find the maximal ideal of the ring ?.
Clash Royale CLAN TAG#URR8PPP
find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$
My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$
Is its correct ?
any hints/solution will be appreciated
$endgroup$
add a comment 
find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$
My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$
Is its correct ?
any hints/solution will be appreciated
$endgroup$

3$begingroup$
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13 
$begingroup$
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15 
1$begingroup$
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19 
2$begingroup$
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31
add a comment 
find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$
My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$
Is its correct ?
any hints/solution will be appreciated
$endgroup$
find the maximal ideal of the ring $$ frac{mathbb{R}[x]}{ (x^2)} $$
My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$
Is its correct ?
any hints/solution will be appreciated

3$begingroup$
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13 
$begingroup$
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15 
1$begingroup$
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19 
2$begingroup$
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31
add a comment 

3$begingroup$
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13 
$begingroup$
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15 
1$begingroup$
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19 
2$begingroup$
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13
No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31
add a comment 
4 Answers
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Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.
$endgroup$
add a comment 
There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.
$endgroup$
add a comment 
Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.
Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It’s pretty clear this is as far as it goes, because if there’s a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).
Let me clarify the last part by an example: so, why wouldn’t $(x6)$ be maximal? Well, multiply $x6$ by $x$. Get $x^26xequiv 6x$. Then we get $x$, then $6$, then $1$.
$endgroup$
add a comment 
Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.
$endgroup$
add a comment 
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.
$endgroup$
add a comment 
Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.
$endgroup$
add a comment 
Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.
$endgroup$
Let $varphi:mathbb{R}[x]rightarrow R:=mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I mapsto varphi (I)$. The only ideals of $mathbb{R}[x]$ that contains $(x^2)$ are $mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(bar x)$ and $(bar x^2)$. Clearly $(0)=(bar x^2)subsetneq (bar x)subsetneq R$, which makes it clear that $(bar x)$ is the only maximal ideal of $R$.
add a comment 
add a comment 
There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.
$endgroup$
add a comment 
There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.
$endgroup$
add a comment 
There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.
$endgroup$
There is a bijective correspondence between the ideals of $x^2$ in $mathbb{R}[x]$ and in $dfrac{mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $dfrac{mathbb{R}}{x^2}$ corresponds to a maximal ideal in $mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $mathbb{R}$ has three ideals which are $mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.
add a comment 
add a comment 
Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.
Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It’s pretty clear this is as far as it goes, because if there’s a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).
Let me clarify the last part by an example: so, why wouldn’t $(x6)$ be maximal? Well, multiply $x6$ by $x$. Get $x^26xequiv 6x$. Then we get $x$, then $6$, then $1$.
$endgroup$
add a comment 
Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.
Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It’s pretty clear this is as far as it goes, because if there’s a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).
Let me clarify the last part by an example: so, why wouldn’t $(x6)$ be maximal? Well, multiply $x6$ by $x$. Get $x^26xequiv 6x$. Then we get $x$, then $6$, then $1$.
$endgroup$
add a comment 
Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.
Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It’s pretty clear this is as far as it goes, because if there’s a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).
Let me clarify the last part by an example: so, why wouldn’t $(x6)$ be maximal? Well, multiply $x6$ by $x$. Get $x^26xequiv 6x$. Then we get $x$, then $6$, then $1$.
$endgroup$
Hint: The maximal ideal in $frac{Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.
Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It’s pretty clear this is as far as it goes, because if there’s a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).
Let me clarify the last part by an example: so, why wouldn’t $(x6)$ be maximal? Well, multiply $x6$ by $x$. Get $x^26xequiv 6x$. Then we get $x$, then $6$, then $1$.
add a comment 
add a comment 
Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.
$endgroup$
add a comment 
Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.
$endgroup$
add a comment 
Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.
$endgroup$
Let $P$ be a maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2in Q$. Hence, $xin Q$. This shows that every maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ frac{mathbb{R}[x]}{ (x^2)} $.
add a comment 
add a comment 
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No: $(x^2)$ is not maximal.
$endgroup$
– Bernard
Jan 6 at 2:13
@Bernard im not getting can u elaborate this ?
$endgroup$
– jasmine
Jan 6 at 2:15
The quotient $mathbf R[x]/(x^2)$ is not a jeld, not even an integral domain, since $x$ is nilpotent ($x^2=0$ in the quotient). But $(x)$ is maximal since the quotient $mathbf R[x]/(x)$ is isomorphic tp $mathbf R$.
$endgroup$
– Bernard
Jan 6 at 2:19
@jasmine, $x^2 = x(x – 0) = 0$. So it has a root, so it is reducible, so it is not prime, so it is not maximal
$endgroup$
– IAmNoOne
Jan 6 at 2:31