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  1. Show that the map $\phi : \mathbb{Z}[x] \rightarrow \mathbb{R}$ given by $\phi(f(x)) := f( \sqrt 2)$ is a ring homomorphism.

  2. Find out the kernel of $\phi $

My attempt : For $1$ we have $\phi(f(x_1 + x_2)) := \phi(f(x_1) + f(x_2))=\phi(f(x_1)) + \phi((x_2))= f( \sqrt 2) + f( \sqrt 2) = 2 f( \sqrt 2)$

$\phi(f(x_1)(f(x_2)) :=\phi(f(x_1)).\phi((x_2)) =f( \sqrt 2)^2$

For $2$ . $\ker \phi = (x- \sqrt 2)(x+ \sqrt 2)= (x^2-2)$

Is my way of answer is correct or not ?

Community
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jasmine
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1 Answers1

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Your analysis is not correct. The elements of your ring are polynomials, not variables.

$\phi(f+g) = (f+g)(\sqrt 2)=f( \sqrt 2)+g(\sqrt 2) = \phi(f(x))+\phi(g(x))$

The analysis for multiplication tracks the analysis for addition:

$\phi(f\cdot g) = (f \cdot g)(\sqrt 2)=f( \sqrt 2) \cdot g(\sqrt 2) = \phi(f(x)) \cdot \phi(g(x))$

Finally, $f \in \ker(\phi) \iff \phi(f)=0 \iff f(\sqrt 2)=0 \iff (x^2-2) \mid f$ so $\ker(\phi) = (x^2-2)\Bbb Z[x]$.

Robert Shore
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  • ..thanks u @Robert shore but one more doubts is that $\ker(\phi) = (x^2-2)\Bbb Z[x]$ or $\ker(\phi) = (x^2-2)$ ....Here im not getting why u have wriiten $\mathbb{Z[x]}$ – jasmine Oct 25 '19 at 10:07
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    @jasmine Because any polynomial that's a multiple of $x^2-2$ is also in $\ker(\phi)$. In other words, $\ker(\phi)$ is the principal ideal generated by $x^2-2$. – Robert Shore Oct 25 '19 at 18:38