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Let $R$ be a commutative ring with identity and let $f: R[x]\to R$ be a ring homomorphism defined by

$$ f(r_0 + r_1 x + \cdots + r_n x^n) = r_0 + r_1 + \cdots + r_n. $$

I need to prove that $\ker f$ is a principal ideal of $R[x]$, and find an element of $R[x]$ that generates $\ker f$.

Apparently, here $f$ is the evaluation map that evaluates every polynomial with coefficients in $R$ at $x = 1$. However, I can't even figure out what the kernel of this homomorphism is, let alone how to show that it is a principal ideal, since I can't figure out when $r_0 + r_1 + \cdots r_n = 0$.

My thoughts thus far are that if $n+1$ is even (i.e., when there is an even number of terms), the image of $f$ is $0$ when $\forall i$ from $i = 0$ to $n$, $\exists j$ from $j = 0$ to $n$ such that $i \neq j$ and $r_i = -r_j$.

On the other hand, if $n+1$ is odd, then $n$ must be even. So supposing that an odd number $k$ of the terms $= 0$, then for the remaining $m:=n+1 - k$ terms, of which there are an even number, if we renumber them as $s_0$ to $s_{m-1}$, then the image of $f$ is $0$ when $\forall i$ from $i = 0$ to $m-1$, $\exists j$ from $j = 0$ to $m-1$ such that $i \neq j$ and $s_i = -s_j$.

I don't, however, know how to write this compactly and in such a way where it would be easy to show that it is principal and to tell what element of $R[x]$ generates it.

Could somebody please assist me with this? First, if I am not on the right track with identifying the kernel of $f$, please help me do so. Secondly, once I have the kernel identified, how do I show that it is a principal ideal of $R[x]$ and (thirdly) go about identifying its generator?

Thank you ahead of time for your help and patience!

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  • $\begingroup$ Since you've already identified that this is basically evaluation at the identity, one can see that the kernel should be the set of polynomials with root $1$. Hence I suppose that $\ker f = (X-1)$. $\endgroup$ Apr 29, 2017 at 22:29

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Clearly $x-1\in\ker f$. Therefore the ideal $I$ generated by $x-1$ is contained in $\ker f$.

Suppose $p(x)\in \ker f$; then $$ p(x)=(x-1)q(x)+r $$ for some $r\in R$ and some $q(x)\in R[x]$. Substituting $1$, we get $$ 0=p(1)=(1-1)q(x)+r $$ and therefore $r=0$, so $p(x)=(x-1)q(x)\in I$.

Summing up, $I=\ker f$.

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  • $\begingroup$ I suppose we are trying to show that $p(x) = x-1$ here? Anyway, if $p(x) \in \ker f$ and $f$ is the evaluation map where $x = 1$, I would need $p(1) = 0$, right? So, I would need $p(1) = (1-1)q(1) + r$, which implies that $r = 0$? $\endgroup$
    – user100463
    Apr 29, 2017 at 22:49
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    $\begingroup$ @ALannister Sure:you want to show $p(x)=(x-1)q(x)$, that is, $r=0$. Polynomial division (with remainder) is always possible when the divisor is monic. $\endgroup$
    – egreg
    Apr 29, 2017 at 22:52
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    $\begingroup$ @ALannister Why? You want to show $p(x)$ belongs to the ideal generated by $x-1$. $\endgroup$
    – egreg
    Apr 29, 2017 at 22:59
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    $\begingroup$ @ALannister No. You're on the wrong track. The ideal generated by $x-1$ consists of all polynomials of the form $(x-1)q(x)$, where $q$ is an arbitrary polynomial. $\endgroup$
    – egreg
    Apr 30, 2017 at 20:04
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    $\begingroup$ @ALannister I added the complete argument. The set of all $(x-1)q(x)$ is an ideal by general theory. $\endgroup$
    – egreg
    Apr 30, 2017 at 20:17

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