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$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\s}{\sigma}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathbb{F}}$

example 1: $ R=\Z, I=n\Z$ with $f(r)=r+I$, ker f =I

example 2: Let $R=\R[x]$ and $\C$ be a ring. So there is an $i\in \C$ and a substitution homomorphism $\s=\s_{i}:\R[x] \rightarrow \C$ by $\s_{i}(A)=A(i)$. What is $I=ker \s$? Surely $x^{2}+1 \in I$, because $i^{2}+1=0$. It follows $(x^{2}+1)A \in I$ for every $A\in R$, so $(x^{2}+1)R\subset I$. But with $L=a+bx(a,b \in \R)$ it is : $\s(L)=a+bi\ne 0$. That gives us : $A=(x^{2}+1)Q+L$ with rest $L=a+bx$. It follows that $\tilde{\s}:R/I\rightarrow \C$ is a ringisomorphism. Especially $R/I = \R[x]/(x^{2}+1)\R[x]$ is a field, because $\C$ is a field.

example 3: $\Q[i]=\Q+\Q i$ (this is even a field)

example 4: $\Z[\frac{1}{2}]=\Z+\Z \frac{1}{2}+\Z(\frac{1}{2})^{2}+\cdots$

example 5: $a=2^{1/3}$, $\Z[a]=\Z+\Z a+ \Z a^{2}$

example 6: $Char(\F_{4}) = 2 $, because 1+1=0.

example 7 : $Char(\F_{2} \times \F_{2})= 2$

Example 1 : The $ker$ is everything that is mapped to 0, is it correct that this is always an ideal of R?

Example 2: I understand that $i^{2}+1=0$ in $\C$, but how can it be concluded from that, that $x^{2}+1 \in I$, because of the substitution homomorphism? How is the step(s??) from $\s(L) = a+bi$ to $A=(x^{2}+1)Q+L$ done?

Example 3: Apart from using the definition of a field, is there a way one can see immediately, that $\Q[i]$ is a field?

Example 4 and Example 5: Why isn't that the same as in Example 3?

Example 6: I think $\F _{4} = \{0,1,2,3\}$ but that is wrong according to this example...

Example 7: $\F _{2} \times \F_{2} = \{0,1\} \times \{0,1 \} = \{0,0\}, \{0,1\}, \{1,0\},\{1,1\}$

Characteristic is the littlest number so when the rest is 0 with r=mq+d, so when it is $\{0,0\}$; but why aren't there two characteristics even if it is the same ring (so the characteristic of this should be $2,2$ and not only 2?

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I don't understand what you're saying in (2). Could you clean it up so I can read it? –  anon Feb 3 '12 at 20:03

2 Answers 2

up vote 6 down vote accepted
  1. Yes, the kernel of a ring homomorphism is always an ideal. In fact:

    Proposition. Let $R$ be a ring, and let $I$ be a subset of $R$. Then $I$ is an ideal of $R$ if and only if there exists a ring $S$ and a ring homomorphism $f\colon R\to S$ such that $I=\mathrm{ker}(f)$.

    Proof. The projection map $\pi\colon R\to R/I$ is a ring homomorphism, and $\mathrm{ker}(\pi)=I$ (this is the example you have). Conversely, if $f\colon R\to S$ is a ring homomorphism, then $\mathrm{ker}(f)$ is nonempty (since $f(0)=0$); if $a,b\in\mathrm{ker}(f)$ then $f(a-b) = f(a)-f(b) = 0$, so $a-b\in I$; and if $a\in I$ and $r\in R$, then $f(ar) = f(a)f(r)=0f(r)=0$, $f(ra)=f(r)f(a)=f(r)0=0$, so $ar,ra\in I$. Thus, $I$ is an ideal. $\Box$

  2. You can conclude that $x^2+1$ is in the kernel of the map by simply noting that $\sigma_i(x^2+1) = 0$. This amounts to plugging in $i$ for $x$ and verifying you get $0$. Now, assume that $p(x)$ is any polynomial in $\mathbb{R}[x]$. Using Long Division, we can divide $p(x)$ by $x^2+1$ with remainder; that is, we can find unique polynomials $q(x)$ and $r(x)$ such that $$p(x) = q(x)(x^2+1) + r(x),\qquad r(x)=0\text{ or }\deg(r)\lt\deg(x^2+1).$$ Then we have that $$\sigma_i(p(x)) = p(i) = q(i)(x^2+1) + r(i) = q(i)0 + r(i) = r(i),$$ so $p(x)\in\mathrm{ker}(\sigma_i)$ if and only if $r(x)\in\mathrm{ker}(\sigma_i)$. Now, since $\deg(x^2+1)=2$, $r(x)$ must be a polynomial of degree at most $1$, so we can write $r(x) = a+bx$ with $a,b\in\mathbb{R}$ (possibly zero). Under what conditions will $r(x)$ be in $\mathrm{ker}(\sigma_i)$? $r(x)\in\mathrm{ker}(\sigma_i)$ if and only if $\sigma_i(r(x))=0$, if and only if $r(i) = a+bi = 0$. But since $a$ and $b$ are real numbers, the only way for $a+ib$ to be equal to $0$ is if $a=b=0$. That is, $r(x)\in\mathrm{ker}(\sigma_i)$ if and only if $r(x)=0$. Going back to $p(x)$, we see that $p(x)\in\mathrm{ker}(\sigma_i)$ if and only if $p(x)$ can be written as $q(x)(x^2+1)$ for some polynomial $q(x)$, if and only if $p(x)$ is a multiple of $x^2+1$.

