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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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Shooting from the hip, perhaps $\mathbb Z[1/4]$? – Alex Becker Mar 28 '12 at 4:24
It's not fairly clear. Why do you think that? $4t$ isn't equal to zero; it's equal to $-3$. – Qiaochu Yuan Mar 28 '12 at 4:25
The leading coefficient of $4t + 3$ is not a unit, so I don't think you have a division algorithm on hand. A suggestion: $t$ is going to have to have to stand for some element of $\mathbf R$ (maybe even $\mathbf Q$, if you're lucky). What might this be? If I handed you $\mathbf Z[x]/(2x - 1)$, for example, I would expect you to say, "ah, $x$ is an inverse for $2$, so I'm getting $\mathbf Z[1/2]$". – Dylan Moreland Mar 28 '12 at 4:25
@DylanMoreland Right...$\mathbb{Z}[t]$ isn't a PID and so it can't be a Euclidean domain. One of my classmates suggested using $t=-\frac{3}{4}$ since we have that $4t+3=0$ but I wasn't entirely sure how to apply that. – chris Mar 28 '12 at 4:29
@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 $1/2$. – Dylan Moreland Mar 28 '12 at 4:33
up vote 7 down vote accepted

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.

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