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 a connected 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 connected set $$S\in\mathbb{C}^{n^2} \ $$. Since $$U(n) \ $$ is connected, 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.

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,

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

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

$$\sum_{m=1}^N \left({ a_mx^m+b_my^m+c_mz^m }\right)= \sum_{m=1}^N \left({ a_my^m+b_mx^m+c_mz^m }\right) = \dots \ $$,

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

$$\sum_{m=1}^N \sum_{n=1}^N \left({ d_{mn}x^my^n +e_{mn}y^mz^n+f_{mn}x^mz^n }\right) = \sum_{m=1}^N \sum_{n=1}^N \left({ d_{mn}z^my^n +e_{mn}y^mx^n+f_{mn}z^mx^n }\right) = \dots \ $$,

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

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

We must show that this is a ring. Call the set of functions in $$\mathbb{C}[x,y,z] \ $$ that fit the above criterion $$A \ $$, and let $$\phi_1, \phi_2 \in A \ $$, and set $$N=\text{max}(\text{deg}(\phi_1), \text{deg}(\phi_2)) \ $$. Then $$\exist a_i, b_{ij}, c_{ijk}, d_i, e_{ij}, f_{ijk}, \ 1\leq i,j,k \leq N \ $$ fulfilling the obvious roles of constants in the above criterion. Then we have

$$(\phi_1+\phi_2)(x,y,z) \ $$

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

$$+ \sum_{m=1}^N \left({ d_m(x^m+y^m+z^m) }\right) + \sum_{m=1}^N \sum_{n=1}^N \left({e_{mn} (x^my^n +y^mz^n+x^mz^n ) }\right) + \sum_{m=1}^N \sum_{n=1}^N \sum_{p=1}^N \left({ f_{mnp}x^my^nz^p }\right) \ $$

$$=\sum_{m=1}^N \left({ g_m(x^m+y^m+z^m) }\right) + \sum_{m=1}^N \sum_{n=1}^N \left({h_{mn} (x^my^n +y^mz^n+x^mz^n ) }\right) + \sum_{m=1}^N \sum_{n=1}^N \sum_{p=1}^N \left({ l_{mnp}x^my^nz^p }\right) \in A \ $$,

where we define $$g_i=a_i+d_i, h_{ij}=b_{ij}+e_{ij}, l_{ijk}=c_{ijk}+f_{ijk} \ $$.


 *  I really don't want to play with the indices as much as needed to show the obvious fact that A is closed under multiplication.