For a polynomial $f(x)$ in Integral Domain $D[x]$ to be irreducible, it has to be
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neither the zero polynomial nor a unit
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if $f(x)$ can be expressed as $f(x) = g(x)h(x)$ with $g(x), h(x) \in D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$
I assume unit here means an element which has a multiplicative inverse. I am unable to understand the significance of having (or not having) a multiplicative inverse in both the conditions.
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Why isn't the condition for being an irreducible polynomial not just "one which cannot be factored into lower degree polynomials"? What extra does having these extra conditions add?
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Also, in a field, these extra conditions don't exist – I can understand why the 2nd condition isn't required in a field – it's because every element of a field has a multiplicative inverse. However, if we do add the first condition, then there can never be a irreducible polynomial in a field – is that the sole reason for not having the first condition in a field or is there some other reason why that condition loses it's significance in a field?
Best Answer
This relates to the criteria that are used to define a unique factorization domain (UFD). In a ring of polynomials, for example, we would like each polynomial to have a unique factorization into irreducible elements, up to multiplication by a unit. If you do not include this last condition ("up to multiplication by a unit"), then you end up having multiple factorizations of the same polynomial, which creates unnecessary confusion. Consider a polynomial $p$ in the integral domain $F[X]$, where $F$ is a field. Then $p(X) = a_0 + a_1X + ... + a_nX^n$, for some $a_0, a_1,..., a_n \in F$. If $p$ appears in the factorization of some other polynomial $q(X) \in F[X]$, then we could define a "new" factorization of $q(X)$ by simply "factoring" $p(X)$ as $a_0(1 + a_0^{-1}a_1X + ... + a_0^{-1}a_nX^n)$. This doesn't really tell us anything new about either $p(X)$ or $q(X)$, so we tend to treat such a distinction as in some sense irrelevant.