Prove that $K \to \frac{K[t]}{\langle m \rangle}: a \mapsto a + \langle m \rangle $ is a simple algebraic extension. I am trying to understand the proof of the following theorem, and will bolden the parts which leave me somewhat confused.
Given a field $K$ and an irreducible polynomial $m$ over $K$, the field extension defined by $\iota: K \to \frac{K[t]}{\langle m \rangle}: a \mapsto \langle m \rangle $ defines a simple algebraic extension.
The proof is in three parts, as follows.
a). Let $f(t) + \langle m \rangle$ denote an element of $\frac{K[t]}{\langle m \rangle}$. Then by dividing $f$ by $m$ to obtain a quotient $q$ with remainder $g$, we can write: $f(t) + \langle m \rangle = q(t)m(t) + g(t) + \langle m \rangle = g(t) + \langle m \rangle$ and since every element of the coset $g + \langle m \rangle$ is of the form $g + hm$ then this is the only element of degree less than $m$.
b). Let $\partial m  = n$ then by part a), we can write each element of $\frac{K[t]}{\langle m \rangle}$ uniquely as $$ a_0 + a_1 t + ... + a_{n-1}t^{n-1}$$ with $a_i \in K$. This is the same as the element $$ a_0 + a_1(t + \langle m \rangle) + ... + a_{n-1}(t + \langle m \rangle)^{n-1} = a_0 + a_1 \alpha + ... + a_{n-1} \alpha^{n-1}$$
where $\alpha = t + \langle m \rangle$.
Therefore, every element of $\frac{K[t]}{\langle m \rangle}$ is in $K(\alpha)$. Since $\alpha \in \frac{K[t]}{\langle m \rangle}$, we have $K(\alpha) \subseteq \frac{K[t]}{\langle m \rangle}$ and hence $\frac{K[t]}{\langle m \rangle} = K(\alpha)$
c). Since $K = \iota(K) \subseteq \frac{K[t]}{\langle m \rangle}$, we can regard the
polynomial $m$ as a function from $\frac{K[t]}{\langle m \rangle}$ to itself, then $m(\alpha) = m(t + \langle m \rangle) = m(t) + \langle m \rangle = \langle m \rangle$. Therefore, $\alpha$ is algebraic over $K$ and therefore $K \to \frac{K[t]}{\langle m \rangle}$ is a simple algebraic extension. This ends the proof.
I think the source of my confusion is somewhat easy to explain -- it stems from the abuse of notation that arises when we say $K = \iota(K)$. Suppose I did not want to abuse notation, then my attempt to correct the statement would be:
'each element of $\frac{K[t]}{\langle m \rangle}$ is in $(\iota (K))(\alpha)$. Since $\alpha \in \frac{K[t]}{\langle m \rangle}$, we have $(\iota(K))(\alpha) \subseteq \frac{K[t]}{\langle m \rangle}$ and hence $\frac{K[t]}{\langle m \rangle} = \iota(K(\alpha))$'
(EDIT: Therefore, the 'actual extension' we desire is $\iota(K) \to (\iota(K))(\alpha) = \frac{K[t]}{\langle m \rangle}$ and not $K \to \frac{K[t]}{\langle m \rangle}$.)
This statement makes more sense to me when I write it out, but I have this aching feeling that it might be complete gibberish and doesn't correspond to what I'm thinking about in my head, mostly because there has been an abuse of notation in everything I've read all the way up to this point and I have just sort of learned to live with it.
If it is better to abuse notation like this, an explanation of the meaning behind the expression "$\frac{K[t]}{\langle m \rangle} = K(\alpha)$ would be very helpful for my understanding of this topic.
Feel free to ask for clarification on anything I've said, since I'm not even sure if I've conveyed my confusion properly.
 A: I understand your discomfort, and I had it as well when starting to learn field theory. It always bothered me that often one writes $\subseteq$ when it isn’t really a subset. But most of the time it makes sense, I’ll try to explain why.
I think one main purpose of the theorem your presenting is the following: given any irreducible polynomial $m$ of $K[t]$, there is a simple algebraic field extension $\iota: K\to L$ such that $m$ admits a root inside $L$. More carefully written, one would have to say that actually $\iota(m)$ has a root in $L$. But as long as it is clear what the embedding of $K\to L$ is, and as long as you are dealing only with one field extension (as opposed to many different ones at a time), it makes sense to identify $K$ with its image in $L$, because after all the intent of all this is that you want to “add” certain roots to a field $K$. Moreover, the embedding of $K$ into $L=K[t]/(m)$ given by $\lambda\mapsto\lambda +(m)$ is so canonical that it really is more of a notational obstacle if you would insist on writing $\iota(K)$ for the image of $K$ inside $L$. But it is true that in general the embedding of $K$ into $L$ isn’t unique, in this case one just agrees that if nothing else is written, one means the canonical embedding.
Now what is true, and what sometimes gets a bit swept under the rug, is that one has to be careful when multiple such field extensions interact. For example, if one doesn’t care at all about distinguishing subsets and injections, one could be tempted to say something like “as all irreducible polynomials $m$ split into linear factors with set of roots $r_m$ in some field extension $K\subseteq L$, we can construct the algebraic closure of $K$ by writing $\overline{K}=K\left(\bigcup_{m\text{ irreducible}}r_m\right)$”. This would be too careless, because you are dealing with infinitely many different extensions at the same time, who don’t really live in sone common set. So in this case it makes sense to be a bit more careful about the different field extensions.
One way to circumvent your discomfort, at least for algebraic field extensions, would be the following: we carefully show that every field admits an algebraic closure, which is unique up to isomorphism. The if we have a field $K$ and are interested in the algebraic extensions, we can fix an algebraic closure $\overline{K}$ of $K$, fix an embedding $K\to\overline{K}$, and identify $K$ with its image in $\overline{K}$. Then any algebraic extension of $K$ embeds into $\overline{K}$, so abstract statements about algebraic extensions of $K$ and their interactions can always be stated in terms of subsets. But it may be a bit akward having to always embed everything into $\overline{K}$.
Now I think what the boldened part means is the following (I will allow myself to write it by abusing language with respect to the inclusions): you have shown that $L:=K[t]/(m)$ is a field containing an element $\alpha$ such that $m(\alpha)=0$. You can then wonder whether there is a strict subfield of $L$ containing $K$ and $\alpha$. One writes $K(\alpha)$ for the smallest subfield of a given extension (in this case $K\to L$) contianing both $K$ and $\alpha$, i.e. $K(\alpha)=\bigcap_{\substack{E\text{ subfield of }L\\ K\cup\{\alpha\}\subseteq E}}$. As it is shown that $1,\alpha,\ldots,\alpha^{n-1}$ generates $L$ as a $K$-vector space, and as surely $1,\alpha,\ldots,\alpha^{n-1}\in K$, we must have $K(\alpha)=L$.
At last, I think you should try to get comfortable which this abuse of notation in field theory, but I also think it is a good thing if one is attentive about it when dealing with many extensions and a time.
