Polynomials and fields · Definition

Polynomial ring

modern algebrapolynomials

Polynomials are the first example of a ring built from another one, and the place where the factorization of integers finds its analogue.

Definition

The polynomial ring \(R[x]\) over a commutative ring \(R\) with identity consists of the finite sums of terms \(a_ix^i\), with coefficients \(a_i\in R\) and an indeterminate \(x\).

\[\Big(\sum_i a_ix^i\Big)\Big(\sum_j b_jx^j\Big)=\sum_k\Big(\sum_{i+j=k}a_ib_j\Big)x^k\]

A polynomial is a formal expression: two polynomials are equal exactly when all their coefficients agree. This sets it apart from the function it defines. In \(\mathbb{Z}_2[x]\), the polynomials \(x^2\) and \(x\) are different, even though as functions they agree at \(0\) and \(1\). That is why we write \(R[x]\) and speak of the indeterminate \(x\), not a variable.

The degree of a nonzero polynomial is the largest exponent with nonzero coefficient. Always \(\deg(f+g)\le\max(\deg f,\deg g)\), and if the leading coefficient of \(f\) times that of \(g\) is nonzero, then \(\deg(fg)=\deg f+\deg g\). This fails when there are zero divisors: in \(\mathbb{Z}_6[x]\), \((2x+1)(3x+1)=5x+1\) has degree 1, not 2.

Properties of \(R\) pass to \(R[x]\): if \(R\) is commutative, so is \(R[x]\); if \(R\) is an integral domain, so is \(R[x]\), and its units are those of \(R\). When \(R\) is not a domain there can be unexpected units: in \(\mathbb{Z}_4[x]\), \(1+2x\) is its own inverse. The ring \(R\) is identified with the constant polynomials, and \(R[x,y]=R[x][y]\) builds several indeterminates.

A polynomial ring is never a field even if \(R\) is one: \(x\) has no inverse, by the degree argument. What does hold, when \(R=F\) is a field, is a division algorithm, the topic of the next concept, and with it polynomials inherit an arithmetic similar to that of \(\mathbb{Z}\). Also, evaluation \(f\mapsto f(a)\) is a ring homomorphism, to which the first isomorphism theorem applies.

Think about it

In \(\mathbb{Z}_2[x]\), are the polynomials \(x^2+x\) and \(0\) equal?

Not as polynomials, since their coefficients differ; yes as functions, because \(a^2+a=0\) for \(a=0,1\) in \(\mathbb{Z}_2\).

Why does \(x\) have no inverse in \(F[x]\) when \(F\) is a field?

If \(xg=1\), degrees would give \(1+\deg g=0\), which is impossible.

Exercises 10

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §9.1
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 16
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §3.9
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §22
  • Serge Lang, Algebra, Revised 3rd edition, Springer, 2002. cap. II §3