Suppose that \(f(x) \in (\ZZ/n\ZZ)[x]\) is a zero divisor. Show that there is a nonzero \(a\in \ZZ/n\ZZ\) with \(af(x) = 0\).
Solution
- Write \(f(x) = \sum_{k=0}^n a_k x^k\), and supposing it’s a zero divisor choose \(g(x) = \sum_{k=0}^m b_k x^k\) of minimal degree so that \(g\neq 0, b_m\neq 0\), and \(f(x)g(x) = 0\).
- The claim is that the top coefficient \(b_m\) will suffice.
- Write the product: \[ 0 = f(x)g(x) = (a_0 + \cdots + a_{n-1}x^{n-1} + a_n x^n) (b_0 + \cdots + b_{m-1}x^{m-1} + b_m x^m) .\]
- Equating coefficients, the coefficient for \(x^{m+n}\) must be zero, so
(importantly) \(a_n b_m =
0\).
- Since \(a_n b_m=0\), consider \(a_ng(x)\). This has degree \(d_1 \leq m-1\) but satisfies \(a_ng(x)f(x) = a_n(g(x)f(x)) = 0\), so by minimality \(a_ng(x) = 0\).
- This forces \(a_n b_0 = \cdots = a_n b_{m-1} = 0\), so \(a_n\) annihilates all of the \(b_k\).
- Now consider the coefficient of \(x^{m+n-1}\), given by \(a_{n-1}b_m + a_{n}b_{m-1}\).
- The second term \(a_n b_{m-1}=0\) since \(a_n\) annihilates all \(b_k\), so (importantly) \(a_{n-1} b_m = 0\).
- Considering now \(a_{n-1}g(x)\):
- The same argument shows this has degree \(d_2 \leq m-1\) but \(a_{n-1}g(x)f(x) = 0\), so \(a_{n-1}g(x) = 0\).
- So \(a_{n-1}\) annihilates all \(b_k\), and allowing this process to continue inductively.
- For good measure, the coefficient of \(x^{m+n-2}\) is \(a_{n-2}b_m + a_{n-1}b_{m-1} +
a_{n}b_{m-2}\).
- Note that \(a_n, a_{n-1}\) annihilate all \(b_k\), so (importantly) \(a_{n-2} b_m=0\), and so on.
- Thus \(a_k b_m = 0\) for all \(0\leq k \leq n\), and by linearity and commutativity, we have \[ b_m f(x) = b_m \sum_{k=0}^n a_k x^k = \sum_{k=0}^n (b_m a_k) x^k = 0 .\]