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If $$g(x) = x^4 + x^3$$

From my understanding, the degree of the above polynomial i.e. $g(x)$ is 4. However, for this polynomial, $$f(x) = (x-1)(x-2) \cdots (x-p+1)$$

What degree does $f(x)$ have? My guess was, by plug in $p = 3$, I have $f(x) = (x-1)(x-2)$, so this is polynomial of degree $p - 1$? Any idea?

Note where $p$ is prime


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$(x-p+1)=(x-(p-1))$, so yes. – quantumelixir Apr 18 '11 at 17:50
Where do you get that $f(x) = (x-1)(x-2)\cdots(x-p+1)$? Are you talking about values of the polynomial modulo some prime? Which prime? The equation is false as written if you refer to the same polynomial, unless $p$ is something very specific. This is rather badly written. – Arturo Magidin Apr 18 '11 at 17:52
@Arturo Magidin: The original problem asked to prove that each coefficient of that polynomial divides $p$, where $p$ is prime. I asked how to get the coefficient before, but I was stuck with that idea. Yesterday, when I read the book, it said each coefficient must divide every prime $p$ if and only if the number of roots is greater than the degree of the polynomial. Now I'm a bit confused because seemed like the question was wrong. Clearly, the number of roots of $f(x)$ is p-1 and its degree is also $p-1$. – Chan Apr 18 '11 at 18:02
@Chan: You (mis)quote out of context, so I cannot comment. What is "that polynomial"? Is it $(x-1)(x-2)\cdots(x-p+1)$? Is it $x^4+x^3$? What is $p$? What book? What did it say, exactly? What you provide only confuses the issue more. – Arturo Magidin Apr 18 '11 at 18:11
@Arturo Magidin: The $x^4 + x^3$ is my own make-up polynomial, to state what I understood about the degree of the polynomial. The $(x-1)(x-2) \cdots (x-p+1)$ is the one that I want to know the degree. $p$ is a prime as I edited. It was not actually in any book, it was from my lecture notes. I will post the exact text later. I really apologize for the confusion, sometimes I'm even confused what I should ask because I don't know exactly what should have been proven. Thank you. – Chan Apr 18 '11 at 18:29
up vote 6 down vote accepted

There are $(p-1)$ factors, so the leading coefficient will be $x^{p-1}$, and hence the degree is indeed $p-1$.

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Thanks a lot. – Chan Apr 18 '11 at 18:04

As a more general guideline, for 'formal' (univariate) polynomials and polynomials over $\mathbb{R}$:

  • The sum of a polynomial of degree $m$ and a polynomial of degree $n$ is a polynomial of degree no more than $\max(m,n)$ (but be aware that it can be less than this if $m=n$; for instance, consider the sum of $p(x) = x^3-2x^2+x+1$ and $q(x) = -x^3+2x^2+x+1$.)
  • Multiplying by a (non-zero) scalar doesn't change the degree of a polynomial; the degree of $4p(x)$ is the same as the degree of $p(x)$ for any polynomial $p$.
  • The product of a polynomial of degree $m$ and a polynomial of degree $n$ is a polynomial of degree exactly $m+n$ (notice how this differs from the result for sums, since no cancellation of the top-order term can happen in a product)

The last case is the one that applies to your problem; you're taking the product of $p-1$ polynomials each of degree $1$, so the degree of the result is just the sum of $p-1$ copies of $1$ - that is, $p-1$.

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I thought I would try to answer a question raised by your comment. A version of the following lemma is true in any field, but I will formulate it for the integers modulo $p$.

Lemma Let $P(x)$ and $Q(x)$ be monic (lead coefficient $1$) polynomials of degree $d$. If $P(a) \equiv Q(a) \pmod{p}$ for $d$ incongruent numbers $a$, then the corresponding coefficients of the two polynomials are congruent modulo $p$.

Now look at the two polynomials $P(x)=(x-1)(x-2)\cdots(x-p+1)$ and $Q(x)=x^{p-1}-1$. They are both monic of degree $p-1$. Clearly $P(a) \equiv 0$ for $a=1,2,\dots,p-1$. Also, $Q(a)\equiv 0$ at the same places, by Fermat's Theorem.

So corresponding coefficients are congruent to each other. Note in particular that if $p$ is an odd prime, then the constant term of $P$ is $(p-1)!$. The constant term of $Q$ is $-1$, and we conclude that $(p-1)! \equiv -1 \pmod{p}$, Wilson's Theorem! Take any other coefficient of $P$ except the coefficient of $x^{p-1}$. The corresponding coefficient in $Q$ is $0$, which tells us that all the coefficients of $P$ are congruent to $0$ except for the first and the last.

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@user6312: Many thanks for your answer. By the way, there was a typo in your post, it should be $(p-1)! \equiv -1 \pmod{p}$. – Chan Apr 19 '11 at 5:50
@Chan: Thank you, I am typo-prone, find it hard to proofread on screen, and don't have a printer at home. The typo is fixed. – André Nicolas Apr 19 '11 at 6:00

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