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Definition: Let $F$ be a field. An ideal in $F[x]$ is a subspace $M$ of $F[x]$ such that $f\cdot g\in M$, $\forall f\in F[x]$, $\forall g\in M$.

Let $A$ be an $n\times n$ matrix over a field $F$. Show that the set of all polynomials $f$ in $F[x]$ such that $f(A)=0$ is an ideal.

My attempt: $A\in M_{n}(F)$. We need to show $M=\{f\in F[x]\ |\ f(A)=0\}$ is ideal in $F[x]$. Since $\dim (M_n(F))=n^2$, we have $I,A,A^2,…,A^{n^2}$ is linearly dependent. So $\exists c_0,c_1,…,c_{n^2}\in F$ such that $c_j\neq 0$, for some $j$ and $\sum_{i=0}^nc_i\cdot A^i=0$. Let $h=\sum_{i=0}^nc_i\cdot x^i\in f[x]\setminus \{0\}$. Then $h(A)=0$. So $h\in M$. Thus $M\neq \emptyset$. Let $f,g\in M$ and $c\in F$. By theorem 2 section 4.2, $(c\cdot f+g)(A)=c\cdot f(A)+g(A)=0$. Thus $c\cdot f+g\in M$. Hence $M$ is subspace of $F[x]$. Let $f\in F[x]$ and $g\in M$. By theorem 2 section 4.2, $f\cdot g(A)=f(A)\cdot g(A)=f(A)\cdot 0=0$. Thus $f\cdot g\in M$. Is my proof correct?

user264745
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  • I just realised, property $M\neq \emptyset$ is equivalent to $0\in M$. By definition $0(\alpha )=0_V$, $\forall \alpha \in V$ ($V$ is linear algebra over $F$). – user264745 Dec 03 '22 at 18:17
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    Yes, the zero polynomial is your friend. – blargoner Dec 03 '22 at 18:19
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    Your proof is correct. As you have now realized, you can show $M$ is nonempty more quickly by using the zero polynomial. However your "longer" proof is still valuable as it shows that $M$ contains more than just the zero polynomial. After all, ${0}$ is an ideal; it could have been the case that $M = {0}$. You have shown that $M$ contains more than just $0$. I don't have this textbook, but I will suggest that you should remember this part of the proof, as there's a good chance it will be an exercise in the near future. :-) – Zach Teitler Dec 03 '22 at 19:30
  • @ZachTeitler I don’t know about exercise but this ideal is building block of minimal polynomial. Can you please help me with this post (division algorithm for polynomial)? I have been stuck on that proof for long time. – user264745 Dec 04 '22 at 07:58

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