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This is more like only well-known fact to me, but no one really coherently explained it to me. I’ve only done questions for proving that a polynomial is an identity: for example, in the case of a quadratic, all the three standard coefficients should be zero for it to be an identity. I wanted more intuition for this. Can someone explain the whole topic and relevant surrounding stuff to me?

Devansh Kamra
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prarabdh shivhare
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    The title of this question has a problem, because polynomials and identities are entirely different kinds of things. Probably you meant with "becomes an identity" that the polynomial has to be zero, in which case, why not say just that? – Marc van Leeuwen Apr 03 '20 at 14:23
  • ...and even if you identify polynomials with functions, "the identity" would be the polynomial $x$, not $0$. – Federico Poloni Apr 03 '20 at 14:30
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    @MarcvanLeeuwen , thank you for bringing this to my knowledge but I'd rather stick with the title. A lot of people identify with this type of speech, even though it's incorrect. I guess I can reach a slightly bigger audience with the original thing. For the people who do get confused, there's always your comment :) – prarabdh shivhare Apr 03 '20 at 14:57

2 Answers2

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We can always write a polynomial in terms of its roots. Suppose it is of degree $n$ and has leading coefficient $a$: $$a(x-r_1)(x-r_2)\cdots(x-r_n)$$ That there are $n$ factors is a result of the fundamental theorem of algebra and the factor theorem.

From this we see that if $a\ne0$, there can only be the roots $r_1,r_2,\dots,r_n$ – everywhere else, all the factors and $a$ are non-zero, which multiply into a non-zero product. Contraposition then gives the desired result: if there are more roots, $a=0$ and the polynomial collapses into the identity.

Parcly Taxel
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    And, if in a ring where nonzero factors may give the product zero, the result does not hold anymore. For example, the polynomial $x^2-1$ taken over $\mathbb{Z}/8\mathbb{Z}$ has more than two distinct (in $\mathbb{Z}/8\mathbb{Z}$) roots. In fact it has four, $x=\pm 1, x=\pm 3$. (Also means that here, the square root is not unique up to a sign/negation.) – Jeppe Stig Nielsen Apr 03 '20 at 14:18
  • @JeppeStigNielsen can you elaborate, did not get you – prarabdh shivhare Apr 03 '20 at 19:27
  • In what way is the zero function the identity? – hkBst Apr 04 '20 at 08:21
  • @prarabdhshivhare Do you now what $\mathbb{Z}/8\mathbb{Z}$ is? Consider the square roots of 1 in that ring. All "odd" members work: $1^2 = 1$ and $3^2 = 9 \equiv 1$ and $5^2 = 25 \equiv 1$ and $7^2 = 49 \equiv 1$, modulo 8. This means that the 2nd degree equation $x^2=1$, or the 2nd degree polynomial $x^2-1$, over $\mathbb{Z}/8\mathbb{Z}$ has more than 2 roots in $\mathbb{Z}/8\mathbb{Z}$. We still have factorizations (but not unique) in this case: $$x^2-1=(x-1)(x-7)$$ and: $$x^2-1=(x-3)(x-5)$$ over $\mathbb{Z}/8\mathbb{Z}$. But we cannot conclude from one of those that there are no more roots. – Jeppe Stig Nielsen Apr 04 '20 at 09:06
  • not really, haven't heard it before---- not a clue @JeppeStigNielsen – prarabdh shivhare Apr 05 '20 at 00:19
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A very important theorem, you have to know is the foundamental theorem of algebra:

A polynomial $P(x)$ of degree $n$ with complex or real coefficents have no more than $n$ zeros each counted with its multiplicity.

Now, it's very easy to see that a polynomial written in the form: $$a_nx^n+\cdots+a_0=0 \rightarrow a(x-x_1)\cdots(x-x_n)$$ can't have more than $n$ real or complex solutions.

Matteo
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