Another way to determine if a polynomial is irreducible over a ring, is to check if the zeros of the polynomial are in the ring, ie, does p have a linear factor.

asked 2017-10-09 06:11:00 +0100

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For Z_5, write a "for" loop to check if p evaluated at every element of Z_5 is zero.

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"The zeros" are all the zeros?

A linear factor is only one factor?

If a (monic) polynomial over a (commutative) ring (with one) has a linear factor, than it is of course reducible. But the converse is obviously false, e.g. $(x^2+2)^2$ over integers, even over reals, or as is over some finite field...

This looks like homework, please insert the own trials and errors.

dan_fulea ( 2017-10-10 03:49:57 +0100 )edit

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Another way to determine if a polynomial is irreducible over a ring, is to check if the zeros of the polynomial are in the ring, ie, does p have a linear factor.

For Z_5, write a "for" loop to check if p evaluated at every element of Z_5 is zero.

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Another way to determine if a polynomial is irreducible over a ring, is to check if the zeros of the polynomial are in the ring, ie, does p have a linear factor. edit

For Z_5, write a "for" loop to check if p evaluated at every element of Z_5 is zero.

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Another way to determine if a polynomial is irreducible over a ring, is to check if the zeros of the polynomial are in the ring, ie, does p have a linear factor.