Why must the ring $K[x]$ have infinitely many irreducible polynomials?

[mathmari]

Hey!

Let $K$ be a field.
I want to show that the ring $K[x]$ has infinitely many irreducible polynomials.I have done the following:

We suppose that there are finite many irreducible polynomials, $f_1(x), f_2(x), \dots, f_n(x)$ with $deg f_i(x)>0$.

Let $g(x)=f_1(x) \cdot f_2(x) \cdots f_n(x)+1$.

So that we can say that $g(x)$ can be written as a product of irreducible polynomials, what does it have to satisfy?? (Wondering)
Do we have to say something about the degree of $g(x)$?? (Wondering)

 

[mathbalarka]

Well, note that $g(x)$ modulo any $f_i(x)$ leaves reminder $1$, so none of them divides $g(x)$. But fundamental theorem of arithmetic over $K[x]$ says any polynomial can be written as a product of irreducibles over $k[x]$ so either $g(x)$ is a prime or it has some prime factor other than those $f_i$s. Both cases induces a contradiction. Thus there are infinitely many primes.

 

[mathmari]

Well, note that $g(x)$ modulo any $f_i(x)$ leaves reminder $1$, so none of them divides $g(x)$. But fundamental theorem of arithmetic over $K[x]$ says any polynomial can be written as a product of irreducibles over $k[x]$ so either $g(x)$ is a prime or it has some prime factor other than those $f_i$s. Both cases induces a contradiction. Thus there are infinitely many primes.

I understand! (Happy)So, could I formulate it as followed?? (Wondering)

We suppose that there are finite many irreducible polynomials, $f_1(x),f_2(x),\dots,f_n(x)$.

We consider $g(x)=f_1(x) \cdot f_2(x)\cdots f_n(x)+1$.

Each polynomial can be written as a product of irreducible polynomials over $K[x]$.
There are two cases:
a) $g(x)$ is an irreducible polynomial
b) $g(x)$ can be analyzed into irreduble polynomials

a) Since $g(x)$ is irreducible, it has to be $g(x)=f_i(x), 1\leq i \leq n$
Since $ f_i (x) \mid g(x)$ and $f_i(x) \mid f_1(x)\cdot f_2(x) \cdots f_n(x)$ we have that $f_i (x) \mid 1$, that cannot be true since $f_i(x)$ is not a constant. b) Since $g(x)$ can be analyzed into irreducible polynomials, it has to be $g(x)=f_i(x)q(x), 1\leq i \leq n$
Since $ f_i (x) \mid g(x)$ and $f_i(x) \mid f_1(x)\cdot f_2(x) \cdots f_n(x)$ we have that $f_i(x) \mid 1$, that cannot be true since $f_i(x)$ is not a constant.

Therefore, there are infinite many irreducible polynomials.

 

[mathmari]

Or is the way I wrote it wrong??

 

[mathbalarka]

Part 2 can be made simpler : $g(x)$ is either an irreducible, or product of irreducibles but $\prod f_i(x) + 1$ always leaves nonzero (in fact, 1) reminder when divided by $f_i(x)$ for any $i = 1, 2, ..., n$. Thus $g(x)$ has an (well, nontrivial) irreducible factor not in $\{f_1(x), f_2(x), \cdots, f_i(x)\}$. This factor (which can in fact be $g(x)$, in case $g(x)$ is a prime) is our desired polynomial to contradict the hypothesis.

A word of caution : to construct $f_1(x)f_2(x) \cdots f_n(x) + 1$ you need to show first that there is at least *one* prime (nonconstant?) in $K[x]$. But this is always possible : take $x$. It's better to mention this part in your proof, although traditionally one just skips it as an obvious fact.

 

[mathmari]

Part 2 can be made simpler : $g(x)$ is either an irreducible, or product of irreducibles but $\prod f_i(x) + 1$ always leaves nonzero (in fact, 1) reminder when divided by $f_i(x)$ for any $i = 1, 2, ..., n$. Thus $g(x)$ has an (well, nontrivial) irreducible factor not in $\{f_1(x), f_2(x), \cdots, f_i(x)\}$. This factor (which can in fact be $g(x)$, in case $g(x)$ is a prime) is our desired polynomial to contradict the hypothesis.

A word of caution : to construct $f_1(x)f_2(x) \cdots f_n(x) + 1$ you need to show first that there is at least *one* prime (nonconstant?) in $K[x]$. But this is always possible : take $x$. It's better to mention this part in your proof, although traditionally one just skips it as an obvious fact.

I got stuck right now... (Worried)

Do you mean that I should formulate it as followed??

We consider $g(x)=f_1(x) \cdot f_2(x) \cdots f_n(x)+1$, with $deg f_i(x)>0, i=1,2, \dots, n$.

$g(x)$ is either an irreducible, or product of irreducibles, that means that $f_i(x) \mid g(x)$, for any $i=1,2, \dots, n$.

But since $f_i(x) \mid g(x)$ and $f_i(x) \mid f_1(x) \cdot f_2(x) \cdots f_n(x)$, it follows that $f_i(x) \mid 1$, which cannot be true, since $deg f_i(x) >0$.

Is this correct?? (Wondering) Or have I understood it wrong?? (Wondering)

 

[mathbalarka]

Yes, that'd be the way to do it. (Yes)

 

[mathmari]

Yes, that'd be the way to do it. (Yes)

Great! (Yes) Thank you very much! (Smirk)