# Abstract algebra rings - How to prove it?

Let $f_p(x)$ be the associated polynomial in $Z_p[x]$ obtained by reducing the coefficients of $f(x)$ mod $p$. If $\deg(f(x)) = \deg(f_p(x))$ and if $f_p(x)$ is irreducible in $Z_p[x]$ then $f(x)$ is irreducible in $Z[x]$.

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Suppose that $f(x)$ is reducible. Let $f(x)$ be represented as $a(x)b(x)$ where $\deg(a(x)) < \deg(f(x))$ and $\deg(b(x)) < deg(f(x))$. Let $f_p(x)$ be represented as $a_p(x)b_p(x)$ where $\deg(a_p(x)) < \deg(f_p(x))$ and $\deg(b_p(x)) < \deg(f_p(x))$. Suppose also that $\deg(f_p(x)) = \deg(f(x))$ this then means that $\deg(a_p(x))$ and $\deg(b_p(x))$ are both less than $\deg(f(x))$, which implies that $f_p(x)$ is reducible in $Z[x]$. Is this right? Please proof-read and correct me –  crazy student Oct 22 '11 at 1:34
That's the right direction. However, upon rereading the question, it seems that you are missing an assumption somewhere in the problem statement. How do you prevent it from concluding (with $p=2$) that $3(X+1)=3X+3$ is irreducible in $\mathbb Z[X]$? –  Henning Makholm Oct 22 '11 at 1:50
Um, no. Linear polynomials can have roots and still be irreducible. Or what would you say its factors were? –  Henning Makholm Oct 22 '11 at 2:20
No, that's not it. (I'm working from the assumption here that you're trying to solve a homework problem and have simply failed to read the problem closely enough. That's not completely certain, of course, but your wording of the question is not exactly forthcoming about why you'd be trying to prove exactly this statement). –  Henning Makholm Oct 22 '11 at 2:39
So much the better. Then you can present a counterexample in your solution! (The missing assumption is that $f$ is primitive, i.e. that it has no non-unit constant factor. Given that, neither $a$ nor $b$ in your proof sketch can be constants, and therefore the factorization in $\mathbb Z_p[X]$ is proper). –  Henning Makholm Oct 22 '11 at 2:53