Showing every prime ideal is maximal.

Question

I need help with this question number 8, at least some hints or approaches to start with. Sorry for posting an image, I am only doing so , because I don't have a pc to type the question in latex and I don't know how to do so using mobile in any other way.

Answer 1

Hint:

Consider a prime ideal $\mathfrak p$ of $R$. Observe that $R/\mathfrak p$ is a finite-dimensional $k$-vector space. It is a consequence of the following result, which you can prove:

Let $A\subset B$ be integral domains such that $B$ is a finitely generated $A$-module. Then $A$ is a field if and only if $B$ is a field.

Edit: You can adapt the proof of the left-to-right implication to your context. Here's how:

All you have to show is that any non-zero element $\bar r\in R/\mathfrak p$ has a reciprocal. So consider multiplication by $\bar r$ in the quotient. Observe it is $k$-linear and injective. Furthermore $R/\mathfrak p$ is finite dimensional. What can you conclude from these elements?

Answer 2

If $D$ is an integral domain which is finite dimensional over some subfield

$k \subset D, \tag 1$

that is,

$\dim_k D = n < \infty, \tag 2$

then $D$ is in fact a field. For given

$0 \ne u \in D, \tag 3$

the set of the first $n$ powers of $u$

$S = \{1_D, u, u^2, \ldots, u^n \} = \{ u^k, \; 0 \le k \le n, \; k \in \Bbb Z \} \subset D, \tag 4$

must contain a linearly dependent subset by virtue of (2); thus there exist

$\alpha_i \in k, \; 0 \le i \le n, \tag 5$

not all $0$, such that

$\displaystyle \sum_0^n \alpha_i u^i = 0; \tag 6$

thus we have

$p(x) = \displaystyle \sum_0^n \alpha_i x^i \in k[x] \tag 7$

with

$p(u) = 0; \tag 8$

now granted that such polynomials exist, we may choose

$m(x) = \displaystyle \sum_0^{\deg m} m_i x^i \in k[x], \; m(u) = 0, \tag 9$

of minimal degree amongst all such $p(x)$; for such $m(x)$,

$m_0 \ne 0 \tag{10}$

lest we write

$m(x) = x \displaystyle \sum_1^{\deg m} m_i x^{i - 1}, \tag{11}$

and hence

$u \displaystyle \sum_1^{\deg m} m_i u^{i - 1} = m(u) = 0, \tag{12}$

which since $u \ne 0$ implies

$\displaystyle \sum_1^{\deg m} m_i u^{i - 1} = 0, \tag{13}$

i.e., $u$ is a zero of

$g(x) = \displaystyle \sum_1^{\deg m} m_i x^{i - 1} \in k[x], \tag{14}$

which contradicts the mininality of the degree of $m(x)$ amongst polynomials satisfied by $u$; thus

$m_0 \ne 0, \tag{15}$

and we have

$\displaystyle \sum_0^{\deg m} m_i u^i = 0, \tag{16}$

or

$u\displaystyle \sum_1^{\deg m} m_i u^{i - 1} = \sum_1^{\deg m} m_i u^i = -m_0, \tag{17}$

whence

$u \left (-m_0^{-1} \displaystyle \sum_1^{\deg m} m_i u^{i - 1} \right ) = 1, \tag{18}$

that is,

$u^{-1} = -m_0^{-1} \displaystyle \sum_1^{\deg m} m_i u^{i - 1}; \tag{19}$

since every $0 \ne u \in D$ is invertible, we conclude that $D$ is a field; indeed, a field extension of $k$.

We apply this result to the present situation as follows: if $P$ is prime in $D$, then $D/P$ is an integral domain, and (2) implies

$\dim_k D/P < \infty, \tag{20}$

as well; thus by what we have done above, $D/P$ is a field and hence $P$ is maximal in $D$. $OE\Delta$.

Nota Bene: I must say in closing that if I were writing this proof as an answer to an exam (see comnents to the question itself), I would certainly find a way to make it shorter! End of Note.