  3. Apart from using the definition of "field"? Not really.

  4. See below.

  5. The difference between 3, 4, and 5, is that the element we are adding in 3 and 5 satisfies a monic polynomial; in 3, $i$ satisfies $x^2+1$ (that is, $i^2+1=0$); so any expression of the form $a_0 + a_1i + a_2i^2 + \cdots + a_ki^k$ can be rewritten using only constant and linear terms. In 5, $a$ satisfies $x^3-2$, so again, any expression of the form $b_0+b_1a+b_2a^2+\cdots +b_ka^k$ can be rewritten so that you only use constant, linear, and quadratic terms. But in 3, the element $\frac{1}{2}$ does not satisfy any monic polynomial with integer coefficients: it is impossible to express $\left(\frac{1}{2}\right)^n$ as a sum of integer multiples of smaller powers of $\frac{1}{2}$, so you do need all powers.

  6. $\mathbb{F}_4$ is the field of 4 elements; $\mathbb{Z}_4$ is not a field. $\mathbb{F}_4 = \{0,1,\alpha,\alpha+1\}$, where $\alpha^2 +\alpha+1=0$.

  7. $\mathbb{F}_2\times\mathbb{F}_2 = \Bigl\{ (0,0), (0,1), (1,0), (1,1)\Bigr\}$ with addition and multiplication componentwise. The characteristic is the smallest positive integer $k$ such that $kr = (0,0)$ for all $r$ in the ring. (Your definition made absolutely no sense to me). $k=1$ doesn't work, because $(1,1)\neq (0,0)$. But $2$ works, because $2(a,b) = (a,b)+(a,b) = (a+a,b+b) = (0,0)$ no matter what $a$ and $b$ are (since in $\mathbb{F}_2$, $r+r=0$ for all $r$).

P.S. You describe all these as "Examples". Examples of what? Surely you mean "statement"?

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Dear Arturo. You are amazingly fast. I am posting my answer, even though your perfect answer is already here. I hope you don't mind. +1 from me. –  Rudy the Reindeer Feb 3 '12 at 20:41
@MattN. Of course I don't mind. –  Arturo Magidin Feb 3 '12 at 20:55
Thank you, Arturo Magidin. –  VVV Feb 4 '12 at 14:37

@Example 1: Yes. Proof: Let $r \in R$ and $x,y \in \operatorname{ker}{f}$. Then $f(rx) = f(r) f(x) = 0$ and $f(xr) = f(x) f(r) = 0$ hence $rx \text{and} xr \in \operatorname{ker}{f}$ and $f(x + y) = f(x) + f(y) = 0$ hence $x + y$ is also in $ \operatorname{ker}{f}$.

The kernel of any ring homomorphism is an ideal.

@Example 2: I've not seen the use of capital letters to denote polynomials. I'll stick to $p \in \mathbb{R}[x]$, hope you don't mind. You have $(x^2 + 1)$ evaluated for $x = i$ is $0$ hence $(x^2 + 1)p(x)=0$ for $x=i$.

@Example 3: What does it look like? Things in $\mathbb{Q}$ all have inverses and $i^2 \in \mathbb{Q}$. A thing in $\mathbb{Q}(i)$ looks like $a + ib$ for $a,b \in \mathbb{Q}$. So what is its inverse $(a + ib)^{-1}$? Well, $\frac{1}{a + ib} =\frac{a - ib}{(a + ib)(a -ib)} = \frac{a - ib}{a^2 + b^2} $ hence $1 = (a + ib) \frac{a - ib}{a^2 + b^2}$, so the inverse of $a + ib$ is $\frac{a - ib}{a^2 + b^2} $.

@Example 4: This cannot be the same as example $3$ because example $3$ is a field and this one isn't. To see why think about what an inverse of for example $1 + \frac12$ would look like: $(1 + \frac12)^{-1} = \frac{1}{1 + \frac12} = \frac23$.

@Example 5: Try to think about why this is not a field.

@Example 6: $\mathbb{Z}/4\mathbb{Z}$ is not a field because $2$ doesn't have an inverse. The field with $4$ elements is $(Z/2Z)[T]/(T^2+T+1)$, see here.

Note that this is an important construction: take a ring and quotient it by a polynomial to get a field of a given number of elements.

@Example 7: No the characteristic is defined to be the smalles $n$ such that $1 + 1 \dots + 1 = 0$ $(n \text{ times})$.

Hope this helps.

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I don't understand your comment in 4 and 5; the original post had $\mathbb{Z}[\frac{1}{2}] = \mathbb{Z}+\mathbb{Z}(\frac{i}{2}) + \mathbb{Z}(\frac{i}{2})^2+\cdots$, I switched the $i$s to $1$s; of course, both the example of adding $\frac{1}{2}$ or adjoining $\frac{i}{2}$ leads to a similar expression... –  Arturo Magidin Feb 3 '12 at 20:53
@ArturoMagidin Yes thank you, that comment is actually misleading as $\mathbb{Z}[\frac{i}{2}]$ is not a field. I'm going to rewrite it. –  Rudy the Reindeer Feb 3 '12 at 21:07
Thanks, Matt N.! –  VVV Feb 4 '12 at 14:37

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