Comments on HW Set 4

Problem 1.

The most interesting part of this problem was to prove that, for $d\ge 1$ and any positive integer $n\ge 2$, the polynomial $P_{n,d}(X_1,\ldots, X_n)=X_1^{d}+X_{2}^{d}+\cdots+X_{n}^{d}-1$ is irreducible. This is done by induction on $n$. When $n=2$, the polynomial in question is $P_{2,d}=X_2^{d}+(X_1^{d}-1)$ which we view as a polynomial in $\Q[X_1][X_2]$. The polynomial $X_1^d-1=(X_1-1)(q(X_1))$ where $q(1)\not=0$ ($q(X_1)=1+X_1+\cdots+X_1^{d-1}$). Therefore $X_1-1$ generates a prime ideal $\mathbf{p}$ in $\Q[X_1]$ and so $P_{2,d}$ is an Eisenstein polynomial in $\Q[X_1][X_2]$ for this prime $\mathbf{p}$ and therefore irreducible.

Now suppose we know that $P_{n,d}$ is irreducible in $\Q[X_1,\ldots, X_n]$. Since this multivariable polynomial ring is a UFD, $P_{n,d}$ generates a prime ideal. Therefore $X_{n+1}^d+P_{n,d}$ is an Eisenstein polynomial for that prime, and is therefore irreducible.

Problem 3.

Let $R=F[x,y_1,y_2,\ldots]/I$ where $I$ is the ideal generated by $(x-y_1^2, y_1-y_2^2,\ldots)$.

The element $\overline{y}_{m}=y_{m}+I\in R$ satisfies $\overline{y}_{m}^{2^{m}}=\overline{x}$ and more generally $\overline{y}_{m}^2=\overline{y_{m-1}}$. We let $R_{n}$ be the subring of $R$ generated by $\overline{y}_{n}$. To simplify matters, we write $\overline{y}_{0}=\overline{x}$.

a. $R=\bigcup_{i=m}^{\infty} R_{m}$.

Proof: Any element $f$ of $R$ is a polynomial in finitely many of the variables $\overline{y}_{i}$. Let $N$ be the largest integer such that $\overline{y}_{N}$ appears in a monomial with nonzero coefficient in $f$. Since $\overline{y}_{m}=\overline{y}_{N}^{2^{N-m}}$ for $m\le N$, we may rewrite $f$ as a polynomial in $\overline{y}_{N}$. Therefore $f$ belongs to $R_{N}$.

b. $R_{n}$ is isomorphic to a polynomial ring in one variable over $F$.

Proof: Let $S_{n}=F[x,y_1,\ldots, y_n]$. We have a ring homomorphism $\phi_{n}:S_{n}\to R$ coming from the inclusion of the polynomial ring $S_{n}$ into the big ring $F[x,y_1,\ldots]$. If $f\in S_{n}$, then $\phi(f)\in R_{n}$ because, as we’ve seen in the argument in part (a), any polynomial in the variables $\overline{y_i}$ for $i\le n$ is equivalent mod $I$ to a polynomial in $\overline{y}_{n}$ and thus lies in $R_{n}$. In fact, $\phi_n$ is a map onto $R_{n}$ since $R_{n}$ is generated by $\overline{y}_{n}$ and $\overline{y}_{n}$ lies in the image of $\phi_n$.

The ideal $J_{n}=(x-y_1^2,y_1-y_2^2,\ldots, y_{n-1}-y_{n}^2)\subset S_{n}$ lies in the kernel of $\phi$. The quotient $S_{n}/J_{n}$ is isomorphic to $F[y_{n}]$. To see this, note that $S_{n}=F[y_1,\ldots,y_n][x]$ so that any polynomial in $f$ can be written uniquely $f=f_{1}+a_1(x-y_1^2)$ where $f_{1}$ belongs to $F[y_1,\ldots, y_n]$. Iterating this process, we can write $f$ uniquely as $f=f_{n}+a_{n}(y_{n-1}-y_{n}^2)+a_{n-1}(y_{n-2}-y_{n-1}^2)+\cdots+a_{1}(x-y_1^2)$ where $f_{n}$ is in $F[y_n]$.

We would like to show that the kernel of the map $\phi_{n}:S_{n}\to R_{n}$ is $J_{n}$, since that would prove that $R_{n}$ isomorphic to $S_{n}/J_{n}=F[y_{n}]$. But if $\phi_n(f)=0$ in $R$ it means that, viewed as an element of the big polynomial ring $F[x,y_1,\ldots]$, the polynomial $f\in S_n$ belongs to $I$. In other words, we can write $f = \sum_{j=1}^{N} b_{j}(y_{j-1}-y_{j}^2).$ This equation holds in $S_{N}$ for some finite $N$ so it says that $f$ lies in $J_{N}$. Since we already know that $f=a_0+a_1 y_n+\cdots +a_k y_n^k+J_{n}$ can be written uniquely as a polynomial in $y_n$ modulo $J_n$, in the ring $S_{N}$ we know that $f$ must be $a_0+a_1y_{N}^{2^{N-n}}+\cdots+a_ky_{N}^{2^{N-n}}$ modulo $J_{N}$ and since $f$ lies in $J_{N}$ all of the coefficients $a_{i}$ must be zero. In other words, $f$ was already zero modulo $J_{n}$ and therefore $\phi_n$ is injective.

