Section I.5.2

Polynomial ring

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Problem I.5.2.1

The polynomials f(X)=X5+3X4+X3+4X2−3X−1,g(X)=X2+X+1f(X)=X^{5}+3 X^{4}+X^{3}+4 X^{2}-3 X-1, g(X)=X^{2}+X+1 can be regarded as belonging to the ring Z[X]\mathbb {Z}{}[X] or, say, to the ring Z5[X]\mathbb {Z}_{5}[X], depending on how their coefficients are interpreted. Applying the division algorithm with remainder, show that in the first case f(X)f(X) is not divisible by g(X)g(X), while in the second it is. Is the opposite situation possible?

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Problem I.5.2.2

Prove, using Theorem 3, that if FF is a field, then the group of all automorphisms of the ring F[X]F[X] that are the identity on FF is isomorphic to the group of transformations X↦aX+bX \mapsto a X+b, where a,b∈Fa, b \in F and a≠0a \neq 0.

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Problem I.5.2.3

Show that a polynomial f∈F[X1,…,Xn]f \in F\left[X_{1}, \ldots , X_{n}\right] is a form of degree mm (see the proof of Theorem 4) if and only if f(tX1,…,tXn)=tmf(X1,…,Xn)f\left(t X_{1}, \ldots , t X_{n}\right)=t^{m} f\left(X_{1}, \ldots , X_{n}\right), where tt is a new variable.

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Problem I.5.2.4

Show that the number of distinct monomials in nn independent variables of total degree mm equals (m+n−1m)\binom {m+n-1}{m}.

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Problem I.5.2.5

Returning to the definitions of item 1, let us consider the collection A[[X]]A[[X]] of so-called formal power series f(X)=∑i⩾0aiXif(X)=\sum_{i \geqslant 0} a_{i} X^{i} in the variable (indeterminate) XX, or, if one likes, of sequences (a0,a1,a2,…)\left(a_{0}, a_{1}, a_{2}, \ldots \right) with any, possibly infinite, number of nonzero coefficients aia_{i}, belonging to a commutative ring AA. Operations with formal power series from A[[X]]A[[X]] are carried out according to the same rules as operations with polynomials:

(∑aiXi)+(∑biXi)=∑(ai+bi)Xi(∑aiXi)⋅(∑biXi)=∑ckXk,ck=∑i+j=kaibj. \begin{gathered} \left(\sum a_{i} X^{i}\right)+\left(\sum b_{i} X^{i}\right)=\sum \left(a_{i}+b_{i}\right) X^{i} \\ \left(\sum a_{i} X^{i}\right) \cdot \left(\sum b_{i} X^{i}\right)=\sum c_{k} X^{k}, \quad c_{k}=\sum _{i+j=k} a_{i} b_{j}. \end{gathered}

Show that the set A[[X]]A[[X]], considered together with these operations, is an associative and commutative ring with identity 1=(1,01=(1,0, 0,…)0, \ldots ).

Since arbitrarily high powers XiX^{i} of the variable XX occur in the power series f=∑aiXif=\sum a_{i} X^{i}, instead of the degree deg⁡f\operatorname {deg} f, which no longer makes sense, it is natural to consider the order ω(f)\omega (f) — an integer equal to the smallest index nn for which an≠0a_{n} \neq 0 (one also sets ω(0)=+∞\omega (0)=+\infty).

Show that: i) ω(f−g)⩾min⁡{ω(f),ω(g)}\omega (f-g) \geqslant \min \left\{ \omega (f), \omega (g)\right\}; ii) ω(fg)⩾ω(f)+ω(g)\omega (f g) \geqslant \omega (f)+\omega (g).

If AA is an integral domain, then ω(fg)=ω(f)+ω(g)\omega (f g)=\omega (f)+\omega (g). In particular, together with AA, the ring A[[X]]A[[X]] is also an integral domain.

Show also that A[X]A[X] is a subring of A[[X]]A[[X]].

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Problem I.5.2.6

Polynomials and power series are often used as generating functions of various numerical quantities. We shall explain the meaning of operating with them using two simple examples.

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(a)

Establish the relation

∑i=0k(mi)(nk−i)=(m+nk) \sum _{i=0}^{k}\binom {m}{i}\binom {n}{k-i}=\binom {m+n}{k}

starting from the binomial formula ∑(ni)Xi=(1+X)n\sum \binom {n}{i} X^{i}=(1+X)^{n} in Z[X]\mathbb {Z}{}[X] and the obvious expansion (1+X)m(1+X)n=(1+X)m+n(1+X)^{m}(1+X)^{n}=(1+X)^{m+n}.

(b)

Find the number lnl_{n} of all possible ways of inserting parentheses in a product of length nn of elements of a set with one binary operation. To this end it is convenient to introduce the generating function — the formal power series

l(X)=∑n⩾1lnXn=X+X2+2X3+… l(X)=\sum _{n \geqslant 1} l_{n} X^{n}=X+X^{2}+2 X^{3}+\ldots

whose initial coefficients were already computed in item 3, § 1, Ch. 4. From the obvious recurrence relation

ln=∑k=1n−1lkln−k l_{n}=\sum _{k=1}^{n-1} l_{k} l_{n-k}

it follows that l(X)2=l(X)−Xl(X)^{2}=l(X)-X. Solving this quadratic equation, we find

l(X)=1−1−4X2 l(X)=\frac{1-\sqrt{1-4 X}}{2}

(the sign before the radical is determined by the condition ln>0l_{n}>0). But if a power series f(X)f(X) is such that fr=1+λX,r∈Nf^{r}=1+\lambda X, r \in \mathbb {N}, then

f(X)=1+∑k=1∞[∏i=0k−1(1r−i)](λX)kk! f(X)=1+\sum _{k=1}^{\infty }\left[\prod _{i=0}^{k-1}\left(\frac{1}{r}-i\right)\right] \frac{(\lambda X)^{k}}{k !}

(the Taylor series expansion, which can for now be taken on faith). In our case r=2,λ=−4r=2, \lambda =-4, and a simple substitution gives the final expression

ln=1n(2n−2n−1) l_{n}=\frac{1}{n}\binom {2 n-2}{n-1}

(note that ln=Cn−1l_{n}=C_{n-1} is the classical Catalan number).

The reader is invited to carry out all the intermediate calculations.