# Danielle Stanton

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11.8.3
Prove that the ring $$\mathbb{F}_2[x]/<x^3+x+1>$$ is a field, but that $$\mathbb{F}_3[x]/<x^3+x+1>$$ is not a field.

Solution.
Let $$\varphi:\mathbb{F}_2[x] \to \mathbb{F}_2[x]/<x^3+x+1>$$ be the canonical homo. The kernel of $$\varphi$$ is the ideal $$<x^3+x+1>$$. By a bunch of theorems in this section, we know that a polynomial ring over a field modded out by an ideal is a field if and only if $$<x^3+x+1>$$ is a maximal ideal in $$\mathbb{F}_2[x]$$. That’s a maximal ideal only if $$x^3+x+1$$ is monic (and it is, obvs) and irreducible. Okay so all we have to do is show $$x^3+x+1$$ is irreducible. I cheated and used the computer. It is irreducible. But if you wanna show it’s irreducible without running screaming into the loving arms of Wolfram Alpha, it’s NDB. There’s only a couple polynomials of degree 2 and 1 in that ring, so you just do the long division and see that none of them work. You conclude that $$x^3+x+1$$ is irreducible, therefore $$\mathbb{F}_2[x]/<x^3+x+1>$$ is a field.

For $$\mathbb{F}_3[x]/<x^3+x+1>$$, you do the same thing, but it turns out that $$x^3+x+1=(x+2)(x^2+x+2)$$. So we have a reducible polynomial generating the ideal, so the quotient ring $$\mathbb{F}_2[x]/<x^3+x+1>$$ isn’t a field.

Ta-da!

11.8.1
Which principal ideals in $$\mathbb{Z}[x]$$ are maximal ideals?

Solution.
Let $$I=<f(x)>$$ be a principal ideal in $$\mathbb{Z}[x]$$. We’ll show that $$I$$ is not maximal by constructing an ideal $$M$$ such that $$I \subset M \subset \mathbb{Z}[x]$$.

If $$f(x)=0$$, take $$M=<x>$$ and $$I$$ is not maximal.
If $$f(x)=1$$, then $$I=\mathbb{Z}[x]$$ and $$I$$ is not maximal.
If $$f(x)=c \in \mathbb[Z]$$, take $$M=<c,x>$$ then we have $$<c> \subset<c,x> \subset \mathbb{Z}[x]$$, and $$I$$ is not maximal.

Okay, now if we have $$f(x)=f_0+f_1 x+…+f_n x^n$$ then take $$M=<p, f(x)>$$ where $$p$$ is a prime in $$\mathbb{Z}$$ such that $$p$$ and $$f_i$$ are relatively prime for all $$i$$. Then $$<f(x)> \subset <p,f(x)> \subset \mathbb{Z}[x]$$.

Conclude there are no principal ideals in $$\mathbb{Z}[x]$$ that are maximal.

11.7.4
Prove that the field of fractions of the formal power series ring $$F[[x]]$$ over a field $$F$$ can be obtained by inverting the element $$x$$. Find a neat description of that field.

Solution.
(“neat”?!?! lol artin you kill me).

Let FF be the fraction field. The elements of FF are $$\frac{f}{g}$$ with $$f,g \in F[[x]]$$. Factor the biggest power of $$x$$ outta $$g$$ that we can, say the biggest power is $$n$$. Then $$g=x^n h$$ for some $$h \in F[[x]]$$. By that terrible exercise 11.2.2, we know everything in $$F[[x]]$$ is a unit, in particular there’s a $$h^{-1} \in F[[x]]$$. So we get
$\frac{f}{g}=\frac{f}{x^n h}=\frac{fh^{-1}}{x^n}= (fh^{-1})x^{-n}$
And we know the little bit of nonsense $$(fh^{-1})$$ is in $$F[[x]]$$, so we know that the whole nonsense $$(fh^{-1})x^{-n}$$ is in $$F[[x]][x^{-1}]$$. Okay so everything in the fraction field FF is in $$F[[x]][x^{-1}]$$.

