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I'm doing a study on module theory and I'm working on the following problem:

Let $K$ be a field and let $f_1, \dots, f_n \in K[X]$ be monic polinomials. Consider the $K[X]$-module $$ V := K[X]/(f_1)\bigoplus\dots\bigoplus K[X]/(f_n) $$

Show that the characteristic polynomials of the endomorphism $v \mapsto X \cdot v$ of the $K$-vector space $V$ equals $\prod_i f_i$.

My attempt:

Question: Am I correct that the act of the endomorphism works via $v = (v_1, \dots, v_n) \mapsto X \cdot v = (X v_1, \dots, X v_n)$?

Question: the exercise does not say that these polynomials must be irreducible. If if this is not the case then we could replace $f_1$ be a possibly smaller degree $\overset{\sim}{f_1}$ irreducible monic polynomial s.t. $(\overset{\sim}{f_1}) = (f_1)$.

In order to find the characteristic polynomial we need to find a basis. For each of the component vector spaces $K[X]/(f_i)$.

Question: am I correct to say that this basis is given by: $\{1, X, X^2, \dots, X^{\deg f_i - 1}\}$? Even if the $f_i$ are not irreducible?

Suppose that this basis is correct. And let $\deg f_1 = d$ and $f_1 = X^d + a_1^{d-1} X^{d-1} + \dots + a_1^0$. (So I use the lower index of the $a_i^j$ to indicate which $f_i$ and the upper to indicate which coeficient in the polynomial. Question: is it true that the endomorphism restricted to $K[X]/(f_1)$ has the following matrix representant: $A = \begin{bmatrix} 0 & 0 & 0 & -a_1^0 \\ 1 & \ddots & 0 & -a_1^2 \\ 0 & \ddots & 0 & \vdots \\ 0 & 0 & 1 & -a_1^{d-1} \end{bmatrix} $. Where I found the last column by noting that $X^d = -X^{d-1}a_1^{d-1} + \dots + -a_1^0$.

And that the full endomorphism would be a block diagonal matrix of matrices of this type?

Now if we compute the characteristic polynomial for $A$. We get $\det(A-tI)=f_1(t) - t^{\deg f_1}$ (up to a sign), but I think that we need to get $f_1$. What is going on here?

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  • $\begingroup$ If $M_1,M_2$ are endomorphisms of $V_1,V_2$ then $M(v_1,v_2) = (M_1v_1,M_2v_2) $ is an endomorphism of $V_1 \oplus V_2$ and $\det(M) = \det(M_1)\det(M_2)$. Then the determinant stays the same in $K[X]/(f_1)$ and $\overline{K}[X]/(f_1)$. In the latter $f_1$ splits, assume it is monic and separable, so it reduces to the determinant of $v \mapsto (t-X)v$ in $\overline{K}[X]/(X-\alpha)$ which is $t-\alpha$. Thus $\det[v \mapsto (t-X)v \in \overline{K}[X]/(f_1)]= f_1(t)$ and $\det[v \mapsto (t-X)v \in \bigoplus_{j=1}^n K[X]/(f_j)] = \prod_{j=1}^n f_j(t)$ $\endgroup$ – reuns Mar 1 at 13:01
  • $\begingroup$ What do you mean by $(\tilde{f_1}) = (f_1)$? Two polynomials generate the same ideal in $K[X]$ is and only if one is a nonzero scalar multiple of the other. In a particular you cannot replace $f_i$ by a smaller degree polynomial giving the same summand as $K[X]/(f_i)$. $\endgroup$ – Marc van Leeuwen Mar 2 at 10:35
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I think your solution is correct you just miscalculated at the very end. You are looking at $$\begin{bmatrix} -x & 0 & 0 & \dots & -a_1^0 \\ 1 & -x & 0 & \dots & -a_1^1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 1 & \dots & -x-a_1^{d-1} \end{bmatrix} $$ so when you start calculating the determinant you get the diagonal term: $(-x)^{d-1}(-x-a_1^{d-1})=(-x)^{d}-a_1^{d-1}(-x)^{d-1}$ which corresponds to the top degree and the top degree -1 terms. Then the off diagonal terms follow accordingly (and the signs work out as well).

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You are basically right. First observe that one can compute the characteristic polynomial as a product of those of the individual summands, so it is sufficient to show that the characteristic polynomial of multiplication by $X$ on $K[X]/(f)$ is $f$, whenever $f\in K[X]$ is monic (no arithmetic properties required). Indeed, choosing the basis of the images of the first $\deg f$ powers of $X$, the matrix of multiplication by $X$ is the companion matrix of $f$, which is your matrix $A$. And as is well known (though you seem to be unaware) the characteristic polynomial of any companion matrix is the polynomial for which it is companion matrix (with its leading term in included; don't know why you dropped it, you should know that the characteristic polynomial of an $n\times n$ matrix is always monic of degree$~n$). Also please note that the characteristic polynomial is $\det(XI_n-A)$, and not $\det(A-tI)$; getting the sign right is essential for the "characteristic polynomial of companion matrix" statement to hold).

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