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Consider the ring homomorphism $ϕ : \mathbb{R}[x] → \mathbb{R}[\sqrt{−3}]$ defined by $ϕ(x) = \sqrt{−3}$.

i) Show that $ϕ$ is surjective.

It seems obvious, so not sure how to show it

ii) Find $\ker ϕ$ and prove your answer is correct. [Hint: use the fact that $\mathbb{R}[x]$ is a principal ideal domain.]

I know the kernel is $x^2 +3$ but dunno how to use the PID to prove it.

Can anyone help?

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  • $\begingroup$ i) How in general do you show a function is surjective? ii) The kernel is an ideal and thus principal, and you know it contains x^2+3. $\endgroup$ – John Brevik Apr 13 '15 at 21:42
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1) Show that both $\mathbb R$ and $\sqrt{-3}$ are in the image. These generate the ring. (Or maybe the result is even more trivial, depending on you rlocal defiinition of $\mathbb R[\alpha]$).

2) The kernel is of the form $(f)$ for some polynomial $f$. We need $f\mid x^2+3$. Why can ew exclude the possibilities $\deg f=1$ and $\deg f=0$?

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Part i) is obvious. The ring $\mathbb{R}[\sqrt{-3}]$ is the set of polynomial expressions of the form $a_0+a_1s+\dots+a_ns^n$, with $a_i\in\mathbb{R}$ and $s=\sqrt{-3}$.

For any ring homomorphism $f\colon R\to S$ and any element $s\in S$, there is a unique ring homomorphism $\hat{f}\colon R[x]\to S$ such that $\hat{f}(r)=f(r)$, for $r\in R$, and $\hat{f}(x)=s$.

The kernel of $\phi$ is indeed the ideal generated by $X^2+3$. Indeed, it's almost obvious that $\phi((X^2+3)f(X))=0$, because $\phi(X^2)=-3$.

Conversely, if $g(X)\in\ker\phi$, then $\sqrt{-3}$ is a root of $g$, so also $-\sqrt{-3}$ is a root (why?). So… (Note that the embedding of $\mathbb{R}$ into the complex numbers is used.)

Alternatively, let $h(X)$ be a generator of $\ker\phi$. Then $h$ divides $X^3+3$, so…

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