Let’s get containment the other way too. By some stoopid theorem in this chapter, $$F[[x]][x^{-1}]$$ has a basis $$\{1, x^{-1}\}$$. So everything looks like

$\sum_{i=0}^{\infty}a_ix^i + x^{-1}\sum_{j=0}^{\infty}b_jx^j=\frac{b_0}{x}+\text{who cares}$

Which is in FF if you do a bunch of algebra but who knows how to add fractions anyway. We’ve got containment both ways so we’re done.

We can’t use the aforementioned stoopid theorem because the polynomial that we’d mod out by to adjoin $$x^{-1}$$ to $$F[[x]]$$ isn’t a monic polynomial. so nvm.

By a different terrible exercise, 11.5.7, $$F[[x]][x^{-1}]$$ is the Laurent polynomials, but infinite ones. ISN’T THAT NEAT.

GAH.

11.7.2
Let $$R$$ be an integral domain. Prove that the polynomial ring $$R[x]$$ is an integral domain, and identify the units in $$R[x]$$.

Solution.
Consider $$p(x), q(x) \in R[x]$$ such that $$p(x)q(x)=0$$, where $$n$$ and $$m$$ are the degrees of $$p(x)$$ and $$q(x)$$, respectively. WTS $$p(x)=0$$ or $$q(x)=0$$. For contradiction, suppose $$p(x) \neq 0$$ and $$q(x) \neq 0$$. Then $$p(x)q(x)$$ has leading term $$p_n q_m x^{(n+m)}$$ and $$p_n q_n \neq 0$$ because $$R$$ is an integral domain. Then the degree of $$p(x)q(x)$$ is $$n+m$$. CONTRADICTION! The degree of $$p(x)q(x)$$ is undefined, because $$p(x)q(x)=0$$! (just zero, not zero factorial) So we can’t have $$p(x) \neq 0$$ and $$q(x) \neq 0$$, so one of ’em has to be zero and we’re done.

Now let’s identify the units. Consider $$p(x), q(x) \in R[x]$$ such that $$p(x)q(x)=1$$, where $$n$$ and $$m$$ are the degrees of $$p(x)$$ and $$q(x)$$, respectively. Since $$p(x)q(x)=1$$, the degree of $$p(x)q(x)$$ is zero. Also the degree of $$p(x)q(x)$$ is $$m+n$$. So we have $$m+n=0$$ and can conlcude that both $$m$$ and $$n$$ are zero. Therefore the units in $$R[x]$$ are the “constants” that are units in $$R$$.

Ta-da!

11.6.6
Describe the ring obtained from $$\mathbb{R} \times \mathbb{R}$$ by inverting the element (2,0).

Solution.
By 11.6.5, $$\mathbb{R} \times \mathbb{R} \cong \mathbb{R}[x]/<x^2-1>$$. (NB: which makes sense, because those are just things that look like ax+b). Then the element (2,0) can be identified with 2x. We want to send (2,0) to (1,1), that’s the same as sending 2x to x+1. In other words, we want to force

\begin{align*}
2x&=x+1\\
x-1&=0
\end{align*}

Okay no problem, we just throw that polynomial into the ring $$\mathbb{R}[x]/<x^2-1>$$ in the usual way and see what we get. We know (hopefully, heck I’m not sure about anything anymore) that the ideal <x^2-1> is contained in the ideal <x-1> because
$x^2-1=(x+1)(x-1)$
So modding out by x^2-1 on top of modding out by x-1 doesn’t change anything. We can say that $$\mathbb{R}[x]/<x^2-1, x-1> \cong \mathbb{R}[x]/<x-1>$$. And that’s just modding out by a monic polynomial of degree 1, so what we’ve got left is just the boring old ordinary ring $$\mathbb{R}$$.

Ta-da!

11.6.3
Classify all rings of order 10.

Solution.
Let R be a ring of order 10 with $$R=\{r_0,r_1,r_2,r_3,…,r_9\}$$. Let $$r_0$$ and $$r_1$$ be the additive and multiplicative identities in $$R$$, respectively. We will construct an isomorphism from $$R$$ to $$\mathbb{Z}_{10}$$.

