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:


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

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

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

Hint is appreciated.

Thanks in advance

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $\varphi(f(x)) = f(1) = 1 - 3 + 2 = 0 \implies f(x) \in \ker \varphi$.


\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 $x - 1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Mathematicing Exactly, that's the definition of "$(x - 1)$ is a factor of $f(x)$". $\endgroup$ – M. Vinay May 30 '16 at 9:17
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.