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Let $I=(4t+3)$ be an ideal in $\mathbb{Z}[t]$. Find a subring of $\mathbb{R}$ isomorphic to $\mathbb{Z}[t]/I$.

If $(4t+3)$ were monic, this question would be easily answered but since it isn't I'm having problems seeing exactly how $\mathbb{Z}[t]/I$ could be isomorphic to a subring of the reals.

Consider $7t+a\in\mathbb{Z}[t]$. The remainder is $3t+(a-3)$.By looking at the remainder of polynomials divided by $4t+3$, it's fairly clear that elements in $\mathbb{Z}[t]/I$ have the form $a+bt$ for $a\in\mathbb{Z}$ and $b\in\mathbb{Z}/4\mathbb{Z} \Rightarrow \mathbb{Z}[t]/I\cong \mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$. But how can this be isomorphic to a subring of $\mathbb{R}$?

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    @chris I think your classmate is on to something; Zev's answer shows how to make this rigorous. Note that adjoining $-3/4$ is the same as adjoining $1/4$ or even just $1/2$.2012-03-28

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Hint: In the ring $\mathbb{Z}[t]/(4t+3)$, let $\overline{t}$ denote the equivalence class of $t$, and identify integers with their equivalence classes. Then $4\overline{t}+3=0.$ Can you think of a real number that has the above property? Now define an injective homomorphism from $\mathbb{Z}[t]/(4t+3)$ to $\mathbb{R}$, sending integers to integers and $\overline{t}$ to this real number; the image of this homomorphism will then be a subring of $\mathbb{R}$ isomorphic to $\mathbb{Z}[t]/(4t+3)$. It will probably help to appeal to the First Isomorphism Theorem when showing that this homomorphism is injective.

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Hint $\: $ Map $\mathbb Z[t]$ into $\mathbb R$ by evaluating $t$ at $-3/4$. The kernel is $(4t+3)$ via Factor Theorem in $\mathbb Q[t],$ combined with Gauss' Lemma.