Recently a few weeks back I was faced with an admission test question. The question goes as follows:
The number of unit preserving ring homomorphisms from the ring $\mathbb{Z}[\sqrt{2}]$ to $\mathbb{Z}_2 \times \mathbb{Z}_7 \times \mathbb{Z}_{17}$ is
a. 0
b. 2
c. 4
d. 8
I am still unable to approach this question. What do we specifically mean by "unit preserving homomorphism" ? Aren't all homomorphism supposed to preserve unit ? Any help is appreciated.
Number of unit preserving ring homomorphism
ring-homomorphismring-theory
Related Solutions
Why do professors go through proof after proof with no rhyme or reason?
One theory is that this is an "easy" way to give a lecture (to be negative about it, a "lazy" way.) This may be true in some cases. But on the other hand, much of the instructor's education might have been this way, and maybe they even think the experience is valuable. So, they might actually be giving the students the best path they know of. Some students might even feel like that is the way they are most comfortable learning. So to be fair, such instruction may be given in good faith, and may have good points.
The fact is that really good exposition requires a really skillful teacher, and it's not easy to do. Incidentally, I found Artin a very good expositor, but I did observe that by doing this, some less dedicated readers might get bored or distracted during his exposition.
One of my books learning abstract algebra was Martin Isaacs' Algebra. At the time I did not like it very much, but looking back on it now I think I do like its exposition. This just goes to show that reasonable exposition is not always easy to evaluate.
My question is How do I prevent this from happening to me in future?
Oh, well that's easy! Go skim through a lot of alternative books on the same topic and soak up whatever you can! Don't pretend like it's your teacher's responsibility to put text on your plate. You already applied this when you picked up Artin's book and learned something from it.
why are textbooks like Gallian's popular in math instruction?
The "like" part here makes this a loaded question, but I could just say that this book is probably considered basic, safe and affordable. It probably also depends upon the teacher's experience with texts too.
How are you supposed to read them?
This varies a lot from person to person. Personally I discovered that I learn best by having three or four texts on the same topic that I can use to cross-reference topics. Usually at least one of the authors is going to say something that makes things click.
And most of all, this sets me up with a big supply of problems. Doing problems does a lot more than plain reading, for me. Of course you have to spend some time reading or you won't know what tools you have at hand, and you won't see the themes in the proofs.
Is math supposed be learnt like you are learning a game where you are given all the rules ( definitions) and you have to play the game according to those rules( theorems, exercises) ?
I guess ideally "no", but for some people, that's how mathematics first begins! Those who persist eventually find their own appreciation for the subject matter, and develop their ability with it. This "game" analogy certainly doesn't paint a pretty picture of pedagogy, but it's very rare to find teachers with enough ability to get the beauty of mathematics across from the very beginning.
Luckily, it sounds like you at least know mathematics is more than a string of memorized definitions and theorems and proofs, so you, my friend, are already well ahead of many other students. The rest are in the even sadder situation of thinking "Yes, that's all mathematics is. Isn't it awful?!"
For any commutative ring $R,$ there is a unique homomorphism $f : \Bbb Z\to R,$ as specifying that $1\mapsto 1$ tells you where each element of $\Bbb Z$ must map to. Concretely, for any positive $n\in\Bbb Z,$ you have $n = 1 + \dots + 1$ ($n$ times,) so that $$f(n) = f(1 + \dots + 1) = f(1) + \dots + f(1) = n\cdot f(1).$$ You also know that $f(0) = 0,$ and if $n\in\Bbb Z$ is negative, then $f(n) = f(-(-n)) = -f(-n).$ Thus, there is exactly one morphism $f : \Bbb Z\to R,$ because the image of any element is determined by where $1$ is sent and the ring homomorphism rules.
Now, let's examine what it takes to define a morphism $\Bbb Z[x]\to R.$ We already know that we don't have a choice for where $\Bbb Z\subseteq\Bbb Z[x]$ is sent. What about $x$? Well, it turns out we can send $x$ to any element of $R$ that we want.
