# Relatively prime monic polynomials (Error in Hoffman?)

I normally don't spend my time attempting to disprove basic theorems in elementary linear algebra, but I found myself today not understand why Theorem 10 on Page 137 of Hoffman's (and Kunze... he always gets left in the dust) Linear Algebra is true.

Let $f$ be a non-scalar monic polynomial over the field $F$ and let $$f = p_{1}^{n_{1}} \cdot\cdot\cdot p_{k}^{n_{k}}$$ be the prime factorization of $f$. For each $j$, $1 \leq j \leq k$, let $$f_{j} = \frac{f}{p_{j}^{n_{j}}} = \prod_{i \neq j} p_{i}^{n_{i}}$$ Then $f_{1},...,f_{k}$ are relatively prime.

It then proceeds to state, "we leave the easy proof of this to the reader". This result seems blatantly false to me! Since we know that $f = p_{1} p_{2} \cdot \cdot \cdot p_{m}$ if and only if $f$ and $f'$ are relatively prime, where $f'$ denotes the standard derivative of $f$. We then have that, $$f' = p_{1}' p_{2}p_{3} \cdot\cdot\cdot p_{m} + p_{1}p_{2}'p_{3} \cdot \cdot\cdot p_{m} + \cdot\cdot\cdot + p_{1}p_{2}p_{3} \cdot \cdot \cdot p_{m}'$$ Thus, we assume without loss of generality that $p_{1} \mid f$, then since we know deg$(p_{1}') <$ deg$(p_{1})$, we have that $p_{1} \nmid p_{1}'$.

Can anyone provide a proof of this result? I'm quite confused as I doubt such an answer would have lasted a good 50 years in reprinting without someone noticing, so I must be wrong.

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"Relatively prime" probably doesn't mean "pairwise relatively prime" but that the $f_i$ don't all have a common factor, which is true. – Qiaochu Yuan Mar 31 '12 at 5:25
For each $j$, $p_j\nmid f_j$, hence $p_j$ does not divide the greatest common divisor of the $f_j$'s. On the other hand the $p_j$'s are the only candidates for the prime divisors of the GCD. – Bruno Joyal Mar 31 '12 at 5:28
There is absolutely nothing wrong with the result. – André Nicolas Mar 31 '12 at 5:28
@QiaochuYuan: Ahhh, I see! Thank you for clearing that up. Care to make that an answer? – Samuel Reid Mar 31 '12 at 5:44

## 1 Answer

Here $f_1,\ldots,f_k$ are relatively prime means $(f_1,\ldots,f_k) = (1)$. This construction is employed in one of the standard proofs of the Chinese Remainder Theorem / Pierce Decomposition. It is applied later in Sec. 6.8, Theorem 12, to obtain primary decompositions of linear operators on finite dimensional vector spaces.

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