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I'm studying for an algebra midterm and I'm really just having a hard time wrapping my head around quotients of polynomial rings, especially ones where the ideal being quotiented by is something non-principle (i.e an ideal of the form $(x^2 - 2, 3$) in an appropriate polynomial ring).

For example this question Set of Ideals of a Polynomial Ring makes use of the fact that

$$\mathbb{Z}[x]/(2,x^3 + 1) \cong \mathbb{Z}_2[x]/(x^3 + 1)$$

to arrive at a solution, but this isomorphism doesn't at all seem obvious to me (hopefully because I'm just not thinking about the quotient in the correct way). Another example, also a question from dummit and foote ($\S 9.1, 13$), is ''Prove that the rings $F[x,y]/(y^2 - x)$ and $F[x,y](y^2 - x^2)$ are not isomorphic for any field $F$ ''. Really I don't even see an obvious direction to proceed, but I think, on a more fundamental level, I really just have no intuitive notion as to what those fields even look like.

So I was hoping for some helpful way(s) of thinking about these spaces. Any insight would be much appreciated.

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    $\begingroup$ How comfortable are you with quotients of rings in general? $\endgroup$ – Pete L. Clark Mar 7 '14 at 6:26
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    $\begingroup$ Hint: $F[x,y]/(y^2-x^2)$ has (obvious) zero divisors, $F[x,y]/(y^2-x)$ doesn't. $\endgroup$ – Gerry Myerson Mar 7 '14 at 6:27
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    $\begingroup$ If you are comfortable with principal ideals but not ideal generated by two elements, first thing you should do is to see what does Ideal generated by two elements mean... For $\mathbb{Z}[X]/(2)$ you would just see that you are actually making $2$ to be zero and for $\mathbb{Z}[X]/(2,x^3+1)$ you want $2$ to be $0$ and then you want $x^3+1$ to be zero in the quotient (in which $2$ is already zero) $\endgroup$ – user87543 Mar 7 '14 at 6:30
  • $\begingroup$ In fact as Mr.Gerry Myerson said one is actually a field and the other is not.. you just have to thank Eisenstein Irreducibility criterion for confirming irreducibility of $y^2-x$ is in $F[x,y]$ $\endgroup$ – user87543 Mar 7 '14 at 6:32
  • $\begingroup$ @PeteL.Clark I wouldn't say I'm comfortable. $\endgroup$ – Andrew Ross Mar 7 '14 at 6:39
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The trick is that you need to think about doing algebra with rings. Not with elements of rings, but with the rings themselves.

The isomorphism you mention can be easily calculated:

$$ \mathbb{Z}[x] / (2, x^3 + 1) \cong \big(\mathbb{Z}[x] / (2)\big) / (x^3 + 1) \cong \big(\mathbb{Z} /(2) \big)[x] / (x^3 + 1) $$

although you might have "seen" it by considering the most natural way to represent elements in the rings: you represent an element by writing down an integer polynomial in $x$, and in both cases, the equivalence relation that two integer polynomials represent the same element is the one generated by $2\equiv0$ and $x^3 + 1\equiv 0$.

Your example $F[x,y] / (y^2 - x)$ is a basic example of another particular sort of simplification: this is isomorphic to the ring $F[y]$ by the evaluation homomorphism $x \to y^2$. That is, the homomorphism $f(x,y) \mapsto f(y^2, y)$.

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  • $\begingroup$ See, regarding the step $\mathbb{Z}[x] / (2, x^3 + 1) \cong \big(\mathbb{Z}[x] / (2)\big) / (x^3 + 1)$, how do you justify 'pulling the 2 out of the ideal'? But what you're saying about $F[x,y] / (y^2 - x)$ seems to make sense. $\endgroup$ – Andrew Ross Mar 7 '14 at 7:01
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    $\begingroup$ I think of it in the terms that modding out by $(a,b)$ is the ring you get by setting $a \equiv 0$ and $b \equiv 0$, so the factorization is obvious. An introductory course would probably mutter something about isomorphism theroems instead: $$R/(I+J) \cong (R/I) / ((I+J)/I)$$ and take care to note that $(I+J)/I$ isn't literally $J$: it's just the ideal of $R/I$ generated by the equivalence classes of the generators of $J$. But IMO, this is more of a "simply stated technical fact" rather than the way to really think about it. $\endgroup$ – Hurkyl Mar 7 '14 at 7:45
  • $\begingroup$ The universal property I mentioned is that if $f: R \to S$ is a ring homomorphism with the property that $f(a) = 0$ and $f(b) = 0$, then there is a unique homomorphism $g : R/(a,b) \to S$ such that $f = g \circ \pi$. (where $\pi$ is the projection $R \to R/(a,b)$). Similar statements are true for any number of generators. So it really is reasonable to think of $R/(a,b)$ as the ring you get by making as little change to $R$ as possible to make $a=b=0$. $\endgroup$ – Hurkyl Mar 7 '14 at 7:52
  • $\begingroup$ That's interesting, I think I see what you mean. $\endgroup$ – Andrew Ross Mar 7 '14 at 8:06
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A quotient of rings is a structure where you add a new equation in the previous ring.

For example, $$\mathbb R[T]/(T^2 + 1)$$ is the ring of polynomials, with the new equation $$T^2 + 1 = 0$$so this is $\mathbb C$. So, making a quotient by an ideal generated by 2 elements gives you two new equations. That's all.

  1. For the first example, the ring is $\mathbb Z[x]$ with additional equations: $$2 = 0 \ \ \& \ \ x^3 = 1$$ so this is $\mathbb Z_2[x]/(x^3 + 1)$ indeed.
  2. For the second, consider an isomorphism $f$ from $R_1$ to $R_2$; $f(1) = 1$ so $f$ leaves $F$ invariant; it remains to find images of $x,y$ so take the relationship $$x^2 = y^2 \ \ \ (R_1)$$ it implies that $(x+y)(x-y) = 0$, so it should be the case for the images of $x,y$ in $R_2$.

Let $f(x) = P(x,y) = P(y^2,y)$ and $f(y) = Q(x,y) = Q(y^2,y)$. We have $$(P(y^2,y)+Q(y^2,y))(P(y^2,y)-Q(y^2,y))=0$$ but this is impossible, because it should be true in $\mathbb Z[y]$ which has no $0$ divisors.

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