# Danielle Stanton

#### Category: Artin Page 1 of 5

Here is Frank Quinn’s article “A Revolution in Mathematics? What Really Happened a Century Ago and Why It Matters Today.” Appearing in Notices of the AMS, January 2012

This article articulates clearly, (in a way I wish I could!), some of the confusion/friction I have encountered between the fields of Mathematics and MathEd. It seems to me that so often that when a professional mathematician says “mathematics” they mean something entirely different than when a mathematics educator says “mathematics”. Quinn points to the mathematical revolution occurring in the late 19th and early 20th centuries, a revolution that in his estimation reached the math research community but did not reach the MathEd community as the source of this disconnect. I have been mostly trained as a proto-research mathematician (what Quinn might call a “core user”), not as a math educator. And as such, this part of Quinn’s article resonated in particular for me:

“The new methodology is less accessible to non-users. Old-style definitions, for instance, usually related things to physical experience so many people could connect with them in some way. Users found these connections dysfunctional, and they can derive effective intuition much faster from precise definitions. But modern definitions have to be used to be understood, so they are opaque to nonusers. The drawback here is that nonusers only saw a loss: the old dysfunctionality was invisible, whereas the new opacity is obvious.”

As a proto-research mathematician, mathematical objects do not have any existence outside of our definitions of them, let alone any sort of empirical existence. They are figments of our collective imaginations (that we can use to predict the future, how cool is that?). So trying to “explore”, or “discover”, or develop Vinner’s concept image, or develop an “intuitive understanding” outside of a definition is literally impossible. One cannot understand a thing that does not exist! One may as well seek to learn facts about unicorns. (There are no facts about unicorns. My apologies to Hannah Gadsby). One develops intuition about a mathematical object by winding up the definition and letting it go to see what it does (aka proving theorems about the object). Proof is the method by which we come to know in mathematics. (Though, like “mathematics”, “proof” seems to mean two different things to the two communities). One can certainly motivate a definition empirically or otherwise, but a motivation does not necessarily constitute mathematical knowledge or learning.

This stands in sharp contrast to the mathematics education perspective, in which one tries to develop a concept image almost exclusively through experiment and intuition, rather than deduction. In my estimation there exist certain areas of math that this approach is better suited to than others. Euclidean geometry comes to mind. Which is ironic, considering this is the area of the school curriculum in which precise definitions and deductive proof are most emphasized.

Anyway. These are some of my initial thoughts. Pelican Brief forthcoming.  I hope I’ve piqued somebody’s interest, I’d be curious to hear your thoughts in the comments or on social media.

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!

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