(f(x)\). It is a fact, which you may assume, that \(I\) is an ideal of \(R\). (a) For any \(g,h\in R\), prove that \(I+g=I+h\) if and only if \(g-h\) is divisible by \(f\). (b) Prove that for any \(g\in R\), there is a unique polynomial \(h\in R\) with \(\deg h<2009\) such that \(I+g=I+h\).
Solution
Proof. (a) Given g,h∈R, we have I +g =I +h if and only if g−h∈I, by the coset relation. But the definition of I implies that g−h∈I if and only if g−h is a multiple of f. (b) Given g∈R: (Existence) The division algorithm yields q,h∈ F[x] such that g =qf +h with deg h< degf = 2009. Thusg−h =qf∈I, so by part (a), I +g =I +h, proving existence. (Uniqueness) Suppose we are givenh,h′∈R withI+g =I+h′ =I+h′, and with degh, degh′ < 2009. Then I +h =I +h′, so by part (a), h−h′ is a multiple of f. That is, h−h′ =kf for some k∈R. If k⁄= 0, then deg(h−h′) = deg(kf ) = degk + degf≥ degf = 2009. However, the fact that deg h, degh′ < 2009 implies that deg(h−h′)< 2009 also, a contradiction. Thus, we must have k = 0, and hence h−h′ = 0. That is, h =h′, proving uniqueness. QED Every Ideal in k[x] is Principal. Recall that a principal ideal in a commutative ring consists of all multiples of a fixed element. Problem 27 features principal ideals in k[x]. This is no accident, since every ideal in k[x] is of the form ⟨f⟩ ={gf|g∈k[x]}. Furthermore,f is unique up to multiplication by a nonzero constant ofk. Since the units of k[x] are precisely the nonzero constants, we can equivalently say that f is unique up to multiplication by a unit. Irreducible Polynomials and Maximal Ideals in k[x]. Know the following: • The definition of reducible and irreducible polynomial in k[x]. • The irreducibility criterion for polynomials f∈k[x] of degree 2 or 3: such a polynomial is irreducible if and only if it has no roots in the field k. (See problem 28 below for a proof of this fact for cubic polynomials; can you do the proof for quadratic polynomials?) • An ideal I⊆k[x] is maximal if and only if I =⟨f⟩, where f∈k[x] irreducible. That is: f is irreducible ⇐⇒ ⟨f⟩⊆ k[x] is maximal ⇐⇒ k[x]/⟨f⟩ is a field. Here some problems about these concepts.