NB: by the same argument in 11.6.2, we have $$\mathbb{Z}_{10}$$ isomorphic to $$\mathbb{Z}_{2} \times \mathbb{Z}_{5}$$.

Let the map $$f: R \to \mathbb{Z}_{10}$$ be defined by $$f(r_n)=n$$.

WTS $$f$$ is a homo. It maps the identities to the identities, so we’re good there. To prove addition is preserved, observe $$f(r_i)+f(r_j)=i+j$$.

<s>Now for contradiction suppose $$f(r_i+r_j) \neq i+j$$. Then

\begin{align*}
f(r_i+r_0) &\neq i+0\\
f(r_i) &\neq i
\end{align*}

This contradicts the definition of $$f$$, therefore $$f(r_i+r_j)=i+j=f(r_i+r_j)$$ and addition is preserved. Do the same thing for multiplication to show multiplication is preserved. Conclude $$f$$ is a homo.</s>

The map is surjective because for all $$i \in \mathbb{Z}_{10}$$, we have $$f(r_i)=i$$. And a surjective map between two finite sets of equal cardinally has to be injective too. We conclude that $$f$$ is a bijective homomorphism, therefore $$R \cong \mathbb{Z}_{10}$$.

Ta-da!

11.6.2
Is $$\mathbb{Z}/<6>$$ isomorphic to the product ring $$\mathbb{Z}/<2> \times \mathbb{Z}/<3>$$? Is $$\mathbb{Z}/<8>$$ isomorphic to $$\mathbb{Z}/<2> \times \mathbb{Z}/<4>$$?

Solution.
Let $$\varphi: \mathbb{Z} \to \mathbb{Z}/<2> \times \mathbb{Z}/<3>$$ be the homo $$z \mapsto (z \text{ mod} 2, z \text{ mod} 3)$$. That homo is clearly surjective. The kernel of $$\varphi$$ is all $$z \in \mathbb{Z}$$ such that $$z \text{ mod} 2=0$$ and $$z \text{ mod} 3=0$$. So it’s all $$z \in \mathbb{Z}$$ that have a factor of 2 and a factor of 3. In other words, the kernel is $$<6>$$. So $$\mathbb{Z}/<2> \times \mathbb{Z}/<3> \cong \mathbb{Z}/<6>$$ by the first isomorphism theorem.

<s>Let $$\varphi: \mathbb{Z} \to \mathbb{Z}/<2> \times \mathbb{Z}/<4>$$ be the homo $$z \mapsto (z \text{ mod} 2, z \text{ mod} 4)$$. The kernel of $$\varphi$$ is all $$z \in \mathbb{Z}$$ such that $$z \text{ mod} 2=0$$ and $$z \text{ mod} 4=0$$. So it’s all $$z \in \mathbb{Z}$$ that have a factor of 2 and a factor of 4. In other words, the kernel is $$<4>$$, not $$<8>$$. So $$\mathbb{Z}/<8>$$ is not isomorphic to $$\mathbb{Z}/<2> \times \mathbb{Z}/<4>$$. Maybe?</s>

The rings $$\mathbb{Z}/<8>$$ and $$\mathbb{Z}/<2> \times \mathbb{Z}/<4>$$ aren’t even isomorphic as additive groups, so there’s no way they can be isomorphic as rings. Here’s a picture:

11.6.1
Let $$\varphi : \mathbb{R}[x] \to \mathbb{C} \times \mathbb{C}$$ be the homo defined by $$\varphi(x)=(1,i)$$ and $$\varphi(r)=(r,r)$$ for $$r \in \mathbb{R}$$. Determine the kernel and the image of $$\varphi$$.

Solution.
Let K be the kernel of $$\varphi$$ and let $$I=<(x-1)(x^2+1)>$$. We claim $$I=K$$. Consider $$l \in I$$ and $$k \in K$$.

