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How to show the following:

Let $T: V_F \to V_F$ be a linear operator and $f(x)$ be the minimal polynomial of $T$ over $F$. Let $$f(x)=g_1(x)g_2(x)\cdots g_n(x)$$ where the $g_i$'s are monic and pairwise relatively prime. Then $g_i(x)$ is the minimal polynomial of the linear operator $T|_{\ker g_i(T)}: \ker g_i(T) \to \ker g_i(T)$ and $V=\oplus_{i=1}^n \ker g_i(T).$

Added: Using the result for $n=2$ is permissible.

i.e. you may assume that

Let $T: V_F \to V_F$ be a linear operator and $f(x)$ be the minimal polynomial of $T$ over $F$. Let $$f(x)=g_1(x)g_2(x)$$ where $g_1$ and $g_2$ are monic and relatively prime. Then $g_i(x)$ is the minimal polynomial of the linear operator $T|_{\ker g_i(T)}: \ker g_i(T) \to \ker g_i(T)$ and $V=\ker g_1(T)\oplus\ker g_2(T).$

Marc van Leeuwen
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Sriti
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  • Reduction to the case $n=2$ by induction is obvious, since $g_1g_2\ldots g_{n-1}$ and $g_n$ are relatively prime, supposing that the "relatively prime" hypothesis is "pairwise relatively prime", as it should be. – Marc van Leeuwen May 19 '13 at 08:13

1 Answers1

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For an induction, the cases $n\leq2$ are either obvious or given. For the inductive case $n>2$, the essential observation is that $g_1g_2\ldots g_{n-1}$ and $g_n$ are relatively prime; this is the reason that it is essential that the hypothesis be that the $g_i$ are pairwise relatively prime (no two have a common irreducible factor, as opposed to the weaker condition of having no common irreducible factor to all of them at once). To obtain from that hypothesis that $g_1g_2\ldots g_{n-1}$ and $g_n$ are relatively prime, one could for contradiction assume a common irreducible factor, and by the repeated form of Euclids lemma see that it is a common factor of $g_n$ and at least one of $g_1,\ldots,g_{n-1}$. With this established, one gets by the admitted case $n=2$: $$ V=\ker(g_1g_2\ldots g_{n-1}(T))\oplus\ker g_n(T). $$ Finally apply the induction hypothesis to further decompose the first summand.

Marc van Leeuwen
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  • How to show $g_1,g_2,...,g_n$ are relatively prime implies $g_1,g_2,...,g_{n-1}$ are relatively prime? – Sriti May 19 '13 at 08:30
  • As I said in the comment to the question, the hypothesis must be that the $g_i$ are pairwise relatively prime (as opposed to the much weaker condition that there is no non-trivial common factor to all of them at once). (I will edit the question in this sense unless you have strong reasons to object.) Then there is no irreducible factor that appears in more than one of the $g_i$ at once, so in particular none of the irreducible factors of $g_n$ occur in the product $g_1g_2\ldots g_{n-1}$, which makes them relatively prime. – Marc van Leeuwen May 19 '13 at 10:12