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I'm trying to prove the following, and I've made some progress on (i) and am having a bit of trouble with (ii)

A polynomial $p(x)$ over a field $k$ is $monic$ if the highest power of $x$ has coefficient $1$. Let $p(x)$ be monic and let $r\in k$.

(i) Show that if $p(x)$ is divided by $x-r$, then the remainder is $p(r)$.

(ii) Show that this remains true if $p(x)$ has coefficients in $\mathbb R$ and $r \in \mathbb C$.

For (i), when I set up polynomial long division, I start to see the pattern of $a_{n-1}+r$, then $r(a_{n-1}+r)+a_{n-2}$, then $r(r(a_{n-1}+r)+a_{n-2})$, etc as the coefficient (and I'm assuming eventually the remainder) from each additional subtraction of two terms at the bottom when we add the next term at the top. I think that this is the key element to the proof, as the sequences goes on to be p(r), but not sure how to word this rigorously into a proof.

For (ii) I'm really not sure how to go about this. It almost seems like the polynomial division algorithm works just as well if r is complex as if it isn't, but not sure how to prove it or if it's true.

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(ii) is just a special case of (i), where $k = \Bbb C$ and $p$ has real coefficients. It is strange that you are requested to show that it "remains true" for a special case. Are you sure that the problem is not misstated? Also strange is that $p$ is assumed monic, since that plays no role. Nor does the hypothesis that the ring $k$ is a field, since the standard simple proof shows that the result is true over any ring $k$. Perhaps you know fields but not rings? –  Math Gems Jan 18 '13 at 18:09
    
Yeah, I find that strange as well, but the problem isn't misstated. Regarding rings, this is from a linear algebra book, so I'm guessing they give us just enough abstract algebra to do what we need (ie no rings). –  user1157605 Jan 19 '13 at 10:07

2 Answers 2

up vote 3 down vote accepted

(i) In general, if you divide a polynomial $f(x)$ by a polynomial $g(x)$, you obtain a quotient $q(x)$ and a remainder $u(x)$ with $ \deg u<\deg g$ such that $$\tag1f(x)=q(x)g(x)+u(x).$$ Here, $u(x)$ must be some constant $c$ because $\deg g=1$. Now set $x=r$ in $(1)$ ...

(ii) Nothing special was used.

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Your $(1)$ is not true in an arbitrary ring. For example consider $x^2 / 2x$ in $(\mathbb Z / 4 \mathbb Z)[x]$. It holds however when dividing by the monic polynomials, which is the case here. –  Marek Jan 18 '13 at 16:06
    
@Marek The problem is about polynomials over a field. Fields are not arbitrary rings. Of course my "in general" was intended to mean "also for other divisors" not "also for other rings". –  Hagen von Eitzen Jan 18 '13 at 16:13
    
I didn't notice that assumption, sorry. Nevertheless, I think it's worth pointing out that the assumption that $R$ be a field is not necessary here and that recalling general division algorithm is quite an overkill when dividing by linear monic. –  Marek Jan 18 '13 at 16:27

Express $p(x) = q(x)(x-r) + a$. Try to prove that this holds in $R[x]$ for any ring $R$ whatsoever. This takes care of $(i)$.

Now, $(ii)$ is just a special case of the above when $R = \mathbb C$ and we restrict coefficients to $\mathbb R$.

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