Editing Polynomial ring (section)

=== Polynomial evaluation ===
{{further|Polynomial evaluation}}
Let {{mvar|K}} be a field or, more generally, a [[commutative ring]], and {{mvar|R}} a ring containing {{mvar|K}}. For any polynomial {{mvar|P}} in {{math|''K''[''X'']}} and any element {{mvar|a}} in {{mvar|R}}, the substitution of {{mvar|X}} with {{mvar|a}} in {{mvar|P}} defines an element of {{math|''R''}}, which is [[Polynomial notation|denoted]] {{math|''P''(''a'')}}. This element is obtained by carrying on in {{mvar|R}} after the substitution the operations indicated by the expression of the polynomial. This computation is called the '''evaluation''' of {{math|''P''}} at {{math|''a''}}. For example, if we have 
:<math>P = X^2 - 1,</math>
we have 
:<math>\begin{align}
      P(3) &= 3^2-1 = 8, \\
  P(X^2+1) &= \left(X^2 + 1\right)^2 - 1 = X^4 + 2X^2
\end{align}</math>

(in the first example {{math|1=''R'' = ''K''}}, and in the second one {{math|1=''R'' = ''K''[''X'']}}). Substituting {{math|''X''}} for itself results in 
:<math>P = P(X),</math>

explaining why the sentences "Let {{mvar|P}} be a polynomial" and "Let {{math|''P''(''X'')}} be a polynomial" are equivalent.

The ''polynomial function'' defined by a polynomial {{mvar|P}} is the function from {{mvar|K}} into {{mvar|K}} that is defined by <math>x\mapsto P(x).</math> If {{mvar|K}} is an infinite field, two different polynomials define different polynomial functions, but this property is false for finite fields. For example, if {{mvar|K}} is a field with {{mvar|q}} elements, then the polynomials {{math|0}} and {{math|''X''<sup>''q''</sup> − ''X''}} both define the zero function.

For every {{math|''a''}} in {{math|''R''}}, the evaluation at {{mvar|a}}, that is, the map <math>P \mapsto P(a)</math> defines an [[algebra homomorphism]] from {{math|''K''[''X'']}} to {{math|''R''}}, which is the unique homomorphism from {{math|''K''[''X'']}} to {{math|''R''}} that fixes {{mvar|K}}, and maps {{mvar|X}} to {{mvar|a}}. In other words, {{math|''K''[''X'']}} has the following [[universal property]]:
:For every ring {{mvar|R}} containing {{mvar|K}}, and every element {{mvar|a}} of {{mvar|R}}, there is a unique algebra homomorphism from {{math|''K''[''X'']}} to {{mvar|R}} that fixes {{mvar|K}}, and maps {{mvar|X}} to {{mvar|a}}.

As for all universal properties, this defines the pair {{math|(''K''[''X''], ''X'')}} up to a unique isomorphism, and can therefore be taken as a definition of {{math|''K''[''X'']}}. 

The [[Image (mathematics)|image]] of the map <math>P \mapsto P(a)</math>, that is, the subset of {{mvar|R}} obtained by substituting {{mvar|a}} for {{mvar|X}} in elements of {{math|''K''[''X'']}}, is denoted {{math|''K''[''a'']}}.<ref>Knapp, Anthony W. (2006), ''Basic Algebra'', [[Birkhäuser]], p. 121.</ref> For example, <math>\Z[\sqrt{2}]=\{P(\sqrt{2})\mid P(X)\in\Z[X]\}</math>, and the simplification rules for the powers of a square root imply <math>\Z[\sqrt{2}]= \{a+b\sqrt 2 \mid a\in \Z, b\in \Z\}.</math>