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The question is, Given MCQ, Which of the following is true?

(a) $Z[x]$ is principal ideal domain.

(b) $Z[x,y]/\langle y+1\rangle$ is a unique factorization domain.

(c) If $R$ is a principal ideal domain and $p$ is a non-zero prime ideal, then $R/P$ has finitely many prime ideal.

(d) If $R$ is principal ideal domain, then any subring of $R$ containing $I$ is again a principle ideal domain.

The correct option are (b) and (c). I got the option (c) is correct. For option (b), it was written in the explanation, that $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}\cong \mathbb{Z[x]}$ and since $\mathbb{Z[x]}$ is Unique Factorization Domain, $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}$ is also unique factorization domain.

My question, how we got that $\frac{\mathbb{Z[x,y]}}{\langle y+1\rangle}\cong \mathbb{Z[x]}$? How just seeing this quotient ring, we get the idea that this quotient ring can be isomorphic to $\mathbb{Z[x]}$?

Gary
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math student
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    Conceptually, $\Bbb Z[x,y]/(y+1)$ is the ring $\Bbb Z[x,y]$ with the added relation $y=-1$. Since $-1$ is already in $\Bbb Z$, $\Bbb Z[x,-1]$ is the same as $\Bbb Z[x]$. You can make this precise without any surprises. – pancini Dec 18 '21 at 09:22
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    This is a very useful way to think about quotients in general, by the way. That's why you will see things like $\Bbb Z[x]/(x^2=-1)$ to mean $\Bbb Z[x]/(x^2+1)$. This emphasizes that $x^2$ is "a thing which squares to $-1$." – pancini Dec 18 '21 at 09:29
  • As in the linked dupe, $R[y]/(y-r)\cong R$ follows via the evaluation hom $,y\mapsto r$ and the First Isomorphism Theorem. Thus $\Bbb Z[x,y]/(y-1)\cong \Bbb Z[x],$ is the special case $R = \Bbb Z[x],,$ by $,\Bbb Z[x,y]\cong (\Bbb Z[x])[y] = R[y]\ \ $ – Bill Dubuque Dec 18 '21 at 17:36

1 Answers1

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Given $\alpha\in \mathbb{Z}[x,y]$ we can regard it as a polynomial in $y$ with coefficients in $\mathbb{Z}[x]$: $$\alpha=\sum_{i=0}^n r_i(x)y^i.$$

Then \begin{eqnarray*}\alpha&=&\sum_{i=0}^n r_i(x)(y^i-(-1)^i)+\sum_{i=0}^n (-1)^ir_i(x)\\\\&=&(y+1)\sum_{i=0}^n r_i(x)(y^{i-1}-y^{i-2}+\cdots -(-1)^i)+\sum_{i=0}^n (-1)^ir_i(x)\end{eqnarray*}

Thus in $\mathbb{Z}[x,y]/\langle y+1 \rangle$, $\alpha$ is equivalent to $\sum_{i=0}^n (-1)^ir_i(x)\in \mathbb{Z}[x]$.

Further $1+y$ does not divide any non-zero element of the subring $\mathbb{Z}[x]$, so distinct elements of this subring remain distinct in the quotient.

The above calculation is just capturing the intuition mentioned in comments.

tkf
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