Vtyjkyuk

I'm working on the following problem in preparation for an exam.

Let $f(x) = x^3 + 2 in mathbbZ[x]$. Let $alpha$ be any complex number satisfying $f(x)$. Let $K = mathbbQ(alpha)$ and suppose $R$ is the collection of element in $K$ with integer coefficients (with basis $1, alpha$, and $alpha^2$).

Next we are asked to show the following two equalities:

1. For every non-zero ideal $I subseteq mathbbZ$, show that $IR cap mathbbZ = I$.


2. Show that if $J subseteq R$ is a nonzero ideal , then $J cap mathbbZ neq emptyset$.


3. Show that every nonzero prime ideal in $R$ is maximal.

Here is my work so far: It is clear that $IR$ contains $I$ and $mathbbZ$ contains $I$. Thus $IR cap mathbbZ supseteq I$. Now suppose $x in IR cap mathbbZ$. Then $x = ir$ for some $i in I$ and $r in R$. Also, $ir in mathbbZ$. Now I am not sure how to proceed. I want to show that $x in I$.

I know that $mathbbZ$ is a UFD, and I believe that saying $x$ has a unique factorization in $mathbbZ$ should help. In fact, since $x in mathbbZ$ and $I subseteq mathbbZ$ then $ir in mathbbZ$ says that $r$ is also in $mathbbZ$. But then $x$ is a multiple of an element in $I$, and so $x in I$. (is this correct?)

Unless I am mistaken, $0 in J$ for every nonempty $Jsubseteq R$.

I'm not sure where to start on part 3. I know what prime ideals are, and what maximal ideals are.

If someone has a better suggestion for the title, please let me know. I wasn't sure how to paraphrase this question.

So you have seen that $I subset IR cap mathbb Z$, and are confused about the other way.

The point is, note that every ideal of $mathbb Z$ is of the form $n mathbb Z = nz : z in Bbb Z$. So, write $I = nBbb Z$.

Now, let $x in IR cap mathbb Z$. Clearly, $x$ is an integer. Furthermore, note that $x = ir$ for some $i in I, r in R$. We know what $I,R$ are : this helps us conclude that $i = jn$ for some integer $j$ and $r = d + ealpha + f alpha^2$ for integers $d,e,f$.

Thus, $x = jnd + jnealpha + jnfalpha^2$. Transposing $x$ to the other side, $jnf alpha^2 + jnealpha + jnd - x = 0$. So $alpha$ satisfies a polynomial of degree smaller than $3$ with integer coefficients. This polynomial must be a factor of $x^3 +2$, but that is irreducible, so this cannot happen. Consequently, the above polynomial must be zero. That is, all the coefficients are zero : $jnf = 0$,$jne = 0$ and ,more importantly, $jnd - x = 0$ or $x = jnd$, that is $x in I$. Hence, the first part is complete.

For the second part, we know $0 in J cap mathbb Z$, but let us make it stronger. Let $p(x)$ be a non-zero polynomial such that $p(alpha) in J$. Since $alpha^3 = 2$, we can assume $1 leq deg p leq 2$ (if $deg p = 0$ then $p$ is a non-zero constant, so we are done). Now, using the division algorithm, write $x^3 + 2 = p(x)q(x) + r(x)$, where $r(x)$ has degree less than $p(x)$, and is non-zero because $x^3 + 2$ is irreducible. Substituting $alpha$, we get $0 = p(alpha) q(alpha) + r(alpha)$. Thus, $r(alpha) = p(alpha) times -q(alpha) in J$.

Now, $deg p = 1 implies deg r = 0$ so $J$ contains a non-zero constant.
Now, if $deg p = 2$ then either $deg r = 0$,whence we are done, or $deg r = 1$, whence we replace $p$ by $r$ in the paragraph above to get the conclusion. Either way, $J$ contains a non-zero constant, so $J cap mathbb Z$ is a set which contains more elements than just zero. In fact, let $n$ be the smallest non-zero positive integer contained in $J cap Bbb Z$. Of course, for any $m$, we have $nm in J cap Bbb Z$. However, if using division algorithm , $m = nq + r in J cap mathbb Z$, then $r in J cap mathbb Z$ and $r < n$, contradiction. Consequently, $J cap mathbb Z = nmathbb Z$ with $n$ as above.

Let $J$ be a prime ideal of $R$. Then, $frac RJ$ is an integral domain. Now, note that $J cap mathbb Z = n mathbb Z$, so therefore, $(n mathbb Z) R subset JR = J$. Therefore, it also follows that $frac RJ subset frac R(n mathbb Z) R$ (the latter is the quotient of $R$ by the ideal generated by the constant polynomial $n$ in $R$).

Now, define the map $R to left(fracmathbb Znmathbb Zright)^3$ by $aalpha^2 + balpha+c to ([a],[b],[c])$, where $[x]$ denotes remainder of $x$ when divided by $n$. Clearly, $(n mathbb Z)R$ is the kernel of this map, so it follows that the cardinality of $frac R(n mathbb Z)$ is finite, since it is less than or equal to $n^3$ (infact, it is equal, you can see surjectivity of the map easily). Finally, it follows that $frac RJ$ is a finite integral domain, and these are known to be fields (the popular pigeonhole argument). Consequently, $frac RJ$ is a field, so $J$ is maximal.