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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?

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    $\begingroup$ 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? $\endgroup$
    – Arthur
    May 8, 2015 at 0:56
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    $\begingroup$ 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? $\endgroup$
    – Arthur
    May 8, 2015 at 1:04

1 Answer 1

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$a)$ $f(0, 0)$ 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., $J = (y − x^2, z − x^3)$

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  • $\begingroup$ $\mathbb{Z}[x]$ is not a PID. How do you know the kernel in $c)$ is principal? $\endgroup$
    – leibnewtz
    Oct 7, 2018 at 15:51

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