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Here is the question I am trying to attempt:

enter image description here

I feel like there is a typo... it says let $p(x)=a_{n}x^{n}$. Shouldn't it be $p(x) = a_{n}^{n}\cdots + a_{1}x+a_{0}$? I feel like I must use the roots of $p(x)$ in my answer, but I am not sure how and where. Any hints on what I should do?

Edit: The duplicate answer did not help me because I am trying to prove injectivity...it has already been assumed that $\eta$ is a ring homomorphism.

2 Answers2

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Hint If ker(η)=0 then η is injective. Note that the only ideals of a field are the zero ideal and the unit ideal.

Unknown
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  • Since $F$ is infinite, does that mean the kernel of $\eta$ is the roots of $p(x)$? So there are at most $n$ elements in $ker(\eta)$, but then why is $ker(\eta)=0$? Because the only ideals of $F$ are itself and zero, so it cannot have a finite list of elements of size greater than zero? So it must then be zero...but why not $ker(\eta)=F$? I guess because not all polynomials evaluate to zero. –  Nov 21 '18 at 06:52
  • This answer makes no sense. What do the ideals of a field have to do with the question. The only useful part is "If $\ker \eta = 0$ then $\eta$ is injective," which is almost what it says in the image in the question. – jgon Nov 21 '18 at 16:43
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Your thoughts on the typo are right. It should be $p(x)=a_nx^n+\cdots+a_0$.

Now it asks you to show that $\eta$ is injective, which means that if a polynomial defines the zero function, then it is the zero polynomial.

Well, suppose $\eta(p)=0$ for a polynomial $p$. Then $p(\alpha)=0$ for all $\alpha\in F$. So $p$ has infinitely many roots if $F$ is infinite, which is impossible unless $p$ is the zero polynomial (nonzero polynomials have at most $\deg p$ roots). Thus we have $p=0$, so $\eta$ is injective.

jgon
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  • I see, and so the kernal of $\eta$ is just the set with $0$? Does that mean the kernel is generated by a polynomial in $F[x]$? May I ask how to find such a polynomial generator? –  Nov 21 '18 at 18:00
  • @numericalorange well $K[x]$ is a pid so every ideal is generated by a single polynomial. The zero ideal is generated by 0. – jgon Nov 21 '18 at 18:12
  • If you're asking about the second part of the question in the image you asked about, then use the fact that if a polynomial has a root $\alpha$, then $(x-\alpha)$ divides that polynomial, and apply that result to every element of a finite field to find a generator for the kernel of $\eta$ when the field is finite. – jgon Nov 21 '18 at 18:14