WTS $$l \in K$$. Since $$l \in I$$ we have $$l=q(x-1)(x^2+1)$$ for some $$q \in \mathbb{R}[x]$$. Applyting $$\varphi$$ to $$l$$ we have
$\varphi(l)=\varphi(q)\varphi((x-1)(x^2+1))=\varphi(q) \cdot 0 = 0$
Therefore $$l \in K$$.

WTS $$k \in I$$. Since $$k \in K$$ we have $$\varphi(k)=(0,0)$$. Also $$(x-1)(x^2+1)$$ is monic so we can divide. So for some $$q \in \mathbb{R}[x]$$ and some quadratic $$r \in \mathbb{R}[x]$$ where $$r=ax^2+bx+c$$, we have

\begin{align*}
k &= q(x-1)(x^2+1)+r\\
\varphi(k) &= \varphi(q)\varphi((x-1)(x^2+1))+\varphi(r)\\
(0,0) &= \varphi(q)(0,0)+\varphi(r)\\
(0,0) &= (0,0)+\varphi(r)\\
(0,0) &= \varphi(ax^2+bx+c)\\
(0,0) &= \varphi(ax^2)+\varphi(bx)+\varphi(c)\\
(0,0) &= (a,a)(1,-1)+(b,b)(1,i)+(c,c)\\
(0,0) &= (a,-a)+(b,bi)+(c,c)\\
(0,0) &= (a+b+c,-a+bi+c)
\end{align*}

This forces $$a+b+c=0$$ and $$-a+bi+c=0$$. The second equation has no $$i$$ on RHS, so $$b=0$$. Then we have $$a+c=0$$ and $$-a+c=0$$, so $$b=a=c=0$$. Since all the coefficients are zero, $$r=0$$. Then $$k \in I$$. We have inclusion both ways so the kernel of $$\varphi$$ is the ideal $$<(x-1)(x^2+1)>$$.

Now let’s look at the image. Consider $$p \in \mathbb{R}[x]$$ such that $$p=a_0+a_ix+…+a_nx^n$$. Define $$\alpha$$ to be one of the following set $$\{1,-1,i,-i\}$$. Now we apply $$\varphi$$ to $$p$$.

\begin{align*}
\varphi(p) &= (a_0,a_0)+(a_1,a_1)(1,i)+(a_2,a_2)(1,-1)+…+(a_n,a_n)(1,\alpha)\\
&= \left(\sum_{i=0}^n a_i, \sum_{i=0}^n a_i\alpha_i\right)
\end{align*}

The first coordinate lives in $$\mathbb{R}$$, the second coordinate lives in $$\mathbb{C}$$. Therefore the image of $$\varphi$$ is contained in the ring $$\mathbb{R} \times \mathbb{C}$$. We have containment the other way too because math. See?

Ta-da!

11.5.7
Let $$\mathbb{F}$$ be a field and let $$R=\mathbb{F}[t]$$ be the polynomial ring. Let $$R’$$ be the ring extension $$R[x]/<tx-1>$$ obtained by adjoining an inverse of $$t$$ to $$R$$. Prove that this ring can be identified as the ring of Laurent Polynomials, which are finite linear combinations of the powers of $$t$$, negative exponents included.

Solution.
I wouldn’t read this if I were you. My subscripts have subscripts.

1. We start with $$\mathbb{F}$$ and adjoin $$t$$ in the usual way to get $$R=\mathbb{F}[t]$$. Let $$f_i \in \mathbb{F}$$. Then elements of $$R=\mathbb{F}[t]$$ look like
$f_0 + f_1t + … + f_m t^m$

2. Now we take $$R=\mathbb{F}[t]$$ and adjoin $$x$$ in the usual way to get the ring $$R[x]=(\mathbb{F}[t])[x]$$. (NB: subscripts are of the form {power of t, power of x}). Then elements of $$R[x]=(\mathbb{F}[t])[x]$$ look like
$(f_{0,0} + f_{1,0}t + … + f_{m_0,0} t^{m_0}) + (f_{0,1} + f_{1,1}t + … + f_{m_1,1} t^{m_1})x + … + (f_{0,n} + f_{1,n}t + … + f_{m_n,n} t^{m_n})x^n$

