By Garrett Birkhoff

This vintage, written by means of younger teachers who turned giants of their box, has formed the certainty of recent algebra for generations of mathematicians and is still a worthwhile reference and textual content for self research and school classes.

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6. We construct a field F with 25 elements using the polynomial x 2 - 2, which is easily seen to be irreducible in F5[X] (there is no square root of 2 in F5). The elements of F we view as two-dimensional vectors (a, b) with a, bE F5 {O, 1, 2, 3, 4}. Addition in F is componentwise: = To define multiplication we view (a, b) as the linear polynomial ax + b, and use the rule (a1x + bt) . (a2x + b2) = (a1 a2x2 = + (a 1b2 + a2 bt}x + b1 b2) (a 1 b2 + a2bt}x + (b 1 b2 + 2ala2), which follows from the ordinary rules for multiplying polynomials and the fact that x 2 == 2 (modx 2 - 2).

Hint: Let a be a prim. root in F73. J 4. 1 we saw that the equation x 2 - 1 = 0 has four roots in the ring Zs, of arithmetic mod 8. a. How many roots does x 2 - 1 = 0 have in the ring Z16? b. How many roots does x 2 - 1 = 0 have in the ring Z2n? 5. Let p be a prime number. What is fj>(pm)? J 6. 6 we asserted that the polynomial x 2 - 2 is irreducible over the field F5. Verify this, and more generally determine which of the polynomials x 2 - a, where a E F5 , is irreducible. For those polynomials which are not irreducible, give the factorization into irreducible factors.

There are p2 elements in F of the form alWI + a2W2. If q = p2, again, QED. 1 sheds considerable light on the structure of the additive group of F. It shows that the elements of F can be viewed as m-tuples of elements from Fp, and that if we take this view, then (al, ... ,am}+(bl , ... , bm ) = (al +bt. , am +bm ). However, it tells us very little about the multiplicative structure of F. We shall now proceed to investigate the multiplicative structure. As we shall see, the central fact is that the multiplicative group of F is cyclic of order q - 1.

### Álgebra Moderna by Garrett Birkhoff

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