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I'm a bit stuck on this one.

Thought about defining $\phi:\mathbb{R}[x]\rightarrow\mathbb{R} $

as $\phi(p(x))=p(5)$. So we can see that $\ker\phi=\langle x-5\rangle$

How could I proceed from here?

2 Answers 2

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I presume you're trying to identify $\mathbb R[x]/(x-5)$?

$\phi$ is a homomorphism of rings. As you mentioned, ${\rm ker} \phi = (x-5)$. It shouldn't be hard to see that ${\rm im} \phi = \mathbb R$. Now use the fact that if $\phi : R \to S$ is a ring homomorphism, then $R/{\rm ker } \phi \cong {\rm im} \phi$.

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    Identifying $Im\phi = \mathbb{R}$ was my problem. How did you get that? – 2017-02-28
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    If I give you a number $c \in \mathbb R$, can you find a polynomial $\phi(x) \in \mathbb R[x]$ such that $\phi(5) = c$? – 2017-02-28
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    How about the constant polynomial, $\phi(x) = c$? – 2017-02-28
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    Right... thanks! – 2017-02-28
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Let $q(x)$ be an element in $\Bbb R[x]$, and let $n=\deg q(x)$. Then $$q(x)=(x-5)g(x)+r(x),$$ but $r(x)\in \Bbb R$, so it is a constant because $\deg g(x)=n-1$.

Thus $\Bbb R[x]/( x-5 ) \cong \Bbb R$, because $\ker\phi=\{p(x) : (x-5)/ p(x)\}\cong \Bbb R_{\ge 1}[x]$.

EDIT: the last isomorphism is the following. First of all choose $\mathcal{B}=\{1,x-5,x^2-25,\dots,x^n-5^n\dots\}$ as basis for $\Bbb R[x]$ and $\ker\phi=\langle \mathcal{B}\setminus \{1\}\rangle$. Then the isomorphism is given by a map sending $x^n-5^n$ to $x^n$.

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    Why a downvote? – 2017-03-09
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    It might be because assuming the $\phi$ is the same as in the question you have described the kernel incorrectly. – 2017-03-09
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    The kernel is still incorrect after your edit. For instance $\phi(x) = 5$, while $deg(x) \geq 1$. – 2017-03-09
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    What is the final isomorphism supposed to be? Did you forget to edit it out? – 2017-03-10
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    What do you mean by $\mathbb{R}_{ \geq 1}[X]$ ? I would read that as polynomials with real coefficients all greater than $1$. Then your isomorphism is false, since $-(x-5) \mapsto -x$ which is not in $\mathbb{R}_{ \geq 1}[X]$. Did you mean to say $\mathbb{R}[X]_{\geq 1}$, i.e. all polynomials of degree greater than $0$? In that case, why are you including that isomorphism? All you are saying is that $ker \phi$ is an infinite-dimensional vector space over $\mathbb{R}$. – 2017-03-10
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    Clearly $\Bbb R[x]_{\ge1}$ include polynomial of degree $1$ and greater than one. I include this isomorphism because you asked me this. – 2017-03-10
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    Yes, I asked you to explain the isomorphism that you initially wrote. However, I don't see why the isomorphism has anything to do with the answer in the first place. To me it seems like you are implying that $\Bbb R[X]/(ker \phi) \cong \Bbb R$ because $ker(\phi) \cong \mathbb{R}[X]_{ \geq 1}$, but this is not what is going on at all. – 2017-03-10