3. Now we take $$R[x]=(\mathbb{F}[t])[x]$$, adjoin the multiplicative inverse of $$t$$, $$t^{-1}$$, in the usual way to get $$R[x]/<tx-1>$$. Elements of this thing look like
$(f_{0,0} + f_{1,0}t + … + f_{m_0,0} t^{m_0}) + (f_{0,1} + f_{1,1}t + … + f_{m_1,1} t^{m_1})t^{-1} + … + (f_{0,n} + f_{1,n}t + … + f_{m_n,n} t^{m_n})t^{-n}$
Then we multiply every negative power of $$t$$ through, collect like terms with respect to $$t$$, and we see that elements of $$R[x]/<tx-1>$$ are finite linear combinations of positive and negative powers of $$t$$ with coefficients in the field $$\mathbb{F}$$.

Ta-da!

11.5.6
Let $$a$$ be an element of a ring $$R$$, and let $$R’$$ be the ring $$R[x]/<ax-1>$$ obtained by adjoining an inverse of $$a$$ to $$R$$. Let $$\alpha$$ denote the residue of $$x$$ (the inverse of $$a$$ in $$R’$$).

(a) Show that every element $$\beta$$ of $$R’$$ can be written in the form $$\beta=\alpha^kb$$ with $$b \in R$$.
(b) Prove that the kernel of the map $$R \to R’$$ is the set of elements $$b \in R$$ such that $$a^n b=0$$ for some $$n > 0$$.
(c) Prove that $$R’$$ is the zero ring if and only if $$a$$ is nilpotent.

Solution.

(a) Show that every element $$\beta$$ of $$R’$$ can be written in the form $$\beta=\alpha^kb$$ with $$b \in R$$.
$$\beta \in R’$$, WTS $$\beta = \alpha^kb$$ for some $$b \in R$$. Let $$\pi: R[x] \to R’=R[\alpha]$$ be the canonical homomorphism where $$\pi(ax-1)=0$$. Then
\begin{align*}
\beta &= \pi(b_0+b_1x+…+b_nx^n)\\
&=b_0+b_1\alpha+…+b_n\alpha^n
\end{align*}
And we want
$\beta=b_0+b_1\alpha+…+b_n\alpha^n=\alpha^kb$
So let’s multiply everything through by $$a^k$$ to find out what $$b$$ has to be (heh). Recall that $$a\alpha=1$$!

\begin{align*}
b_0+b_1\alpha+…+b_n\alpha^n&=\alpha^kb\\
b_0 a^k + b_1 \alpha a^k +…+ b_n\alpha^n a^k&= b
\end{align*}

The only restriction on $$k$$ is that it has to be in $$\mathbb{N}$$, so let $$k=n$$, then
\begin{align*}
b_0 a^n + b_1 \alpha a^n +…+ b_n\alpha^n a^n&= b\\
b_0 a^n + b_1 a^{n-1} + …+ b_n&= b
\end{align*}

And we have
\begin{align*}
\alpha^n b &= \alpha^n (b_0 a^n + b_1 a^{n-1} + …+ b_n)\\
&= b_0(a\alpha)^n+b_1(a\alpha)^{n-1}\alpha +…+ b_n\alpha^n\\
&= b_0+b_1\alpha+…+b_n\alpha^n\\
&= \beta
\end{align*}

So every element $$\beta$$ of $$R’$$ can be written in the form $$\beta=\alpha^kb$$ for some $$k \in \mathbb{N}$$.

(b) Prove that the kernel of the map $$R \to R’$$ is the set of elements $$b \in R$$ such that $$a^n b=0$$ for some $$n > 0$$.
(NB: I’m real iffy on this one, not sure what the map $$R \to R’$$ is… Is it the same as the map $$\pi$$ above, but restricted to the constants?)
Let $$\varphi$$ be our map, let the kernel of the map be $$K$$ and let $$B=\{b \in R | a^nb=0 \text{ for some } n \in \mathbb{N}\}$$. Consider $$k \in K$$ and $$b \in B$$. WTS $$b \in K$$ and $$k \in B$$.

