Beauty of Ring Theory

Problem

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This is an interesting question from my BMath Ring Theory course, which was taught by Prof. B. Sury. The course had way too many interesting problems and clever ideas, tbh some of them are really wild ones, but this question is special, beacuse you can explain it to even high schoolers, and yet, the solution is super slick! Thanks again to Prof. B. Sury for sharing this in class.

\( \textbf{Question:} \) Let \( f \in \mathbb{Z}[x] \) be a polynomial of degree \( n > 0 \). Pick \( n+1 \) distinct prime numbers, say \( p_1, \dots, p_{n+1}\). Show that there exists a polynomial \( g \in \mathbb{Z}[x] \) such that

\[ fg = \sum\limits_{i=1}^{n+1} c_i x^{p_i} \quad \text{for }c_i \text{ in } \mathbb{Z} \]

I hope that the problem statement is clear! Please give it a honest try, and once you are done, the solution is all yours!

Solution

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Alright, so what’s the trick? The first thing to notice is that the specific prime numbers \( \{ p_1, \dots, p_{n+1} \} \) don’t actually matter; they’re just there to distract you! The statement holds for any \( n+1 \) distinct numbers, so don’t overthink about those primes.

The way to attack this problem is to consider the quotient ring:

\[ R = \mathbb{Q}[x] \big/ (f) \]

where \( (f) \) is the ideal generated by \( f \). Why are we looking at this? Because \( R \) is a \( \mathbb{Q} \)-vector space! (That’s why we take \( \mathbb{Q} \) if we worked over \( \mathbb{Z} \), we wouldn’t get a vector space, and things would be messier, if we bring modules etc.)

Now, since \( f \) has degree \( n \), the space \( R \) has dimension \( n \). The reason for that is simple: any power of \( x \) that’s \( \geq n \) can be rewritten in terms of smaller powers, using the fact that we’re in \( R \), and that we have in some sense an “extra” zero, namely \( f \), a polynomial of degree \( n \). So eventually, any polynomial in \( \mathbb{Q}[x] \) reduces to a polynomial of degree at the most \( n \) in \( R \).

Thus, a natural basis for \( R \) is \( \{ \overline{1}, \overline{x}, \dots, \overline{x^{n-1}} \} \). Now, look at the set \( \{ \overline{x^{p_1}}, \overline{x^{p_2}}, \dots, \overline{x^{p_{n+1}}} \} \), this is an \( n+1 \) element set inside an \( n \)-dimensional vector space, that’s one more than the \( \dim R \), so it has to be linearly dependent set! That means there exist some \( u_i \in \mathbb{Q} \) such that

\[ \sum\limits_{i=1}^{n+1} u_i \overline{x^{p_i}} = 0 \quad \text{in } R, \]

\[ \implies \sum\limits_{i=1}^{n+1} u_i \overline{x^{p_i}} = \overline{0}. \]

But what does that actually mean? Well, if we lift the equation we got in \(R\) to \( \mathbb{Q}[x] \), we get that there’s some polynomial \( g \in \mathbb{Q}[x] \) such that

\[ \sum\limits_{i=1}^{n+1} u_i x^{p_i} = f g. \]

Now, there’s just one last detail: we need to make sure everything is in \( \mathbb{Z}[x] \). To do that, just multiply through by the least common multiple of the denominators of the coefficients of \( g \) and \(u_i\), and boom we get:

\[ fg = \sum\limits_{i=1}^{n+1} c_i x^{p_i}, \]

for some integers \( c_i \). And that’s it! A super clean solution, but the trick is sneaky. Hope you enjoyed this one! See you in my next post! Bye :)