Remark: I think this is what is meant by the comment in DF about algebraic relations holding in $S_{N}$ for some $N\ge n$. Essentially we have proved that if we view $S_{n}$ as a subring of $F[x,y_1,\ldots,]$ then $I\cap S_{n}=J_{n}$; but to prove this we had to deal with the possibility that if $f$ involves the variables only up to $y_{n}$, it couldn’t still somehow be an element of $I$ involving higher numbered variables.

b’. $R$ is a Bezout domain.

Proof: Suppose that $a$ and $b$ are two elements of $R$. Then they lie in $R_{n}$ for some finite $n$. Since $R_{n}$ is a polynomial ring in one variable, there is a polynomial $h,x,y\in R_{n}$ such that $ax+by=h$ and such that $a=uh$ and $b=vh$. Then if $z=ra+sb$ for $r,s\in R$, we have $z=(ru+sv)h$ so $z$ belongs to $hR$; and conversely $hR$ belongs to $(a,b)$ since $h=ax+by$. Thus any ideal generated by two elements (or, more generally, any finitely generated ideal) is principal.

c. The ideal generated by $x,y_1,y_2,\ldots$ in $R$ is not finitely generated.

Proof: If it were, it would be principal and generated by some $h$ belonging to $R_{n}$ for some finite $n$. This would mean that for each $m\ge n$ there would be an $N\ge m$ for which we would have an equation $\overline{y}_{m}=hr_{N}$ for $r_{N}\in R_{N}$. Since $h\in R_{n}$, it can be written as a polynomial in $\overline{y}_{n}$ and, in $R_{N}$, as a polynomial in $\overline{y}_{N}^{2^{N-n}}$. Similarly $\overline{y}_{m}=\overline{y}^{2^{N-m}}$. Since $2^{N-m}$ is smaller than $2^{N-n}$, such a relation can only hold in $R_{N}=F[y_{N}]$ if $h$ is a constant polynomial, which would mean that the ideal generated by $x,y_1,\ldots$ would contain a unit and hence be all of $F[x,y_1,\ldots]$. But this is not true; the quotient of the big polynomial ring by this ideal is $F$.

Problem 5

Let $V$ be a vector space of dimension $n$ over a field $F$. A complete flag in $V$ is a sequence of subspaces $Z: W_{0}=(0)\subset W_1\subset W_2\subset \cdots \subset W_{n-1}\subset W_{n}=V$ where $W_{i}$ has dimension $i$. The group $\GL(V)$ acts on the flags by permuting the subspaces.

Given a subspace $W_{i}\subset W_{i+1}$ where $W_{i+1}$ is of one higher dimension, then, given a basis of $W_{i}$, you can find a vector in $W_{i+1}$ to add to this basis to get a basis of $W_{i+1}$. Using this inductively, given a flag, you can construct an ordered basis of $V$. Conversely, given an ordered basis $v_1,\ldots, v_n$ you get a flag by taking the span of $v_1$, then the span of $v_1, v_2$, and so on.

Now suppose you have two flags. Using the above process you get two ordered bases for $V$. Choose an element of $GL(V)$ carrying one basis to the other; it will carry one flag to the other. This proves the action on flags is transitive.

Now fix a flag and choose a compatible ordered basis $v_1,\ldots, v_n$. Suppose $g$ is an element of $\GL(V)$ that fixes this flag. This means that $gW_i\subset W_i$ for each $i$, and since $W_i$ is spanned by $v_1,\ldots, v_i$ we must have $g v_i = \sum_{j=1}^{i} b_{ji}v_{j}$ with $b_{ji}=0$ if $j>i$. This means the matrix of $g$ is upper triangular. Since it’s invertible, the diagonal entries can’t be zero.

Finally, over a finite field, we know that $\GL(V)$ has $\prod_{i=0}^{n-1}(q^{n}-q^{i})$ elements. An upper triangular matrix can have arbitrary entries in the off diagonal upper spots, and must have nonzero entries on the diagonal. So the there are $\prod_{i=1}^{n}(q-1)q^{n-1}$ elements of this type. The quotient gives the number of flags:

\[(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^{n-1})\]

If $q=2$ and $n=3$ this gives $3\cdot 7=21$ flags.