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Let $\mathbb R[x]$ denote the ring of all polynomials with real coefficients. The mapping $f(x)\rightarrow f(1)$ is a ring homomorphism from $\mathbb R[x]$ onto $\mathbb R$.

Question: Describe the kernel of this ring homomorphism.

Recall the definition of kernel of a $\left ( ring \right )$ homomorphism: $\ker\left ( \varphi \right )=\left \{ f\left ( x \right ) \in\mathbb R\left [ x \right ] \mid \left [ f\left ( x \right ) \right ]\varphi=0 \right \}$

Now, we define:

Define

$\varphi:R\left [ x \right ]\rightarrow R$

$f\left ( x \right ) \mapsto f\left ( 1 \right )$

I was almost certain the kernel is the trivial kernel but my solution sheet dictates otherwise.

Hint is appreciated.

Thanks in advance

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  • $\begingroup$ The only possible hint that doesn't give away the answer is a suggestion to apply the definition of kernel that you have written, to this particular homomorphism. It's like putting two and two together. $\endgroup$ – M. Vinay May 30 '16 at 8:38
  • $\begingroup$ I'm not sure how to think about this. I need to find a polynomial f(x)=0. This way, by the Kernel, this polynomial is mapped to the zero element. $\endgroup$ – Mathematicing May 30 '16 at 8:53
  • $\begingroup$ You're missing something. You need a polynomial which, when the homomorphism is applied to it, becomes zero. Now, what the homomorphism does is evaluate the given polynomial at $x = 1$. $\endgroup$ – M. Vinay May 30 '16 at 8:56
  • $\begingroup$ You seem to be self-learning ring theory, and at a good pace too. Keep it up! $\endgroup$ – M. Vinay May 30 '16 at 9:29
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    $\begingroup$ technically I'm in University. But yes, I self-learn using books instead of relying on lectures. $\endgroup$ – Mathematicing May 30 '16 at 9:32
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Hint: This is primarily a counterexample that shows the kernel to be non-trivial, but it also serves as a strong hint. Let $f(x) = x^2 - 3x + 2$. Then $\varphi(f(x)) = f(1) = 1 - 3 + 2 = 0 \implies f(x) \in \ker \varphi$.


Solution:

\begin{align*} \ker \varphi & = \{\, f(x) \in \mathbb R[x] \mid \varphi(f(x)) = 0 \in \mathbb R \,\}\\ & = \{\, f(x) \in \mathbb R[x] \mid f(1) = 0 \,\}\\ & = \{\, f(x) \in \mathbb R[x] \mid (x - 1)\ \text{is a factor of}\ f(x) \,\}\\ & = \{\, (x -1)g(x) \mid g(x) \in \mathbb R[x] \,\}\\ & = \langle x - 1 \rangle \end{align*}

That is, $\ker \varphi$ is the ideal generated by $x - 1$.

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  • $\begingroup$ How were you able to arrive at x-1 being a factor of f(x)? $\endgroup$ – Mathematicing May 30 '16 at 9:13
  • $\begingroup$ @Mathematicing Factor theorem. $\endgroup$ – M. Vinay May 30 '16 at 9:15
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    $\begingroup$ @Mathematicing Exactly, that's the definition of "$(x - 1)$ is a factor of $f(x)$". $\endgroup$ – M. Vinay May 30 '16 at 9:17
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    $\begingroup$ @Mathematicing If you're familiar with vector spaces, then an ideal generated by a set is elements of a (commutative) ring is almost exactly like a subspace spanned by a set of elements of a vector space. And of course, principal ideals (ideals generated by a single element), have an even simpler structure — they consist exactly of all multiples (by all ring elements) of the generator. $\endgroup$ – M. Vinay May 30 '16 at 9:21
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    $\begingroup$ @Mathematicing To be precise, $\langle x - 1 \rangle$ is the set of all multiples of $x - 1$, or in other words, $\{\ (x - 1) g(x) \mid g(x) \in \mathbb R[x] \,\}$ — i.e., the (principal) ideal of $\mathbb R[x]$ generated by $x - 1$. $\endgroup$ – M. Vinay May 30 '16 at 9:22

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