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On reading Axler's Linear Algebra Done Right, Chapter 4, I tried to prove the following theorem:

Theorem. Suppose $a_{0},\dots,a_{m}\in\mathbb{F}$. If $a_{0}+a_{1}z+\cdots+a_{m}z^{m}=0$ for every $z\in\mathbb{F}$, then $a_{0}=\cdots=a_{m}=0$.

Note that, following the convention of the book, $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}.$

This is my attempt:

Proof. Suppose $a_{0}+a_{1}z+\cdots+a_{m}z^{m}=0$ for all $z\in\mathbb{F}$ ($\dagger$). Let $f\left(z\right)=a_{0}+a_{1}z+\cdots+a_{m}z^{m}$. Then $f\left(z\right)=0$ for all $z\in\mathbb{F}$, that is, $f=0$. Hence, by definition, $\deg f=-\infty$. Suppose there is $j\in\left[0..m\right]$ such that $a_{m}\neq0$. We can choose the maximum of such $j$, let it be $m'\in\left[0..m\right]$. Then $\deg f=m'$. Since $m'\in\left[0..m\right]$, $\deg f\geq0$, which contradicts $\deg f=-\infty$. Therefore, $a_{0}=\cdots=a_{m}=0$.

But it is quite different from the proof in the book and seems simpler. relevant portion of the book

I suspect there might be a circular argument in my proof, but I could not find a clue. Question: Is this proof correct? If not, what is wrong with this proof?

Git Gud
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akarin64
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    A polynomial (not the polynomial function) is defined to be zero if all it's coefficients vanish. Correspondingly the degree is defined to be $-\infty$ provided all the coefficients vanish. So essentially your argument boils down to ''all the coefficients vanish because all the coefficients vanish ". – peek-a-boo Apr 05 '22 at 09:26
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    Stripping it down to its essentials, your proof reads like the following. "We have $P$. Suppose $\neg P$ holds. This is a contradiction." This isn't a valid method of proof. – Git Gud Apr 05 '22 at 09:40
  • @peek-a-boo I checked the book and this author equates functions with polynomials functions, which works for $0$ characteristic fields. – Git Gud Apr 05 '22 at 09:41
  • Note that this is wrong in a finite field. In an infinite field , we can use that a polynomial with degree $n>0$ has at most $n$ roots , but it must have infinite many roots. Hence the polynomial must be constant and therefore identically $0$. – Peter Apr 05 '22 at 09:43
  • Needs a definition of the notion polynomial. – Wuestenfux Apr 05 '22 at 09:45
  • Avoid using pictures for large portions of your question. I will edit the question accordingly. – K.defaoite Apr 05 '22 at 09:50
  • See this question: [Two questions about a proof that "if a polynomial is the zero function, all of its components are zero](https://math.stackexchange.com/questions/3169442/two-questions-about-a-proof-that-if-a-polynomial-is-the-zero-function-all-of-i?rq=1). – J.-E. Pin Apr 05 '22 at 09:51
  • @GitGud interesting, I didn't realize the author defined it this ("reversed") way – peek-a-boo Apr 05 '22 at 09:52
  • @K.defaoite It was me who included the screenshot. – Git Gud Apr 05 '22 at 09:55
  • @Peter That's not precise enough, I don't think. You can take an infinite field with non-zero characteristic. Example [here](https://math.stackexchange.com/a/58425/55235). – Git Gud Apr 05 '22 at 10:09
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    @GitGud Thanks for the feedback. I have (therefore) deleted my answer. I assumed that comments following the posting had resolved the issue of whether the OP's work was valid, so that it only remained to furnish a proof of the underlying assertion. Also, I have never studied Field Theory so, I am clearly out of my depth. Again, thanks for the feedback. – user2661923 Apr 05 '22 at 11:38
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    @GitGud It is correct. The map from the polynomial ring to the ring of functions is injective if and only if field is infinite. Characteristic zero of course is sufficient, but is not necessary. See, e.g., [here](https://math.stackexchange.com/a/304888/742). Argument works for positive characteristic, as long as the field is infinite. Peter is correct. – Arturo Magidin Apr 05 '22 at 13:08

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