Let’s show $$b \in K$$ first. Let $$\varphi(b)=\beta$$, WTS $$\beta = 0$$. Since $$b \in B$$, we have $$a^n b = 0$$ for some $$n$$. Then $$\varphi$$ it:
\begin{align*}
a^n b &= 0\\
\varphi(a^n b) &=\varphi( 0)\\
\varphi(a^n)\varphi(b) &= 0\\
a^n \beta &= 0\\
\alpha^n a^n \beta &= \alpha^n 0\\
\beta &= 0
\end{align*}
Therefore $$b \in K$$.

Now let’s show $$k \in B$$. Since $$k \in K$$, $$\varphi(k)=0$$. For contradiction, suppose $$k \notin B$$. Then $$a^n k \neq 0$$ for all $$n \in \mathbb{N}$$ and we have

\begin{align*}
a^n k &\neq 0\\
\varphi(a^n k) &\neq 0\\
\varphi(a^n) \varphi(k) &\neq 0\\
\varphi(a^n) \cdot 0 &\neq 0\\
0 &\neq 0
\end{align*}
Which is not true. So we have to have $$k \in B$$.

We’ve got inclusion in both directions, so $$K=B$$ and the kernel of the map $$R \to R’$$ is the set of elements $$b \in R$$ such that $$a^n b=0$$ for some $$n > 0$$.

(c) Prove that $$R’$$ is the zero ring if and only if $$a$$ is nilpotent.
Let $$\pi: R[x] \to R’=R[\alpha]$$ be the canonical homomorphism where $$\pi(ax-1)=0$$.
Let’s go forwards first. Suppose $$R’$$ is the zero ring. WTS there exists $$n \in \mathbb{N}$$ such that $$a^n=0$$. Since $$R’$$ is the zero ring, $$1=0$$ in there. So for all $$p \in R[x]$$ we have $$\pi(p)=0$$ because $$p$$ just doesn’t have anywhere else to go. Let $$p= b_0+b_1 x+ …+ b_m x^m$$ where $$b_i \in R$$. All the $$b_i$$ are in the kernel because everything is in the kernel. So by (b) above we have $$a^k b_i = 0$$ for some $$k \in \mathbb{N}$$. Let $$n$$ be the biggest $$k$$ such that $$a^k b_i=0$$ for all $$i$$. Then we have $$a^n b_i=0$$ for all $$i$$. Now watch this fancy move:

\begin{align*}
p &= b_0+b_1 x+ …+ b_m x^m\\
a^n p &= a^nb_0+a^nb_1 x+ …+ a^nb_m x^m\\
\end{align*}

Each term on the right hand side is zero, so $$a^n p =0$$ for all $$p \in R[x]$$. Set $$p=1$$, then $$a^n =0$$ for some $$n$$ and $$a$$ is nilpotent. (I now am almost positive there was something much easier to do… forget about $$p$$ altogether and just look at $$1 \in R[x]$$. blarg.)

Now backwards. Suppose $$a$$ is nilpotent. Then there exists an $$n$$ such that $$a^n=0$$. We want to show that $$R’$$ is the zero ring. It suffices to show that 1 is in the kernel. By (b), 1 is in the kernel if there exists $$a^i \cdot 1 = a^i \in R$$. And this thing does exist, by assumption. So in $$R’$$ 1=0, and $$R’$$ is the zero ring. Fine.

We went forwards, we went backwards, so the iff is proved. And we’re done.

P.S. Does this mean if we throw multiplicative inverses into a field everything gets squished? That doesn’t make sense. We should just get the field back. Fields don’t have nilpotent elements. An element can be either nilpotent or a unit, and everything in a field is a unit. Suppose $$\beta$$ is nilpotent and a unit. Then $$\beta^n=0$$ and there exists a $$\beta^{-1}$$ such that $$\beta^{-1}\beta=1$$. Then

\begin{align*}
\beta^n&=0\\
\beta^{-n}\beta^n&=\beta^{-n}\cdot 0\\
1&=0
\end{align*}
and that’s not a field. cool, can i haz extra credit plz.

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