Algebra of polynomials

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Question. What is a polynomial? In schools we are taught that a polynomial is a function of the form

$\displaystyle p(x)=a_nx^n+a_{n-1}x^{n-1}\dots +a_1x+a_0,$     (1)

where $ a_i$ are real numbers. In particular we can draw the graph of a polynomial, compute values of a polynomial, find derivatives, etc. In contrast to that, in this notes we will adopt a purely algebraic point of view.


Definition:. A ring of polynomials is just the set of objects of the form $ ($1$ )$, which can be added and multiplied.


We can think of polynomials as of a kind of a game. We have pieces called $ 1,x,x^2,x^3,\dots$. Let us take 3 pieces ``$ x^3$'', 5 pieces ``$ 1$'' and half of the piece $ x$. So, we have a ``polynomial'' $ x^3+0.5x+5$.


Remark:We should not worry to much about polynomials like $ \pi x-\sqrt3$. Minus just means that we give out some pieces instead of taking and there is nothing wrong with taking $ \pi$ fraction of ``$ x$'', after all it is just a little more than 3.14 $ x$ and a little less than 3.15 $ x$.


Now, we can add polynomials in the usual way: take polynomials $ p_1$ and $ p_2$ and form the polynomial $ p_1+p_2$ just putting all the pieces together. For example, if we had the polynomials $ x^3+0.5x+5$ and $ 2x^2+3x+1$, the result of addition as you may expect would be $ x^3+2x^2+2.5x+6$.

A more sophisticated operation allowed in our game is multiplication. It works as follows. First let us explain what multiplication by a number means. To multiply a polynomial $ x^3+0.5x+5$ by, say, 4 means take 4 copies of it and put everything together, so $ 4\times(x^3+0.5x+5)=4x^3+2x+20$. Note that, in particular, the multiplication by 1 does nothing to a polynomial.

Next, what is multiplication by $ x^k$? It is just an operation of trading: we say we multiply by $ x^k$ and we just trade all pieces of type ``$ x^n$'' to pieces ``$ x^{n+k}$''. We get $ x^k\times(x^3+0.5x+5)=x^{k+3}+0.5x^{k+1}+5x^k$.

Finally, if we want to multiply by polynomial which consists of several terms (pieces) we multiply by each piece separatly and add the results. Here is an example:

$\displaystyle (x+a)\times(x-a)=x\times(x-a)+a\times(x-a)=(x^2-ax)+(ax-a^2)=x^2-a^2. $

Thus we obtain our usual operations of addition and multiplicaton.


Remark:So far it looks that we did not discover much. However, we will see that such a point of view can lead to some interesting mathematics.

Sets of objects where we have two operations: addition and multiplication with properties similar to the ones we described are called rings (in some cases a ring is also called an algebra, in fact we do have an algebra).

The set of pieces $ \{1,x,x^2,x^3,\dots\}$ is called a basis of our ring.

Notice multiplying $ x$ to itself many times, we generate all the ``pieces'' $ x^k$. This is why $ x$ is called a generator of our ring.

Notice also that our multiplication is commutative, it does not matter in what order we multiply: $ p_1(x)p_2(x)=p_2(x)p_1(x)$. Later we will deal with noncommutative rings where this is no longer true.


Let us use a similar approach to differention.


Definition:A differentiation $ d=\frac{d}{dx}$ is another operation which given a polynomial $ p$ produces another polynomial $ dp$. This operation has the properites:

$\displaystyle d(a_1p_1(x)+a_2p_2(x))=a_1dp_1(x)+a_2dp_2(x),$     (2)

called linearity (or the sum rule),
$\displaystyle d(p_1(x)p_2(x))=(dp_1(x)) p_2(x)+p_1(x)(dp_2(x)),$     (3)

called Leibnitz rule (the product rule) and $ dx=1$. Here $ p_i(x)$ are polynomials and $ a_i$ are real numbers.


Lemma 1   We have $ da=0$, for any real number $ a$, also $ dx^n=nx^{n-1}$. More generally, for $ k\leqslant n$, we have

$\displaystyle d^k(ax^n)=n(n-1)\dots(n-k+1)ax^{n-k}=a\frac{n!}{k!}x^{n-k}, $

and $ d^kx^n=0$ if $ k>n$.

Proof. Let us compute the derivative of $ x^2$.

$\displaystyle d(x^2)=d(xx)=(dx)x+xdx=x+x=2x. $


Exercise:Finish the proof of the lemma.


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Now we have two different operations: multiplication by $ x$ and differentiation $ d$. What if we apply them in different orders?

Lemma 2   $ dx-xd=1$. In other words, for any polynomial $ p(x)$, we have $ d(xp(x))-xd(p(x))=p(x)$.


Exercise:Prove the lemma.