Show an ideal is not principal in $mathbbF[x,y]$ [duplicate]

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The ideal $I= langle x,y ranglesubset k[x,y]$ is not principal [closed]

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Let $mathbbF$ be a field. I want to show that the ideal $ xg + yh $
is not a principal ideal in $mathbbF[x, y]$. Do I work by contradiction first, i.e. assume that it is principal?

edited Aug 23 at 1:21

Andrew Tawfeek

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asked Aug 23 at 1:09

Homaniac

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marked as duplicate by rschwiebÂ abstract-algebra
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This question already has an answer here:

The ideal $I= langle x,y ranglesubset k[x,y]$ is not principal [closed]

6 answers

Let $mathbbF$ be a field. I want to show that the ideal $ xg + yh $
is not a principal ideal in $mathbbF[x, y]$. Do I work by contradiction first, i.e. assume that it is principal?

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

asked Aug 23 at 1:09

Homaniac

48019

marked as duplicate by rschwiebÂ abstract-algebra
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This question already has an answer here:

The ideal $I= langle x,y ranglesubset k[x,y]$ is not principal [closed]

6 answers

Let $mathbbF$ be a field. I want to show that the ideal $ xg + yh $
is not a principal ideal in $mathbbF[x, y]$. Do I work by contradiction first, i.e. assume that it is principal?

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

asked Aug 23 at 1:09

Homaniac

48019

This question already has an answer here:

The ideal $I= langle x,y ranglesubset k[x,y]$ is not principal [closed]

6 answers

Let $mathbbF$ be a field. I want to show that the ideal $ xg + yh $
is not a principal ideal in $mathbbF[x, y]$. Do I work by contradiction first, i.e. assume that it is principal?

This question already has an answer here:

The ideal $I= langle x,y ranglesubset k[x,y]$ is not principal [closed]

6 answers

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

asked Aug 23 at 1:09

Homaniac

48019

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

edited Aug 23 at 1:21

Andrew Tawfeek

1,7901722

asked Aug 23 at 1:09

Homaniac

48019

asked Aug 23 at 1:09

Homaniac

48019

asked Aug 23 at 1:09

Homaniac

48019

marked as duplicate by rschwiebÂ abstract-algebra
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1 Answer
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Yup, a proof by contradiction seems to be the right way to go. Suppose that it is principal. Show that $x, y in I$. What does this tell you about the supposed generator? Why is this a contradiction?

answered Aug 23 at 1:21

4-ier

5989

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1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

up vote
1
down vote

Yup, a proof by contradiction seems to be the right way to go. Suppose that it is principal. Show that $x, y in I$. What does this tell you about the supposed generator? Why is this a contradiction?

answered Aug 23 at 1:21

4-ier

5989

add a commentÂ |Â

up vote
1
down vote

Yup, a proof by contradiction seems to be the right way to go. Suppose that it is principal. Show that $x, y in I$. What does this tell you about the supposed generator? Why is this a contradiction?

answered Aug 23 at 1:21

4-ier

5989

add a commentÂ |Â

up vote
1
down vote

Yup, a proof by contradiction seems to be the right way to go. Suppose that it is principal. Show that $x, y in I$. What does this tell you about the supposed generator? Why is this a contradiction?

answered Aug 23 at 1:21

4-ier

5989

Yup, a proof by contradiction seems to be the right way to go. Suppose that it is principal. Show that $x, y in I$. What does this tell you about the supposed generator? Why is this a contradiction?

answered Aug 23 at 1:21

4-ier

5989

answered Aug 23 at 1:21

4-ier

5989

answered Aug 23 at 1:21

4-ier

5989

answered Aug 23 at 1:21

4-ier

5989

add a commentÂ |Â

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