# 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?

• For the first one, you have pointed out that the kernel contains the ideal $(x,y)$. Now, say you had a polynomial outside this ideal. What can you say about that polynomial? Alternatively, $(x,y)$ is a maximal ideal, so if the kernel is any bigger, it would have to be the whole ring. Is that possible? May 8, 2015 at 0:56
• Another trick you can use, especially for the last one: for a ring homomorphism $f:R\to S$, find as big an ideal $I$ as you can that is contained in the kernel of $f$. Now look at $f':R/I\to S$. This makes sense since $I\subseteq \ker(f)$. Can you prove, using clever representatives of elements in $R/I$, that this new homomorphism $f'$ is injective? May 8, 2015 at 1:04

$$a)$$ $$f(0, 0)$$ 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., $$J = (y − x^2, z − x^3)$$
• $\mathbb{Z}[x]$ is not a PID. How do you know the kernel in $c)$ is principal? Oct 7, 2018 at 15:51