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I can't see it.

$\Large{\frac{\mathbb{Z}[x]}{(x^2-29)}}$

I know you take general polynoimals with coefficients in $\mathbb{Z}$ and add $(x^2-29)$. However, I'm a bit confused on what it is. Like I image it as a big coset.

Also, how do you show that it is isomorphic to a subring of $\mathbb{R}$? wouldn't you need to do the subring test.

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    You're just appending a square root of 29 to the ring of integers and calling it $x$, and closing under addition, subtraction, and multiplication.2011-10-12

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Well, do you have any ideas as to what subring of $\mathbb{R}$ it is isomorphic to? For example, what real number(s) does the element $x$ "act like"? Specifically, what is $x^2$ in the ring $\mathbb{Z}[x]/(x^2-29)$?
That should then suggest a homomorphism $f:\mathbb{Z}[x]/(x^2-29)\to \mathbb{R}$ that is an isomorphism onto its image (which will be a subring of $\mathbb{R}$).

The ring $\mathbb{Z}[x]/(x^2-29)$ consists of equivalence classes of polynomials in $\mathbb{Z}[x]$, modulo $x^2-29$. What does that mean? Well, given any two polynomials $f=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in\mathbb{Z}[x]$ $g=b_0+b_1x+b_2x^2+\cdots+b_mx^m\in\mathbb{Z}[x]$ we say that $f\equiv g\bmod (x^2-29)$ when $f-g=(x^2-29)h,\;\;\text{ for some }h\in\mathbb{Z}[x].$ Consider, by analogy, the ring $\mathbb{Z}/(n)$, which consists of equivalence classes $[a]$ (where $a\in\mathbb{Z}$) such that $[a]=[b]$ iff $a\equiv b\bmod n$, and satisfying $[a]+[b]=[a+b]$ and $[a][b]=[ab]$. When does $[a]=[0]$?

Similarly, the ring $\mathbb{Z}[x]/(x^2-29)$ consists of equivalence classes $[f]$ such that $[f]=[g]$ iff $f\equiv g\bmod (x^2-29)$, and $[f]+[g]=[f+g]$ and $[f][g]=[fg]$. When does $[f]=[0]$ in $\mathbb{Z}[x]/(x^2-29)$? Can you characterize that via the roots of $f$? Consider the Factor Theorem.

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    Your second statement is backwards - you want to say that if $f$ is$a$*multiple* of $(x^2-29)$, then $[f]=[0]$. This then implies that $[x^2]=[29]$, because $x^2-29$ is a multiple of, well, $x^2-29$ :) So, can you describe a homomorphism $\phi:\mathbb{Z}[x]/(x^2-29)\to\mathbb{R}$? Note that because $[x^2-29]=[0]$, we must have that $\phi([x^2-29])=\phi([x])^2-\phi(29)=\phi([x])^2-29=0=\phi([0])$ because we want $\phi$ to be a ring homomorphism, and we have to send integers to themselves. That only leaves two possibilities for what $\phi([x])$ could be...2011-10-12