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I started Ring theory recently and I came across this statement while reading polynomial rings..

If $F$ is an infinite field and let $f(x)\in F[x]$ . If $f(a)=0$ for infinitely many elements $a$ of $F$, then $f(x)=0$

I have the following doubt. Take $\Bbb{Q}$ as the infinite field. Now define $$f(x)=\prod_{i=1}^{\infty} (x-i)$$

Since (as far as I know , but I may be wrong!) the degree of polynomials need not necessarily be finite, $f(x)$ is polynomial $\in \Bbb{Q}[x]$ and $f(a)=0$ for infinitely many elements in $\Bbb{Q}$. So where am I making a mistake?

Watson
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Qwerty
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    The degree has to be finite: a polynomial has only finitely many non-zero coefficients. However you can see this. – Watson Jun 23 '16 at 09:20
  • @Watson Is it in the definition of a poynomial? Google doesn't say so I think! link – Qwerty Jun 23 '16 at 09:21
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    Then google is inaccurate in this case. I don't recommend using Google to look up math definitions. The mathematical definition of a polynomial requires finitely many nonzero terms. –  Jun 23 '16 at 09:47
  • @tilper Okay .. I see.. i thought Google was always perfect.. – Qwerty Jun 23 '16 at 09:49
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    @Qwerty Well, I don't think there is any problem with Googling math definitions. In fact, Google can and should be an indispensable tool for learning lots of things. The mistake you made was to only read the first thing at the top and assume it was correct. Reading the next five links should have cast light on the problems with the first thing you read, and convinced you that you need to keep reading to find a clear definition. – rschwieb Jun 23 '16 at 12:55
  • What is the definition of the polynomial ring $,\Bbb Q[x],$ in your textbook? – Bill Dubuque Jun 23 '16 at 16:11
  • @BillDubuque The set of formal symbols $R[x]={a_nx^n+\cdots +a_1x+a_0|a_i\in R,\ n$ is a non-negative integer $}$ is called the ring of polynomials over $R$ – Qwerty Jun 23 '16 at 16:15

2 Answers2

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The degree must be finite, otherwise $\sum \limits _{n = 0} ^\infty \frac {x^n} {n!}$ would be a polynomial - which clearly it isn't, being $\Bbb e^x$.

In fact, a polynomial is just a function from $\Bbb N$ to $F$ with finite support.

Alex M.
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  • I asked google the definition of a polynomial and it didnt mention anything about its degree..link – Qwerty Jun 23 '16 at 09:24
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    @Qwerty: Google says "sum of several terms" and sums, by definition, have a finite number of terms. When there are infinitely many summands, those are series. – Alex M. Jun 23 '16 at 09:26
  • @Qwerty I see four hits for "degree" on the page you linked, so your claim that "it didn't mention anything about degrees" looks pretty silly. – rschwieb Jun 23 '16 at 13:12
  • @rschwieb Okay. I am sorry to have judged Google superficially. – Qwerty Jun 23 '16 at 13:18
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Your $f(x)$ is not a polynomial, which must have a fixed, but arbitrary degree $n\ge 0$. It is a theorem then, that every polynomial of degree $n$ over a field $F$ has at most $n$ zeros. This is no longer true over, say, a skew-field. Consider the polynomial $x^2+1$ over the quaternion algebra, as an example - see here. It has infinitely many roots.

Community
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Dietrich Burde
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