$X^2$ irreducible but not prime
$begingroup$
Let $K$ be a field. I'd like to show that in $R={sum a_i X^iin K[X]mid a_1=0}$, the element $X^2$ is irreducible, but not prime.
Irreducibility is checked easily, but I can't see why it's not prime.
$X^2$ divides every polynomial with constant term $= 0$, but none of the nonzero ones. Thus, I have to multiply 2 polynomials with constant nonzero terms, but then the resulting polynomial also has constant term which isn't zero, so $X^2$ doesn't divide it?
ring-theory integral-domain
edited Dec 30 '18 at 17:22
amWhy
1
asked Jan 11 '15 at 17:33
user207173user207173
233
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field. I'd like to show that in $R={sum a_i X^iin K[X]mid a_1=0}$, the element $X^2$ is irreducible, but not prime.
Irreducibility is checked easily, but I can't see why it's not prime.
$X^2$ divides every polynomial with constant term $= 0$, but none of the nonzero ones. Thus, I have to multiply 2 polynomials with constant nonzero terms, but then the resulting polynomial also has constant term which isn't zero, so $X^2$ doesn't divide it?
ring-theory integral-domain
edited Dec 30 '18 at 17:22
amWhy
1
asked Jan 11 '15 at 17:33
user207173user207173
233
$endgroup$
$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49
add a comment |
$begingroup$
Let $K$ be a field. I'd like to show that in $R={sum a_i X^iin K[X]mid a_1=0}$, the element $X^2$ is irreducible, but not prime.
Irreducibility is checked easily, but I can't see why it's not prime.
$X^2$ divides every polynomial with constant term $= 0$, but none of the nonzero ones. Thus, I have to multiply 2 polynomials with constant nonzero terms, but then the resulting polynomial also has constant term which isn't zero, so $X^2$ doesn't divide it?
ring-theory integral-domain
edited Dec 30 '18 at 17:22
amWhy
1
asked Jan 11 '15 at 17:33
user207173user207173
233
$endgroup$
Let $K$ be a field. I'd like to show that in $R={sum a_i X^iin K[X]mid a_1=0}$, the element $X^2$ is irreducible, but not prime.
Irreducibility is checked easily, but I can't see why it's not prime.
$X^2$ divides every polynomial with constant term $= 0$, but none of the nonzero ones. Thus, I have to multiply 2 polynomials with constant nonzero terms, but then the resulting polynomial also has constant term which isn't zero, so $X^2$ doesn't divide it?
ring-theory integral-domain
ring-theory integral-domain
edited Dec 30 '18 at 17:22
amWhy
1
asked Jan 11 '15 at 17:33
user207173user207173
233
edited Dec 30 '18 at 17:22
amWhy
1
asked Jan 11 '15 at 17:33
user207173user207173
233
edited Dec 30 '18 at 17:22
amWhy
1
edited Dec 30 '18 at 17:22
amWhy
1
edited Dec 30 '18 at 17:22
amWhy
1
1
asked Jan 11 '15 at 17:33
user207173user207173
233
asked Jan 11 '15 at 17:33
user207173user207173
233
asked Jan 11 '15 at 17:33
user207173user207173
233
233
$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49
add a comment |
$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49
$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49
$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49
add a comment |
1 Answer
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$begingroup$
Hint $ X^2mid, X^3cdot X^3,:$ but $, X^2nmid X^3,$ since $,X^3/X^2 = Xnotin R$
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
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$begingroup$
Hint $ X^2mid, X^3cdot X^3,:$ but $, X^2nmid X^3,$ since $,X^3/X^2 = Xnotin R$
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
$begingroup$
Hint $ X^2mid, X^3cdot X^3,:$ but $, X^2nmid X^3,$ since $,X^3/X^2 = Xnotin R$
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
add a comment |
$begingroup$
Hint $ X^2mid, X^3cdot X^3,:$ but $, X^2nmid X^3,$ since $,X^3/X^2 = Xnotin R$
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
$endgroup$
Hint $ X^2mid, X^3cdot X^3,:$ but $, X^2nmid X^3,$ since $,X^3/X^2 = Xnotin R$
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
answered Jan 11 '15 at 17:45
Bill DubuqueBill Dubuque
211k29194648
211k29194648
add a comment |
add a comment |
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$begingroup$
"$X^2$ divides every polynomial with constant term $= 0$" -- this is not true, $X^2$ does not divide $X^3$ (in $R$).
$endgroup$
– Alexey
Dec 30 '18 at 14:49