Mixing
irreducibility of polynomial over field of rational functions
0
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Here
This is an exercise from Dummit and Foote where the following hint is also given : $mathbf{(K[X])(Y)=(K[Y])(X)}$.
Does this mean that we can consider our polynomial over $mathbf K[Y]$ with variable X now which will imply that it is irreducible being a linear polynomial and hence irreducible in the required field $mathbf K(X)$? If I have misunderstood please guide me through the next step.
abstract-algebra ring-theory field-theory extension-field irreducible-polynomials
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
$endgroup$
add a comment |
0
$begingroup$
Here
This is an exercise from Dummit and Foote where the following hint is also given : $mathbf{(K[X])(Y)=(K[Y])(X)}$.
Does this mean that we can consider our polynomial over $mathbf K[Y]$ with variable X now which will imply that it is irreducible being a linear polynomial and hence irreducible in the required field $mathbf K(X)$? If I have misunderstood please guide me through the next step.
abstract-algebra ring-theory field-theory extension-field irreducible-polynomials
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
$endgroup$
$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55
add a comment |
0
0
0
$begingroup$
Here
This is an exercise from Dummit and Foote where the following hint is also given : $mathbf{(K[X])(Y)=(K[Y])(X)}$.
Does this mean that we can consider our polynomial over $mathbf K[Y]$ with variable X now which will imply that it is irreducible being a linear polynomial and hence irreducible in the required field $mathbf K(X)$? If I have misunderstood please guide me through the next step.
abstract-algebra ring-theory field-theory extension-field irreducible-polynomials
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
$endgroup$
Here
This is an exercise from Dummit and Foote where the following hint is also given : $mathbf{(K[X])(Y)=(K[Y])(X)}$.
Does this mean that we can consider our polynomial over $mathbf K[Y]$ with variable X now which will imply that it is irreducible being a linear polynomial and hence irreducible in the required field $mathbf K(X)$? If I have misunderstood please guide me through the next step.
abstract-algebra ring-theory field-theory extension-field irreducible-polynomials
abstract-algebra ring-theory field-theory extension-field irreducible-polynomials
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
asked Jan 29 at 15:34
ShanghaikidShanghaikid
669
669
$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55
add a comment |
$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55
$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55
add a comment |
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$begingroup$
Like egreg says in his answer. We can apply the Gauss Lemma to the polynomial ring $R[Y]$, because $R=K[X]$ is a PID, since $K$ is a field. If you want, rename $R[Y]$ to $R[x]$, if this looks more familiar to you.
$endgroup$
– Dietrich Burde
Jan 29 at 15:56
$begingroup$
@DietrichBurde Yes I understand that Gauss Lemma can be applied but how to proceed further in order to conclude its irreducibilty. Gauss Lemma only says irreducibility in one implies irreducibility in another. I am sure I'm missing something trivial.
$endgroup$
– Shanghaikid
Jan 29 at 16:22
$begingroup$
@DietrichBurde In another answer it is given that we have to use the fact that P and Q are relatively prime in order to show that the linear polynomial in X is irreducible as K[Y] is not a field. But how do I use it?
$endgroup$
– Shanghaikid
Jan 29 at 19:14
$begingroup$
Understood. Thanks
$endgroup$
– Shanghaikid
Jan 30 at 3:55