Mg Problem 28 29

PRELIMINARY:

29. a. Example of an ideal of $\mathbb{Q}[x,y]$ that is prime but not maximal, and make as simple as possible.

$\{0\}$

For any $a*b \in I$, either $a$ or $b$ must be zero, but clearly this is not a maximal ideal, as there are others larger.

b. Prove that every ideal of $\mathbb{Z}$ is principle.

Suppose the contrary. Then the would exist an ideal generated by more than on integer, say $a$ and $b$. If $b$ is a multiple of $a$ or vice versa, clearly the ideal is actually principle. Therefore, $b$ can't be a multiple of $a$ or vice versa. If $a$ and $b$ are relatively prime, then their sum cannot be a multiple of either, and therefore the ideal is not closed under addition. Not sure where to go from here.

c. Show that not every ideal of $\mathbb{Q}[x,y]$ is principle by giving a counterexample.

One can generate an ideal of $\mathbb{Q}[x,y]$ using $x^2$ and $y^2$. This ideal is the set of all rational polynomials that are either zero or have x degree greater than or equal to 2 and y degree greater than or equal to 2.

28. Prove that an ideal I of a ring R is maximal if and only if the quotient ring R/I is a field.

Not sure where to go with this, but I'm guess we'll use the fact that if R/I is a field, then every element has a multiplicative inverse and no zero divisors. Other than that…

FINALS:

28. Prove that an ideal $I$ of a ring $R$ is maximal if and only if the quotient ring $R/I$ is a field.

Hey Andrew, I'm going to have to take the L on this one because there was a proof of it presented in class, but I don't for the life of me understand it, and it's not right to just write down a proof without even being able to explain it. Can we say the statement is true just because?

29. a. Example of an ideal of $\mathbb{Q}[x,y]$ that is prime but not maximal, and make as simple as possible.

The ideal generated by $x^2$ is prime because every element of the ideal can be written as $a*x^2$ for some polynomial, $a$, in $\mathbb{Q}[x,y]$. And since $x^2$ is in the ideal, every element of the ideal is the product of an element of the ideal and something else. However, it is not maximal because no terms other than $0$ of such an ideal have x-degree less than 2, and an ideal generated by $x$ would have more elements while still being proper.

b. Prove that every ideal of $\mathbb{Z}$ is principle.

Suppose there exists an ideal generated by multiple integers, $a$ and $b$. The elements of such an ideal would have the form $na + mb$ for all $n,m \in \mathbb{Z}$. However, $a$ and $b$ can be written as $a = cd$ and $b=ce$, where $c = GCD(a,b)$ and $d,e$ are coprime integers (REMARK: if $a=b$, then $d=e=1$, in which case $GCD(d,e) = 1$, so $d$ and $e$ are coprime, albeit trivially). Therefore, every element of such an ideal would have the form $na+mb = (cd)n+(ce)m = c(dn+em)$. Since $d$ and $e$ are relatively prime, there is some linear combination of $d$ and $e$ that equals 1. As such, there is a linear combination of $d$ and $e$ that equals every possible integer. Therefore, $c(dn+em) = c\mathbb{Z}$, which is a principle ideal.

QED

c. Find an ideal of $\mathbb{Q}[x,y]$ that is not principle.

The ideal generated by $x$ and $y$ is not principle because it is generated by two elements. This can be proven to be an ideal because every polynomial times x added to every polynomial times y is still in the ideal, and it is closed under addition.