Are these ideals principal, proper, maximal and prime?

Question

I am trying to show that the two following ideals are principal, proper, maximal and prime.

a) ${f \in \mathbb{Q}[A,B] : f(0,B)=0}$

and

b) $(4+1\sqrt{-2}, -6+9\sqrt{-2}) \subseteq \mathbb{Z}[\sqrt{-2}]$

For a) I think it is all polynomials that are multiples of x, but this doesn't quite feel like the right way of saying it. Would saying all polynomials of x be more correct? I also know that this one is proper because it doesn't contain 1. But I'm not sure about principal, maximal or prime so if someone can guide me through it I would appreciate it.

For b) I know that it is a PID, so it is principal. I also know that is is proper because it doesn't contain 1. But I don't know how to show it is maximal or prime so I'd appreciate someone guiding me through it.

Answer 1

For a), the ideal $I = \{ f(X,Y) \in \mathbb{Q}[X,Y] : f(0,Y) = 0 \}$ has $I = (X)$, which is not maximal, but is prime, as we can see from $\frac{\mathbb{Q}[X,Y]}{(X)} \cong \mathbb{Q}[Y]$: the quotient by $I$ is an integral domain, so $I$ is prime, but is not a field, so $I$ is not maximal. More concretely, $I \subsetneq (X,Y)$, so $I$ is not maximal.

Showing $I = (X)$ first, notice, if $f(a,Y) = 0$, then $(X-a) | f(X,Y)$. Proof: divide $f(X,Y)$ by $X-A$, giving $f(X,Y) = q(X,Y) (X-A) + r(X,Y), deg( r) < deg(X-A) = 1$. Then $r(X,Y) = k$ is a constant, in particular, 0: evaluating both sides at $X=A$ gives $0 = q(A,Y)(A-A)+k = k$. Hence our claim.

Then if $f(0,Y) = 0$, $X$ divides $f$ and $f\in I$, so $I\subset (X)$. The other direction is easy: if $g \in (X)$, $g(X,Y) = X* h(X,Y)$ for some h, and so $g(0,Y) = 0 *h(0,Y) = 0$, and $g\in I$. So $I= (X)$, and $I$ is prime and principal. It is proper clearly, because e.g. $f(X,Y) = Y \in \mathbb{Q}[X,Y]$ but $f \not \in I$.

As you identified, $\mathbb{Z}[\sqrt{-2}]$ is a PID, most easily seen by noting the norm $N(a+bi\sqrt{2}) = a^2 +2b^2$ makes it a Euclidean domain, and that Euclidean domains are also PIDs. In a PID, gcd's exist (PID is overkill, gcd's exist in UFDs, and PIDs are in particular UFDs) and $(a,b) = (gcd(a,b))$. This is essentially the definition of gcd. My computation skills are rusty, but I believe we have $gcd(3*(-3+2i\sqrt{2}),1+4i\sqrt{2}) = 1+i\sqrt{2}$. Since the norm of $x = 1+i\sqrt{2}$ is $N(x) = 3$, any expression $x=uv$ would mean $N(x)=N(uv)=N(u)N(v)=3$, since the norm is a ring homomorphism. As 3 is prime, u or v must have been exactly $\pm 1$, i.e. a unit, so $x$ is irreducible. Since irreducible=prime in PID's, (x) is a prime ideal, and is therefore also maximal, since prime=maximal in a PID, as well. Hence our desired $(3*(-3+2i\sqrt{2}),1+4i\sqrt{2})=(x)$ is proper, principal, and maximal, and also prime, since maximal ideals are always prime, and since we noted it along the way.

Answer 2

For a):

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Here's two polynomials. One is a multiple of $x$ and one is a polynomial in $x$: $1 + x + x^2$, $x(1 + xy + y^2)$ what happens when you substitute $x = 0$ in each one? So the conclusion is that the ideal is all multiples of $x$. This is by definition the principal ideal $(x)$.

Recall that an ideal $I$ is maximal if $R/I$ is a field and $I$ is prime if $R/I$ is an integral domain. To understand this quotient, notice that every polynomial can be written as $f(x,y) = xg(x,y) + h(y)$ where $h(y)$ collects all the terms not divisible by $x$. So in the quotient $\mathbb Q[x,y]/(x)$ we have $f + (x) = h + (x)$. So $\mathbb Q[x,y]/(x) \cong \mathbb Q[y]$.

Considering the decomposition $f(x,y) = xg(x,y) + h(y)$, prove that:

1. the ideal in part a) is exactly equal to $(x)$.
2. $\mathbb Q[x,y]/(x) \cong \mathbb Q[y]$ and that this is an integral domain but not a field.

Or if you like: $(x,y)$ is a maximal ideal containing $(x)$.

For b):

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You would like to write $(1+4\sqrt{-2}, -9+6\sqrt{-2})$ as a principal ideal for this. To do this consider the norm $$N(a + b\sqrt{-2}) = (a + b\sqrt{-2})(a - b\sqrt{-2}) = a^2 + 2b^2.$$ You can show that if $\alpha \mid \beta$ then $N(\alpha) \mid N(\beta)$. Since $N(1+4\sqrt{-2}) = 33$ and $N(-9+6\sqrt{-2}) = 153$, whatever generates $(1+4\sqrt{-2}, -9+6\sqrt{-2})$ must have a norm of $3$. How many elements of $\mathbb{Z}[\sqrt{-2}]$ have norm $3$? That should give you a finite set of candidates. I claim that $1 + \sqrt{-2}$ works.

Then follow the following chain of isomorphisms: \begin{align*} \mathbb{Z}[\sqrt{-2}]/(1 + \sqrt{-2}) &\cong \frac{\mathbb{Z}[X]/(X^2 + 2)}{(1 + X)} \\ &\cong \frac{\mathbb{Z}[X]}{(X^2 + 2, 1 + X)} \\ &\cong \frac{\mathbb{Z}[X]/(1 + X)}{(X^2 + 1)} \\ &\cong \frac{\mathbb{Z}}{((-1)^2 + 1)} \qquad (1 + X \equiv 0 \iff X \equiv -1) \\ &= \mathbb{Z}/(3). \end{align*}