User:J D Bowen/Math899 HW1

1.1.2) In order for $U(n) \ $ to be a subvariety, it must be a variety in its own right. It is not, and we can prove it.

Assume to the contrary $\exists \ $ a polynomial $f:\mathbb{C}^{n^2}\to\mathbb{C} \ $ such that $U(n)= \mathbb{V}(f) \ $.

If we let $x \ $ be an $n\times n \ $ matrix representing a point in $\mathbb{C}^{n^2} \ $, we know that setting $g(x)=\Delta(x^{-1}-x^\dagger) \ \text{if} \Delta(x)\neq 0, 1 \ \text{else} \ $ gives us a function whose zero set includes $U(n) \ $. Therefore, $x\in U(n) \implies f(x)=g(x) \ $. But as a polynomial, $f \ $ is holomorphic everywhere, and as a function involving complex conjugates, $g \ $ is holomorphic nowhere. A holomorphic function is defined globally by its values on an open subset of its domain. Since this means that $f|_S=g|_S \implies f=g \ $, and we know $f\neq g \ $, we cannot have $f|_S = g|_S \ $ on any open set $S\in\mathbb{C}^{n^2} \ $.

Note that $S\subset U(n) \ $...


 *  this needs some major work 

...no such polynomial $f \ $ exists.

Now let's show that $U(n) \ $  is  a variety in $\mathbb{R}^{2n^2} \ $.

Define three maps $\alpha:\mathbb{C}^{n^2}\to M_{n\times n}(\mathbb{C}), \ \beta:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C}), \ \gamma:M_{n\times n}(\mathbb{C})\to\mathbb{R}^{+}\cup \left\{{0}\right\}\subset\mathbb{C} \ $ as

$\alpha(\vec{x})=\alpha(x_1,\dots,x_{n^2})=\begin{pmatrix} x_1 & x_2 & \cdots & x_n \\ x_{n+1} & x_{n+2} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{(n-1)n+1} & x_{(n-1)n+2} & \cdots & x_{n^2} \\ \end{pmatrix} \ $

$\beta(\mathbf{M}) = \mathbf{I}_n - \mathbf{M}\mathbf{M}^\dagger \ $

$\gamma(\mathbf{N}=(n_{ij}))= \sum_{1\leq i,j \leq n} n_{ij}\overline{n_{ij}} \ $

Observe that $\beta(\mathbf{M})=\mathbf{0} \ $, the matrix having 0 in all entries, if and only if $\mathbf{I}_n=\mathbf{M}\mathbf{M}^\dagger \ $. If $\mathbf{M} \ $ is invertible, then we can multiply this equation by the inverse to get $\mathbf{M}^{-1}=\mathbf{M}^\dagger \ $, which is satisfied by the unitary matrices $U(n) \ $, and only those matrices. If $\mathbf{M} \ $ is not invertible, then there is no matrix $\mathbf{A} \ $ such that $\mathbf{M}\mathbf{A}=\mathbf{I}_n \ $, and so we can be sure that $\beta(\mathbf{M})\neq\mathbf{0} \ $.

Further observe that since $n_{ij}\overline{n_{ij}}=|n_{ij}|^2\geq 0 \ $, we must have $\gamma(\mathbf{N})=0 \iff (\forall i,j, \ n_{ij}=0) \ $.

Consider the function $\gamma\circ\beta\circ\alpha:\mathbb{C}^{n^2}\to\mathbb{R} \ $. Since this function involves complex conjugates, it is certainly not a polynomial in $\mathbb{C}^{n^2} \ $. However, if we define the function $\text{realify}:\mathbb{C}^{n^2} \to \mathbb{R}^{2n^2} \ $ defined $\to(\text{Re}(x_1), \text{Im}(x_1), \dots, \text{Re}(x_n),\text{Im}(x_n) ) \ $, then we can immediately see $\text{realify} \ $ is a bijection, and so $f=\gamma\circ\beta\circ\alpha\circ\text{realify}^{-1}:\mathbb{R}^{2n^2} \to \mathbb{R} \ $ is well-defined, and since the complex conjugates have now just turned into negatives, it certainly is a polynomial.

