Find kernel generators for ring maps This is the textbook question:
Q: Find generators for the kernels of the following maps:


*

*$\mathbb{R}[x,y] \to \mathbb{R}$ defined by $f(x,y) \rightsquigarrow f(0,0)$

*$\mathbb{R}[x] \to \mathbb{C}$ defined by $f(x) \rightsquigarrow f(2+ i)$

*$\mathbb{Z}[x] \to \mathbb{R}$ defined by $f(x) \rightsquigarrow f(1+\sqrt{2})$

*$\mathbb{Z}[x] \to \mathbb{C}$ defined by $x \rightsquigarrow \sqrt{2}+\sqrt{3}$

*$\mathbb{C}[x,y,z] \to \mathbb{C}[t]$ defined by $x \rightsquigarrow t, y \rightsquigarrow t^2, z \rightsquigarrow t^3$


My work on the first three:


*

*Any polynomial that satisfies $f(0,0)=0$ will be in the kernel. Intuitively, the two polynomials $f(x,y)=x$ and $f(x,y)=y$ should generate this. How can I prove that?

*The kernel will have root $(2+i)$ and the conjugate $(2-i)$ which multiply to $(x-(2+i))(x-(2-i)) = x^2-4x+5$. The coefficients are real and the polynomial is irreducible in $\mathbb{R}$. That polynomial is clearly in the kernel. How can I show that it generates the kernel?

*I find polynomial $f(x) = x^2-2x-1$ that is in the kernel and is irreducible in $\mathbb{Z}$. How can I show that it generates the kernel?
 A: 
$a)$ $f(0, 0)$ is the constant summand of the polynomial $f(x, y)$ Thus, the kernel of $φ$ consists of the polynomials with zero constant term, i.e., the kernel is $x\mathbb{R}[x, y] + y\mathbb{R}[x, y]$
$b)$The kernel consists of the polynomials for which $2 + i$, which is generated by the minimal polynomial that has $2 + i$ as a root. That polynomial is
  $f(x) = (x − (2 + i))(x − (2 − i) = x^2 − ((2 + i) + (2 − i))x + (2 + i)(2 − i) = x^2− 4x + 5.$
  Thus, the kernel equals $(x^2 − 4x + 5)$, the principal ideal generated by $x^2 − 4x + 5.$
For $c)$As in (b), the minimal polynomial equals
  $f(x) = (x−(1+√2))(x−(1−√2)) = x^2−((1+√2)+(1−√2))x+(1+√2)(1−√2) = x^2−2x−1.$
  Thus, the kernel equals the principal ideal generated by $x^2 − 2x − 1.$
for $d)$Consider the signed combinations $±√2 + ±√3$, and the polynomial $f(x) = (x −(√2+ √3))((x − (√2 −√3))(x − (−√2 + √3))((x − (−√2 −√3))$
  that has those combinations as its roots. Multiplying the first two factor together, and the second two factors together, we find that
  $f(x) = (x^2 − 2√2x − 1)(x^2 + 2√2x − 1) = x^4 − 10x^2 + 1$.
  Check that no polynomial of smaller degree has $√2 + √3$ as a root. Then, we conclude that
  the kernel equals the principal ideal $(x^4 − 10x^2 + 1).$
For $e)$$x = t, y = t^2$ and $z = t^3$ are related by $y = x^2$ and $z = x^3$
  . Hence, $x^2 − y$ and $x^3 − z$ are in the kernel. Hence, we get a homomorphism $φ′
: C[x, y, z]/(x^2 − y, x^3 − z) = C[x] → C[t]$
  that takes $f(x)$ to $f(t)$. Since the kernel of that homomorphism is just $0$, and the kernel  is also $J/(y − x^2, z − x^3), J/(y − x^2, z − x^3) $ must be $0$, i.e., $J = (y − x^2, z − x^3)$