Suppose that $r\in R.$ If $f : \Bbb Z[x]\to R$ is a ring homomorphism sending $x$ to $r,$ then the ring homomorphism properties imply that we must have \begin{align*} f\left(\sum_{i = 0}^n a_i x^i\right) &= \sum_{i = 0}^nf\left( a_i x^i\right)\\ &= \sum_{i = 0}^n f(a_i) f(x^i)\\ &= \sum_{i = 0}^n f(a_i) f(x)^i\\ &= \sum_{i = 0}^n f(a_i) r^i. \end{align*} Every element of $\Bbb Z[x]$ is of the form $\sum_{i = 0}^n a_i x^i$ for some $n$ and some collection of integers $a_i,$ so we see that specifying where $x$ is sent determines the entire homomorphism. In particular, it is given by $$ p(x) = \sum_{i = 0}^n a_i x^i \mapsto \sum_{i = 0}^n f(a_i) r^i = p(r). $$
Conversely, setting $f(x) = r$ and extending in the above way is always a ring homomorphism. That is, let $r\in R$ and define \begin{align*} f : \Bbb Z[x]&\to R\\ \sum_{i = 0}^n a_i x^i &\mapsto \sum_{i = 0}^n g(a_i) r^i, \end{align*} where $g : \Bbb Z\to R$ is the unique ring homomorphism from paragraph one. We still have $f(1) = g(1) = 1$ and for any $n,m\in\Bbb Z\subseteq\Bbb Z[x],$ we have $$f(n + m) = g(n + m) = g(n) + g(m) = f(n) + f(m).$$ Suppose we have two arbitrary polynomials $\sum_{i = 0}^n a_i x^i,$ $\sum_{i = 0}^m b_i x^i.$ Then without loss of generality $m\leq n,$ and we can say that $\sum_{i = 0}^m b_i x^i = \sum_{i = 0}^n b_i x^i,$ where we define $b_j = 0$ for $j > m.$ Then we have \begin{align*} f\left(\sum_{i = 0}^n a_i x^i + \sum_{i = 0}^n b_i x^i\right) &= f\left(\sum_{i = 0}^n (a_i + b_i) x^i\right)\\ &= \sum_{i = 0}^n g(a_i + b_i)r^i\qquad\textrm{(by definition)}\\ &= \sum_{i = 0}^n \left(g(a_i) + g(b_i)\right)r^i\\ &= \sum_{i = 0}^n (g(a_i)r^i + g(b_i)r^i)\\ &= \sum_{i = 0}^n g(a_i)r^i + \sum_{i = 0}^n g(b_i)r^i\\ &= f\left(\sum_{i = 0}^n a_i x^i\right) + f\left(\sum_{i = 0}^n b_i x^i\right). \end{align*} You can check similarly that $$f\left(\left(\sum_{i = 0}^n a_i x^i\right)\cdot\left(\sum_{i = 0}^n b_i x^i\right)\right) = f\left(\sum_{i = 0}^n a_i x^i\right)\cdot f\left(\sum_{i = 0}^n b_i x^i\right),$$ so that the map defined is indeed a ring homomorphism.
The case of two variables is similar - a ring homomorphism is completely determined by where each variable is sent, and any choice of $r,s\in R$ gives a ring homomorphism with $x\mapsto r$ and $y\mapsto s.$
What's happening is that $\Bbb Z[x,y]$ is the free commutative ring on two generators $x$ and $y,$ which means essentially what I stated above - a ring homomorphism from $\Bbb Z[x,y]$ is given by a choice of where $x$ and $y$ will be sent, and any choices will work. To make sure that this defines a map on the quotient you want, you need to further specify that the images of $x$ and $y$ satisfy the given relation - i.e., because $x^3 + y^2 - 1 = 0$ in $\Bbb Z[x,y]/(x^3 + y^2 - 1),$ we must have $f(x)^3 + f(y)^2 - 1 = 0$ as well.
Best Answer
Four.
We need $1\mapsto1$. And we know $\varphi (\sqrt2) ^2=\varphi (2)=2$. So does $2$ have a square root (is it a quadratic residue) in $\Bbb Z_{238}$?
So, since $2$ has $2$ square roots $\pmod 7$, $2 \pmod {17}$ and $1\pmod 2$, we get $2\cdot 2\cdot 1=4$ unital homomorphisms. (I used Euler's criterion. For instance, $2^{\frac{7-1}2}\equiv1\pmod 7$. There are then $1+\genfrac(){}{2}{2}{7}=2$ roots $\pmod7$).
That's there's four distinct choices for $\varphi (\sqrt2) $. But $\Bbb Z[\sqrt2] =\langle 1,\sqrt2 \rangle $. So those are the possibilities for $\varphi $.