Polynomials over a field

olivierd

Aug 20, 2018 6 min read

Polynomials over a field

In this context, we consider a commutative field (simply: field) $\K$ . The polynomials will be constructed as a certain associative unital algebra over $\K$ (“ $\K$ -aua”), together with a distinguished element called the formal variable.

$\K$ itself, as a vector space over itself and the usual multiplication in $\K$ as the third law, is a $\K$ -aua,

Morphisms between two $\K$ -auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.

All that will be said of the polynomials over $\K$ will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.

Definition

The polynomials over $\K$ is any $\K$ -aua $A$ together with a distinguished element $X$ of $A$ such that the following universal property holds:

For any $\K$ -aua $A'$ and any $a \in A$ , there exists exactly one morphism $\phi: A \to A'$ such that $\phi(X) = a$ .

Unicity up to an isomorphism

Let $(A_1, X_1)$ and $(A_2, X_2)$ be two sets of polynomials over $\K$ . The universal property implies the existence of a morphism $\phi_1: A_1 \to A_2$ such that $\phi(X_1) = X_2$ , and of a morphism $\phi_2: A_2 \to A_1$ such that $\phi(X_2) = X_1$ .

We then have $(\phi_2 \circ \phi_1)(X_1) = X_1$ . Hence $\phi_2 \circ \phi_1$ is a morphism $A_1 to A_1$ that maps $X_1$ to $X_1$ . Now $\Id_{A_1}$ is another such morphism. By virtue of the universal property of $(A_1, X_1)$ , it follows that $\phi_2 \circ \phi_1 = \Id_{A_1}$ , that is, is the $\K$ -aua category identity on $A_1$ .

The same reasoning shows that $\phi_1 \circ \phi_2$ is the $\K$ -aua category identity on $A_2$ .

Hence $\phi_1$ is an isomorphism $A_1 \to A_2$ that maps the formal variable of $A_1$ to that of $A_2$ .

Characterization

Let $A$ be an associative unital algebra over $\K$ , and $X$ an element of $A$ . Then $(A, X)$ represents the polynomials over $\K$ if and only if the family $(X^n)_{n \in \N}$ , with $X^0 = 1_A$ and $X^{n+1} = X X^n$ , is a family-base of the $\K$ -vector space $A$ .

Let us first suppose that the latter condition holds on $(A, X)$ , and prove the universal property.

Let $A'$ be a $\K$ -aua and $a'$ an element of $A'$ .

Let us suppose that $\phi$ is a morphism $A \to A'$ such that $\phi(X) = a'$ . Then for all $n \in \N$ , we have $\phi(X^n) = a'^n$ . Since $\phi$ as a $\K$ -aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and $(X^n)_{n \in \N}$ is a basis, $\phi$ necessarily is the unique linear mapping $A \to A'$ such that $\forall n \in \N, \phi(X^n) = a'^n$ . Now it is easy to check that this linear mapping does preserve the multiplication in $A$ , and is hence a $\K$ -aua morphism.

Thus the universal property is proven for $(A, X)$ .

Let us now suppose the universal property for $(A, X)$ , and show that $(X^n)_{n \in \N}$ is a family-base of the $\K$ -vector space $A$ .

Let us first show that $(X^n)_{n \in \N}$ is linearly independent.

Let $(A', \alpha)$ be a free $\K$ -vector space on set $\N$ . This entails that $(\alpha(n))_{n \in \N}$ is a family-base of $A'$ . To make $A'$ into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.

A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in $A'$ by:

$\forall n_1, n_2 \in \N, \alpha(n_1) \cdot \alpha(n_2) = \alpha(n_1 + n_2)$

We then have $(\alpha(n_1) \cdot \alpha(n_2)) \cdot \alpha(n_3) =$ $\alpha(n_1 + n_2) \cdot \alpha(n_3) =$ $\alpha(n_1 + n_2 + n_3) =$ $\alpha(n_1) \cdot \alpha(n_2 + n_3) =$ $\alpha(n_1) \cdot (\alpha(n_2) \cdot \alpha(n_3))$ .

Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes $A'$ an associative algebra.

$\alpha(0)$ is a unit; hence $A'$ is an associative unital algebra.

Applying the universal property of $(A, X)$ , we find that there exists exactly one $\K$ -aua morphism $\phi: A \to A'$ such that $\phi(X) = \alpha(1)$ .

We then have $\phi(X^0) = \phi(1_A) = 1_{A'} =$ $\alpha(0)$ ; $\phi(X^1) =$ $\phi(X) = \alpha(1)$ ; and, for any $n > 1$ , $\phi(X^n) =$ $\phi(X \cdot \ldots \cdot X) =$ $\phi(X) \cdot \ldots \cdot \phi(X) =$ $\alpha(1) \cdot \ldots \cdot \alpha(1) =$ $\alpha(1 + \ldots + 1) = \alpha(n)$ . Hence for all $n \in \N$ , $\phi(X^n) = \alpha(n)$ .

Let us consider a linear combination of the family $(X^n)_{n \in \N}$ that vanishes; that is, a finite sequence of values of $\K$ , $(a_0, a_1, \ldots a_n)$ , such that $\sum_{i = 0}^n a_i X^i = 0_A$ . Taking the image of this linear combination by $\phi$ , we obtain $\sum_{i = 0}^n a_i \alpha(i) = 0_{A'}$ . Since the $\alpha(i)$ form a basis, all the $a_i$ must be zero.

Thus the family $(X^n)_{n \in \N}$ is linearly independent.

Lastly, we must show that this family generates the $\K$ -vector space $A$ .

Let $B$ be the sub-aua of $A$ generated by $\{X\}$ .

There exists a unique aua morphism $\phi: A \to B$ such that $\phi(X) = X$ .

Let then $\phi'$ be $\phi$ , but with all of $A$ as codomain; in other words, $\phi': A \to A, x \mapsto \phi(x)$ . It is clear that $\phi'$ too is an aua morphism, and it is such that $\phi'(X) = X$ .

But there is exactly one aua morphism $A \to A$ that maps $X$ to $X$ ; and $\Id_A$ is such.

Hence $\phi' = \Id_A$ .

The image of $\phi'$ is that of $\phi$ , and is a subset of $B$ . But the image of $\Id_A$ is $A$ itself. Hence $B = A$ .

Thus the sub-aua of $A$ generated by $\{X\}$ is $A$ itself. This sub-aua is the set of linear combinations of all the powers of $X$ ; hence the sub-vector space generated by $X^0, X^1, X^2, \ldots$ is $A$ .

Thus $(X^n)_{n \in \N}$ is a family-base of the $\K$ -vector space $A$ .