By Madhu Sudan

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**Example text**

JS j = qd To see that S is a eld, it su ces to check that for all ; 2 S, + ; straightforward: ; ; 1 2 S { which is fairly 1. ( + )qd = qd + qd + multiples of p = qd + qd = + . Therefore, + 2 S. 2. ( )q d = q d q d = : Therefore, 2 S. 3. ( )qd = 1qd qd = 1qd = : The last equality follows from the previous one because qd is odd whenever q is odd, and q is even only if the underlying prime p is 2, in which case, 1 = 1 for that eld. Therefore, 2 S. 4. ( 1)qd = ( qd ) 1 = 1. Therefore, 1 2 S. To compute the cardinality of S observe that S cannot have more than qd elements since xqd x has at most qd roots (since its degree is qd ).

APPLICATION: DECODING REED-SOLOMON CODES 39 jth step we nd gj ; hj such that f gj hj ( mod p2k ). The rest of the argument will also be analogous to the bivariate case, for instance instead of computing resultants over R(y) (as was done in the case of R x; y]) we will now compute resultants over R(y1 ; y2 ; : : :; yn ), etc. We omit the details which can be easily lled in. The second approach is more e cient than the rst one we suggested, but still both the methods perform poorly for one non-circumventable (at least apparently) reason.

Why does the existence of g~ and lk yield anything in Step 3(b)? 3. THE FACTORING ALGORITHM 35 We answer these questions in turn. To do so we have to study the properties of the polynomials g0 and gk (relative to f). Notice that g0 does not necessarily correspond to a factor of f. However, g0 does divide (modulo y) an irreducible factor of f. In other words, there exist polynomials g(x; y); h(x; y) such that f(x; y) = g(x; y)h(x; y), g(x; y) is irreducible and g(x; y) = g0 (x; y)l0 (x; y) (mod y) and h0 (x; y) = l0 (x; y)h(x; y) (mod y).

### Algebra and Computation by Madhu Sudan

by Daniel

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