We have $\mathbb{V}(f)=(\text{realify}\circ\alpha^{-1}) ( U(n)) \ $. Hence, $U(n) \ $ is an affine algebraic variety in $\mathbb{R}^{2n^2} \ $.

1.2.1) Suppose $X_1, X_2 \in \mathbb{A}^k \ $ are affine algebraic varieties. Then there exist polynomials $f_1, f_2:\mathbb{A}^k \to \mathbb{C} \ $ such that $X_j=\mathbb{V}(f_j) \ $.  Consider the polynomial $g:\mathbb{A}^k \to \mathbb{C} \ $ defined $g(z)=f_1(z)f_2(z) \ $, where $z\in\mathbb{A}^k \ $.  Then this polynomial will be zero precisely when $f_1 \ $ is zero, or $f_2 \ $ is zero, or both are.

Hence $X_1 \cup X_2 = \mathbb{V}(g) \ $, and so $X_1 \cup X_2 \ $ is an affine algebraic variety.

1.2.3) The twisted cubic curve is defined in Figure 1.5 as $V=\mathbb{V}(x^2-y,x^3-z)=\mathbb{V}(x^2-y)\cap\mathbb{V}(x^3-z) \ $ in $\mathbb{R}^3 \ $. Therefore, for any point $\vec{r}=(x,y,z)^t \in V \ $, we have $x^2-y=x^3-z=0 \ $.  Therefore, $x^2=y, x^3=z \ $, and so we can write any point $\vec{r}\in V \ $ as $(x,x^2,x^3) \ $.

2.1.2,

2.1.3)

Let $I\subset S \ $ be an ideal. We aim to show that any ring homomorphism $\sigma:R\to S \ $ induces an injective homomorphism $\hat{\sigma}:R/\sigma^{-1}(I) \to S/I \ $, and that $\sigma^{-1}(I) \ $ is prime whenever $I \ $ is.

Since we have $x\in R/\sigma^{-1}(I) \implies x=r+\sigma^{-1}(I) \ $, where $r\in R \ $, consider the function defined $\hat{\sigma}(x)= \sigma(r)+I \ $.

If we let $x=r_x+\sigma^{-1}(I), y=r_y+\sigma^{-1}(I) \ $, then we have

$\hat{\sigma}(x+y) = \hat{\sigma}(r_x+\sigma^{-1}(I)+r_y+\sigma^{-1}(I))= \sigma(r_x+r_y)+I=\sigma(x)+\sigma(y)+I=\hat{\sigma}(x)+\hat{\sigma}(y) \ $,

and

$\hat{\sigma}(xy)=\hat{\sigma}((r_x+\sigma^{-1}(I))(r_y+\sigma^{-1}(I)) = \hat{\sigma}(r_xr_y+\sigma^{-1}(I)(r_x+r_y)+\sigma^{-1}(I))= \hat{\sigma}(r_xr_y + \sigma^{-1}(I))=\sigma(r_xr_y)+I=\hat{\sigma}(r_x)\hat{\sigma}(r_y) \ $

since $(r_x+r_y)\sigma^{-1}(I)\in\sigma^{-1}(I) \ $. Hence, $\hat{\sigma} \ $ is a ring homomorphism.

Now suppose $x\in\text{ker}(\hat{\sigma}) \ $. Then we have $\hat{\sigma}(r_x+\sigma^{-1}(I))=0+I\in S/I \implies \sigma(r_x)=0 \implies r_x =0 \ $, so $\text{ker}(\hat{\sigma})=0 \ $, and hence $\hat{\sigma} \ $ is injective.

