Show that $\mathbb{K}$ is a field

Question

Let $\mathbb{F}$, a field and let $f$, a monic polynomial above $\mathbb{F}[x]$ where $\deg (f)=d$. Let $\mathbb{K}$ to be the following set:$$\mathbb{K} = \left\{ g\in\mathbb{F}[x]\ |\ \deg g\lt \deg f \right\}$$ Prove that $\mathbb{K}$ is a field.
EDIT: Addition is define as one would expect, but multiplication is defined as the remainder of $g_1 \cdot g_2 = s\cdot f + r$. Meaning, $g_1 \cdot_\mathbb{K} g_2 = r$.
EDIT2: $f$ is irreducible.

It is easy to see that our "zero" is $f_0 = 0$ and our "one" is $f_1 = f+1$.

Hence, I need to prove that for every $g_1$ there's an inverse, $g_2$ such that $g_1\cdot g_2 = f+1$, but it doesn't quite working.

Also, I can rely on the fact that for every polynomial in $\mathbb{K}$ there're $s,r$ such that $g= s\cdot f + r$ where $\deg r < \deg g$.

Note: When I wrote $1$ I used the $1$ of $\mathbb{F}$ (I think that's a correct use)

I don't understand your inverse, or the definition of $K$. Suppose that $deg(f)=2$. Then $f(x)=x$ is in $K$. An inverse $g(x)$ to it should verify $f(x)g(x)=1$, or not? So $g(x)=1/x$ is not a polynomial. — Dietrich Burde, Mar 17 '17 at 10:24
Why do you define $d$ to be $\deg(f)$, but then never use the symbol $d$? — Git Gud, Mar 17 '17 at 10:32
@GitGud, it is used (sort of) when looking at the remainder (which $\deg r \le d-1$) — OliOliver, Mar 17 '17 at 10:33
What you have is an alternative way of viewing ${\mathbb F}[x]/(f)$. This is a field if and only if $f$ is irreducible. — Magdiragdag, Mar 17 '17 at 10:33
@Magdiragdag, actually you're right, I forgot to mention that. $f$ is irreducible. — OliOliver, Mar 17 '17 at 10:35
I don't understand $\cdot_{\mathbb K}$. A remainder requires a sort of division. What is, for instance, $x\cdot_{\mathbb K} x$? — Git Gud, Mar 17 '17 at 10:35
@OliOliver You need to use the fact that $f$ is irreducible to prove that every non-zero element of ${\mathbb K}$ has an inverse. — Magdiragdag, Mar 17 '17 at 10:36
@GitGud, for every $g\in \mathbb{F}[x]$ there're unique $s,r\in\mathbb{F}[x]$ such that $g = fs + r$, where $\deg r \lt \deg g$, so we define multiplication as follows: $g_1\cdot_\mathbb{K} g_2 = r$ where $g_1\cdot g_2 = sf + r$ — OliOliver, Mar 17 '17 at 10:39
Possible duplicate of Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$. — Dietrich Burde, Mar 17 '17 at 10:42
@OliOliver Yes, and if you define $g\cdot_{\mathbb K}f=r$, where $r$ is the unique entity you mention, I'll know what you mean. But you wrote that "multiplication is defined as the remainder of" $g\cdot f$. Did you perhaps mean to say remainder of $\frac{g}{f}$? — Git Gud, Mar 17 '17 at 10:42
@GidGud What the OP means is that $g_1 \cdot_{\mathbb K} g_2$ is the remainder of dividing $g_1 \cdot g_2$ by $f$. — Magdiragdag, Mar 17 '17 at 10:47
Why would anyone formulate such an exercise? There is not a more horrible way to avoid the notion of a quotient of a ring by an ideal... — MooS, Mar 17 '17 at 10:48
Good lucking showing associativity and trying not to vomit over your notes :) — MooS, Mar 17 '17 at 10:50
I'm not sure if I can use claims regarding quotient ring since we haven't defined it. I think I'm expected to use a straight-forward approach — OliOliver, Mar 17 '17 at 14:16

score 2 · Answer 1 · answered Mar 17 '17 at 11:15

2

What you have defined is an alternative way of looking at the quotient ring ${\mathbb F}[x]/(f)$. The map ${\mathbb K} \to {\mathbb F}[x]/(f)$, $g \mapsto g + (f)$ is a ring isomorphism. In fact, the easiest way to show that ${\mathbb K}$ is a indeed ring, is probably to show that this map is a bijection of sets and preserves the operations.

Because $f$ is irreducible, ${\mathbb F}[x]/(f)$ is a field and therefore ${\mathbb K}$ is too.

answered Mar 17 '17 at 11:15

Magdiragdag

15,049

3

I would not only call it the easiest way, I would call it the only reasonable way :) +1 – MooS Mar 17 '17 at 11:19
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@MooS Agreed :-), but it does assume the OP has seen the abstract notion of a quotient ring before. And I have seen texts where that notion is, for some reason, avoided. For the OP's sake, I can just hope that the text the OP has is not one of those. – Magdiragdag Mar 17 '17 at 11:22
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I hope so, too. I can understand when texts avoid the notion of a quotient ring when they introduce $\mathbb Z/n\mathbb Z$ and rather do it ad hoc. But for $K[x]/(f)$? Not so much...At some point you have to introduce some abstract techniques anyway. – MooS Mar 17 '17 at 11:27
Could you explain "Because ff is irreducible, [x]/(f)F[x]/(f) is a field and therefore K is too."? – OliOliver Mar 17 '17 at 14:09
@OliOliver What is your question? Why ${\mathbb F}[x]/(f)$ is a field? For that: see, for instance, http://math.stackexchange.com/questions/631057/let-k-be-a-field-and-fx-in-kx-prove-that-kx-fx-is-a-field-if-an?noredirect=1&lq=1 as given in the comments by Dietrich Burde. Or why this implies that ${\mathbb K}$ is a field? That is because ${\mathbb K}$ is isomorphic to ${\mathbb F}[x]/(f)$. – Magdiragdag Mar 17 '17 at 16:10

Show that $\mathbb{K}$ is a field

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