Polynomials and fields · Definition

Irreducible element

modern algebrarings

Before you start you need to know

Irreducibles are the atoms of factorization: the pieces that cannot be broken further and from which everything else is built.

Definition

A nonzero nonunit element \(p\) of an integral domain \(D\) with no proper factorization: in every product of two factors equal to \(p\), one of them is a unit.

\[p=ab\ \Rightarrow\ a\ \text{ or }\ b\ \text{is a unit}\]

In \(\mathbb{Z}\) the irreducibles are the prime numbers and their negatives. In \(F[x]\), with \(F\) a field, they are the polynomials that are not a product of two polynomials of lower degree. Being irreducible depends on the ring: \(x^2+1\) is irreducible in \(\mathbb{R}[x]\) but factors as \((x-i)(x+i)\) in \(\mathbb{C}[x]\), and \(x^2-2\) is irreducible in \(\mathbb{Q}[x]\) but not in \(\mathbb{R}[x]\).

Multiplying by a unit changes nothing: if \(p\) is irreducible and \(u\) is a unit, \(up\) is irreducible too (they are called associates). That is why one speaks of factoring "up to associates". A polynomial of degree 2 or 3 over a field is irreducible if and only if it has no roots in that field; for higher degrees this test is no longer enough, since \(x^4+4=(x^2+2x+2)(x^2-2x+2)\) has no rational roots and is still reducible.

In \(\mathbb{Z}\) an irreducible that divides a product divides one of the factors, but this is not automatic. In \(\mathbb{Z}[\sqrt{-5}]\), 2 is irreducible (its norm 4 does not split into norms 2) and divides \(6=(1+\sqrt{-5})(1-\sqrt{-5})\) without dividing either factor. Telling irreducible from prime is what is needed to understand when factorization is unique.

Think about it

Why can an irreducible element not be a unit?

By definition it must be a nonunit: this way the "trivial" factors (the units) do not count as pieces of a factorization.

Is \(2x\) irreducible in \(\mathbb{Q}[x]\)? And in \(\mathbb{Z}[x]\)?

In \(\mathbb{Q}[x]\) yes, because 2 is a unit. In \(\mathbb{Z}[x]\) no: \(2x=2\cdot x\) with both factors nonunits.

Exercises 10

  1. Which of these elements is irreducible in \(\mathbb{Z}\)?
  2. Which polynomial is irreducible in \(\mathbb{Q}[x]\)?
  3. Which polynomial is reducible in \(\mathbb{Q}[x]\)?
  4. Which polynomial is irreducible in \(\mathbb{R}[x]\)?
  5. In \(\mathbb{Z}[i]\) with norm \(N(a+bi)=a^2+b^2\), which of these elements is irreducible?
  6. In \(\mathbb{Z}[\sqrt{-5}]\), with \(N(a+b\sqrt{-5})=a^2+5b^2\), which element is irreducible?
  7. In \(\mathbb{Z}[\sqrt{-5}]\), the element 2 divides \(6=(1+\sqrt{-5})(1-\sqrt{-5})\) but divides neither factor. What follows?
  8. In \(\mathbb{Z}\), how many elements are associates of 5 (including 5 itself)?
  9. A polynomial \(f\in\mathbb{Q}[x]\) of degree 3 is irreducible if and only if:
  10. Which statement about a field \(F\) is true?

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §8.3
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 18
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §3.8
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §45
  • Serge Lang, Algebra, Revised 3rd edition, Springer, 2002. cap. II §5