Consider our result from 2.1.2, which states that an ideal $I \ $ in a commutative ring $S \ $ is prime if and only if $S/I \ $ is a domain. This implies that $R/\sigma^{-1}(I) \ $ is an integral domain because our mapping $\hat{\sigma} \ $ is injective. If it were not a domain, then it would contain $x,y \ $ such that $xy=0, x\neq 0\neq y \ $. But this cannot be, since we have shown $\text{ker}(\sigma)=0 \ $, and so $I \ $ is prime whenever $\sigma^{-1}(I) \ $ is.

2.1.5,

2.2.2) Since $\mathbb{C}[x_1, \dots, x_n] \ $ is Noetherian, the ideal $\mathbb{I}(V) \ $ of polynomials vanishing on any affine algebraic variety $V \ $ is finitely generated. Hence

$\mathbb{V}(\mathbb{I}(V)) = \mathbb{V}(\langle f_1, \dots, f_n \rangle ) = \mathbb{V}(f_1, \dots, f_n) \ $

and so $V \ $ is the intersection of the zero sets of finitely many polynomials.

working on
2.2.3) Let $\phi \in \mathbb{C}[x,y,z] \ $ be a polynomial.  Then it has some largest power, $N \ $, and is composed entirely of powers of x,y,z, and multiples of those powers.  Hence,

$\phi(x,y,z)=C+\sum_{m=1}^N \left({ a_mx^m+b_my^m+c_mz^m }\right) + \sum_{m=1}^N \sum_{n=1}^N \left({ d_{mn}x^my^n +e_{mn}y^mz^n+f_{mn}x^mz^n }\right) + \sum_{m=1}^N \sum_{n=1}^N \sum_{p=1}^N \left({ g_{mnp}x^my^nz^p }\right) \ $.

Now suppose $\phi(x,y,z)=\phi(\sigma(x,y,z)) \ \forall \sigma \in S_3 \ $. Then we must have

$a_mx^m+b_my^m+c_mz^m = a_my^m+b_mx^m+c_mz^m  = \dots \ $,

and so $a_i=b_i=c_i \ \forall i \ $. Similarly, we must have

$ d_{mn}x^my^n +e_{mn}y^mz^n+f_{mn}x^mz^n = d_{mn}z^my^n +e_{mn}y^mx^n+f_{mn}z^mx^n  = \dots \ $,

and so $d_{ij}=e_{ij}=f_{ij} \ \forall i,j \ $. Finally, we have

Moreover, these conditions are sufficient to guarantee that $\phi(\vec{v})=\phi(\sigma(\vec{v})) \ \forall \sigma \in S_3 \ $, and so stating that $\phi \ $ is stable under $S_3 \ $ is equivalent to stating

$\phi(x,y,z)=C+\sum_{m=1}^N \left({ a_m(x^m+y^m+z^m) }\right) + \sum_{m=1}^N \sum_{n=1}^N \left({b_{mn} (x^my^n +y^mz^n+x^mz^n ) }\right) + \sum_{m=1}^N \left({ c_{mnp}x^my^nz^p }\right) \ $.

Hence we have shown that the set $A \ $ of functions in $\mathbb{C}[x,y,z] \ $ is not empty.

Now let us show it is a subring.

Let $f,g\in A \ $ be functions. Then if we define $h(x,y,z)=f(x,y,z)+g(x,y,z), i(x,y,z)=f(x,y,z)g(x,y,z) \ $. If $\sigma\in S_3 \ $, then

Then

$h(\sigma(x,y,z)) = f(\sigma(x,y,z))+g(\sigma(x,y,z)) =f(x,y,z)+g(x,y,z)=h(x,y,z) \ $ and

$i(\sigma(x,y,z)) = f(\sigma(x,y,z))g(\sigma(x,y,z)) =f(x,y,z)g(x,y,z)=i(x,y,z) \ $.

The constant maps $(x,y,z)\mapsto 1, (x,y,z)\mapsto 0 \ $ are clearly in $A \ $.

So $A \in \mathbb{C}[x,y,z] \ $ is a subring.

The structure of an element $\phi \ $ suggests $A \ $ may be generated by the functions $\phi_1=x+y+z, \ \phi_2=xy+yz+xz, \ \phi_3=xyz \ $.